AC and DC circuits Presentation

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AC and DC circuits Presentation

  1. 1. CIRCUITS AND NETWORKS CIRCUITSCIRCUITS Parameters  The various elements of an electric circuit, like resistance, inductance, and capacitance which may be lumped or distibuted. Parameters  The various elements of an electric circuit, like resistance, inductance, and capacitance which may be lumped or distibuted. CIRCUITS  A closed conducting path through which an electric current flows or is intended to flow CIRCUITS  A closed conducting path through which an electric current flows or is intended to flow
  2. 2. CIRCUITSCIRCUITS Linear Circuit  Is one whose parameters are constant (i.e. They do not change with voltage and current. Non-Linear Circuit  Is that circuit whose parameters change with voltage and current. Bilateral Circuit  Is one whose properties or characteristics are the same in either direction. Unilateral Circuit  Is that circuit whose properties or characteristics change with the direction of its operation. Linear Circuit  Is one whose parameters are constant (i.e. They do not change with voltage and current. Non-Linear Circuit  Is that circuit whose parameters change with voltage and current. Bilateral Circuit  Is one whose properties or characteristics are the same in either direction. Unilateral Circuit  Is that circuit whose properties or characteristics change with the direction of its operation. TYPES CIRCUITS AND NETWORKS
  3. 3. ELECTRICAL NETWORKSELECTRICAL NETWORKS Passive Network With no source of emf. Active Network Contains one or more than one sources of emf. Passive Network With no source of emf. Active Network Contains one or more than one sources of emf. ELECTRICAL NETWORK  Connection of various electric elements in any manner ELECTRICAL NETWORK  Connection of various electric elements in any manner TYPES CIRCUITS AND NETWORKS
  4. 4. ELECTRICAL NETWORKSELECTRICAL NETWORKS Node  A junction in a circuit where two or more circuit elements and/or branches are connected together. Branch  Part of a network which lies netween two junctions. Loop  A closed path in a circuit in which no element or node is encountered more than once. Mesh  A loop that contains no other loop within it. Node  A junction in a circuit where two or more circuit elements and/or branches are connected together. Branch  Part of a network which lies netween two junctions. Loop  A closed path in a circuit in which no element or node is encountered more than once. Mesh  A loop that contains no other loop within it. PART S CIRCUITS AND NETWORKS
  5. 5. OHM’S LAWOHM’S LAW OHM’S LAW  One of the most fundamental law in electrical circuits relating voltage, current and resistance  Developed in 1827 by German physicist Georg Simon Ohm OHM’S LAW  One of the most fundamental law in electrical circuits relating voltage, current and resistance  Developed in 1827 by German physicist Georg Simon Ohm CIRCUITS AND NETWORKS
  6. 6. OHM’S LAWOHM’S LAW  According to Ohm’s Law, the current (I) flowing in an electrical circuit is directly is directly proportional to the applied voltage (E) and inversely proportional to the equivalent resistance (R) of the circuit and mathematically expressed as:  According to Ohm’s Law, the current (I) flowing in an electrical circuit is directly is directly proportional to the applied voltage (E) and inversely proportional to the equivalent resistance (R) of the circuit and mathematically expressed as: CIRCUITS AND NETWORKS
  7. 7. SERIES CIRCUITSSERIES CIRCUITS SERIES circuits  A circuit connection in which the components are connected to form one conducting path SERIES circuits  A circuit connection in which the components are connected to form one conducting path SERIES/PARALLEL CIRCUITS
  8. 8. SERIES CIRCUITSSERIES CIRCUITS Where: EX – voltage across the resistor concerned ET – total voltage across the circuit RX – the resistor concerned RT – the sum of all resistances in the circuit Where: EX – voltage across the resistor concerned ET – total voltage across the circuit RX – the resistor concerned RT – the sum of all resistances in the circuit Voltage Division for Series Circuit: EX = ET • RX RT SERIES/PARALLEL CIRCUITS
  9. 9. PARALLEL CIRCUITSPARALLEL CIRCUITS PARALLEL circuits  A circuit connection in which the components are connected to form more than 1 conducting path PARALLEL circuits  A circuit connection in which the components are connected to form more than 1 conducting path SERIES/PARALLEL CIRCUITS
  10. 10. PARALLEL CIRCUITSPARALLEL CIRCUITS Where: IX – current concerned flowing through resistor Rx IT – total current of the circuit Req – equivalent resistance of the parallel circuit except Rx RT – the sum of all resistances in the circuit Where: IX – current concerned flowing through resistor Rx IT – total current of the circuit Req – equivalent resistance of the parallel circuit except Rx RT – the sum of all resistances in the circuit Voltage Division for Parallel Circuit: IX = IT • Req RT SERIES/PARALLEL CIRCUITS
  11. 11. KIRCHHOFF’S LAWKIRCHHOFF’S LAW “In any electrical network, the algebraic sum of the current meeting at a point (or junction) is zero.” “In any electrical network, the algebraic sum of the current meeting at a point (or junction) is zero.” KIRCHHOFF’S LAW  More comprehensive than Ohm’s Law and is used in solving electrical  Termed as “Laws of Electric Networks”  Formulated by German physicist Gustav Robert Kirchhoff KIRCHHOFF’S LAW  More comprehensive than Ohm’s Law and is used in solving electrical  Termed as “Laws of Electric Networks”  Formulated by German physicist Gustav Robert Kirchhoff Kirchhoff’s Current Law (KCL) NETWORK ANALYSIS
  12. 12. KIRCHHOFF’S CURRENT LAWKIRCHHOFF’S CURRENT LAW  In short the sum of currents entering a node equals the sum of currents leaving the node ⁻ Current towards the node, positive current ⁻ Current away from the node, negative current IB + IC + ID = IA (IB + IC + ID) - IA = 0  In short the sum of currents entering a node equals the sum of currents leaving the node ⁻ Current towards the node, positive current ⁻ Current away from the node, negative current IB + IC + ID = IA (IB + IC + ID) - IA = 0 NETWORK ANALYSIS
  13. 13. KIRCHHOFF’S VOLTAGE LAWKIRCHHOFF’S VOLTAGE LAW “The algebraic sum of the products of currents and resistances in each of thr conductors in any closed path (or mesh) in a network PLUS the algebraic sum of the emfs in the path is zero.” “The algebraic sum of the products of currents and resistances in each of thr conductors in any closed path (or mesh) in a network PLUS the algebraic sum of the emfs in the path is zero.” Kirchhoff’s Voltage Law (KVL) NETWORK ANALYSIS
  14. 14. KIRCHHOFF’S VOLTAGE LAWKIRCHHOFF’S VOLTAGE LAW  In short, the sum of the voltages around the loop is equal to zero ⁻ For voltage sources, if loops enters on minus and goes out on plus, positive voltage and if loops enters on plus and goes out on minus, negative voltage. ⁻ For voltage drops, if the loop direction is the same as current direction, negative voltage drop and if the loop direction is opposite to the current direction, positive voltage drop.  In short, the sum of the voltages around the loop is equal to zero ⁻ For voltage sources, if loops enters on minus and goes out on plus, positive voltage and if loops enters on plus and goes out on minus, negative voltage. ⁻ For voltage drops, if the loop direction is the same as current direction, negative voltage drop and if the loop direction is opposite to the current direction, positive voltage drop. NETWORK ANALYSIS
  15. 15. MESH ANALYSISMESH ANALYSIS Loop Analysis Procedure: 1. Label each of the loop/mesh currents. 2. Apply KVL to loops/meshes to form equations with current variables. a. For N independent loops, we may write N total equations using KVL around each loop. Loop currents are those currents flowing in a loop; they are used to define branch currents. b. Current sources provide constraint equations. 3. Solve the equations to determine the user defined loop currents. Loop Analysis Procedure: 1. Label each of the loop/mesh currents. 2. Apply KVL to loops/meshes to form equations with current variables. a. For N independent loops, we may write N total equations using KVL around each loop. Loop currents are those currents flowing in a loop; they are used to define branch currents. b. Current sources provide constraint equations. 3. Solve the equations to determine the user defined loop currents. NETWORK ANALYSIS MESH analysis  A sophisticated application of KVL with mesh currents. MESH analysis  A sophisticated application of KVL with mesh currents.
  16. 16. NODAL ANALYSISNODAL ANALYSIS  Consist of finding the node voltages at all principal nodes with respect to the reference node. PRINCIPAL node – a node with three or more circuit elements joined together. Reference node – the node from which the unknown voltages are measured.  Consist of finding the node voltages at all principal nodes with respect to the reference node. PRINCIPAL node – a node with three or more circuit elements joined together. Reference node – the node from which the unknown voltages are measured. NETWORK ANALYSIS NODAL analysis  A systematic application of KCL at a node and after simplifying the resulting KCL equation, the node voltage can be calculated. NODAL analysis  A systematic application of KCL at a node and after simplifying the resulting KCL equation, the node voltage can be calculated.
  17. 17. SUPERPOSITION THEOREMSUPERPOSITION THEOREM In general:  Number of network to analyze is equal to number of independent sources.  To consider effects of each source independently, sources must be removed and replaced without affecting the final result:  All voltage sources >> short circuited  All current sources >> open circuited In general:  Number of network to analyze is equal to number of independent sources.  To consider effects of each source independently, sources must be removed and replaced without affecting the final result:  All voltage sources >> short circuited  All current sources >> open circuited NETWORK ANALYSIS SUPERPOSITION theorem “ The current through or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the current or voltages produced independently in each source. ” SUPERPOSITION theorem “ The current through or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the current or voltages produced independently in each source. ”
  18. 18. COMPENSATION THEOREMCOMPENSATION THEOREM NETWORK ANALYSIS COMPENSATION theorem  If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount dZ, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IdZ into that branch with all other voltage sources replaced by their internal impedances. COMPENSATION theorem  If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount dZ, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IdZ into that branch with all other voltage sources replaced by their internal impedances.
  19. 19. RECIPROCITY THEOREMRECIPROCITY THEOREM Simply mean,  E and I are mutually transferable, or  The receiving point and the sending point in a network are interchangeable, or  Interchange of an IDEAL voltage source and an IDEAL ammeter in any network will not change the ammeter reading,  Interchange of an IDEAL current source and an IDEAL voltmeter in any network will not change the voltmeter reading Simply mean,  E and I are mutually transferable, or  The receiving point and the sending point in a network are interchangeable, or  Interchange of an IDEAL voltage source and an IDEAL ammeter in any network will not change the ammeter reading,  Interchange of an IDEAL current source and an IDEAL voltmeter in any network will not change the voltmeter reading NETWORK ANALYSIS RECIPROCITY theorem “If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch. ” RECIPROCITY theorem “If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch. ”
  20. 20. MILLMAN’S THEOREMMILLMAN’S THEOREM  In Millman’s Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination.  Millman’s theorem is applicable only to those cicuits which can be re-drawn accordingly.  In Millman’s Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination.  Millman’s theorem is applicable only to those cicuits which can be re-drawn accordingly. NETWORK ANALYSIS MILLMAN’S theorem “ A special case of the application of Thevenin’s Theorem/or Norton’s Theorem used for finding the COMMON voltage (VAB) across any network which contains a number of parallel voltage sources. ” MILLMAN’S theorem “ A special case of the application of Thevenin’s Theorem/or Norton’s Theorem used for finding the COMMON voltage (VAB) across any network which contains a number of parallel voltage sources. ”
  21. 21. MAXIMUM POWER TRANSFER THOREM MAXIMUM POWER TRANSFER THOREM NETWORK ANALYSIS MAXIMUM POWER TRANSFER theorem  For loads connected directly to a DC voltage supply, maximum power will be delivered to the load when the resistance is equal to the internal resistance of the source.  For maximum power transfer: RS = RL MAXIMUM POWER TRANSFER theorem  For loads connected directly to a DC voltage supply, maximum power will be delivered to the load when the resistance is equal to the internal resistance of the source.  For maximum power transfer: RS = RL
  22. 22. THEVENIN’S THEOREMTHEVENIN’S THEOREM where: VTH – the open circuit voltage which appears across the two terminals from where the load resistance has been removed. RTH – the resistance looking back into the network across the two terminals with all the voltage sources shorted and replaced by their internal resistances (if any) and all current sources by infinite resistance. where: VTH – the open circuit voltage which appears across the two terminals from where the load resistance has been removed. RTH – the resistance looking back into the network across the two terminals with all the voltage sources shorted and replaced by their internal resistances (if any) and all current sources by infinite resistance. NETWORK ANALYSIS THEVENIN’S theorem “ Any two-terminal of a linear, active bilateral network of a fixed resistances and voltage source/s may be replaced by a single voltage source (VTH) and a series of internal resistance (RTH). ” THEVENIN’S theorem “ Any two-terminal of a linear, active bilateral network of a fixed resistances and voltage source/s may be replaced by a single voltage source (VTH) and a series of internal resistance (RTH). ”
  23. 23. NORTON’S THEOREMNORTON’S THEOREM where: IN– the current which would flow in a short circuit placed across the output terminals. RN – the resistance of the network when viewed from the open circuited terminals after all voltage sources being replaced by open circuits. where: IN– the current which would flow in a short circuit placed across the output terminals. RN – the resistance of the network when viewed from the open circuited terminals after all voltage sources being replaced by open circuits. NETWORK ANALYSIS THEVENIN’S theorem “ Any two-terminal active network containing voltage sources and resistances when viewed from its output terminals, is equivalent to a constant-current source (IN) and a parallel internal resistance (RN). ” THEVENIN’S theorem “ Any two-terminal active network containing voltage sources and resistances when viewed from its output terminals, is equivalent to a constant-current source (IN) and a parallel internal resistance (RN). ”
  24. 24. THEVENIN-NORTON TRANSFORMATION THEVENIN-NORTON TRANSFORMATION NETWORK ANALYSIS
  25. 25. NORTON-THEVENIN TRANSFORMATION NORTON-THEVENIN TRANSFORMATION NETWORK ANALYSIS
  26. 26. EQUIVALENT THREE-TERMINAL NETWORKS EQUIVALENT THREE-TERMINAL NETWORKS NETWORK ANALYSIS DELTA to WYE  The equivalent resistance of each arm to the wye is given by the PRODUCT of the two delta sides that meet at its end divided by the SUM of the three delta resistances. DELTA to WYE  The equivalent resistance of each arm to the wye is given by the PRODUCT of the two delta sides that meet at its end divided by the SUM of the three delta resistances.
  27. 27. EQUIVALENT THREE-TERMINAL NETWORKS EQUIVALENT THREE-TERMINAL NETWORKS NETWORK ANALYSIS WYE to DELTA  The equivalent delta resistance between any two twrminals is given by the SUM of a star resistance between those terminals PLUS the PRODUCT of these two star resistances DIVIDED by the third resistance. WYE to DELTA  The equivalent delta resistance between any two twrminals is given by the SUM of a star resistance between those terminals PLUS the PRODUCT of these two star resistances DIVIDED by the third resistance.
  28. 28. REVIEW QUESTIONSREVIEW QUESTIONS 1. A battery with a rating of 9 volts has an internal resistance of 20 ohms. What is the expected resistance of the bulb that is connected across the battery to attain a maximum power transfer? a. 20 ohms b. 10 ohms c. 5 ohms d. 2 ohms 2. In a sireis ciscuit a resistors 2200 and 4500 with an impressed voltage of 10, what is the circuit current in mA? a. 1.49 b. 6.67 c. 4.34 d. 1.34 1. A battery with a rating of 9 volts has an internal resistance of 20 ohms. What is the expected resistance of the bulb that is connected across the battery to attain a maximum power transfer? a. 20 ohms b. 10 ohms c. 5 ohms d. 2 ohms 2. In a sireis ciscuit a resistors 2200 and 4500 with an impressed voltage of 10, what is the circuit current in mA? a. 1.49 b. 6.67 c. 4.34 d. 1.34
  29. 29. REVIEW QUESTIONSREVIEW QUESTIONS 3. The current needed to operate a soldering iron which has a rating of 600 watts at 110 volts is. a. 5.455 A b. 66,000 A c. 18,200 A d. 0.182 A 4. The ammeter reads 230 ampere while the voltmeter is 115 volts. What is the power inKW at the time of reading a. 264.5 b. 2645 c. 264500 d.26.45 3. The current needed to operate a soldering iron which has a rating of 600 watts at 110 volts is. a. 5.455 A b. 66,000 A c. 18,200 A d. 0.182 A 4. The ammeter reads 230 ampere while the voltmeter is 115 volts. What is the power inKW at the time of reading a. 264.5 b. 2645 c. 264500 d.26.45
  30. 30. REVIEW QUESTIONSREVIEW QUESTIONS 5. What is the type of circuit whose parameters are constant which do not change with voltage or current? a. Lumped b. Tuned c. Reactive d. Linear 6. What is the resistance of two equal valued resistor series? a. Twice as one b. The sum of their reciprocal c. The difference of both b. The product of both 5. What is the type of circuit whose parameters are constant which do not change with voltage or current? a. Lumped b. Tuned c. Reactive d. Linear 6. What is the resistance of two equal valued resistor series? a. Twice as one b. The sum of their reciprocal c. The difference of both b. The product of both
  31. 31. REVIEW QUESTIONSREVIEW QUESTIONS 7. What do you expect when you use two 20 kohms, 1 watts resistors in parallel instead of one 10 kohms, 1 watt? a. Provide more power b. Provide lighter current c. Provide less power d. Provide wider tolerance 8. The voltage applied in DC circuit having a power of 36 watts and a total resistance of 4 ohms. a. 6 V b. 9V c. 12 V d. 24 V 7. What do you expect when you use two 20 kohms, 1 watts resistors in parallel instead of one 10 kohms, 1 watt? a. Provide more power b. Provide lighter current c. Provide less power d. Provide wider tolerance 8. The voltage applied in DC circuit having a power of 36 watts and a total resistance of 4 ohms. a. 6 V b. 9V c. 12 V d. 24 V
  32. 32. REVIEW QUESTIONSREVIEW QUESTIONS 9. When resistor are connected in series, what happens? a. The effective resistance b. Nothing c. The tolerance d. The effective resistance is increased 10. Find the thevenin’s impedance equivalent across R2 of a linear close circuit having 10-V supply in series with the resistors (R1=100 ohms and R2=200 ohms) a. 666 ohms b. 6.66 ohms c. 66.6 ohms d. 6666 ohms 9. When resistor are connected in series, what happens? a. The effective resistance b. Nothing c. The tolerance d. The effective resistance is increased 10. Find the thevenin’s impedance equivalent across R2 of a linear close circuit having 10-V supply in series with the resistors (R1=100 ohms and R2=200 ohms) a. 666 ohms b. 6.66 ohms c. 66.6 ohms d. 6666 ohms
  33. 33. REVIEW QUESTIONSREVIEW QUESTIONS 11. How much power does electronic equipment consume, assuming a 5.5A current flowing and a 120-V power source. a. 60 W b. 66 W c. 660 W d. 125.5 W 12. How many nodes are needed to completely analyze a circuit according to Kirchoffs Current Law. a. One b. Two c. All nodes in the circuit d. One less than the total number of nodes in the circuit 11. How much power does electronic equipment consume, assuming a 5.5A current flowing and a 120-V power source. a. 60 W b. 66 W c. 660 W d. 125.5 W 12. How many nodes are needed to completely analyze a circuit according to Kirchoffs Current Law. a. One b. Two c. All nodes in the circuit d. One less than the total number of nodes in the circuit
  34. 34. REVIEW QUESTIONSREVIEW QUESTIONS 13. A common connection between circuit elements or conductors from different branches. a. Node b. Junction c. Ground d. Mesh 14. It is used to denote a common electrical point of zero potential. a. Short circuit b. Reference point c. Open circuit d.ground 13. A common connection between circuit elements or conductors from different branches. a. Node b. Junction c. Ground d. Mesh 14. It is used to denote a common electrical point of zero potential. a. Short circuit b. Reference point c. Open circuit d.ground
  35. 35. REVIEW QUESTIONSREVIEW QUESTIONS 15. Loop currents should be assumed to flow in which direction? a. Straight b. Clockwise c. Counterclockwise d. Either b or c 16. In mesh analysis, we apply: a. KCL b. KVL c. Source d. Millman’s theorem 15. Loop currents should be assumed to flow in which direction? a. Straight b. Clockwise c. Counterclockwise d. Either b or c 16. In mesh analysis, we apply: a. KCL b. KVL c. Source d. Millman’s theorem
  36. 36. REVIEW QUESTIONSREVIEW QUESTIONS 17. Which of the following is not a valid expression of Ohms Law a. R = PI b. E = IR c. I = E/R d. R = E/I 18. Using ohms Law, what happens to the circuit current if the applied voltage increases? a. Current doubles b. Current increases c. Current remians constant d. Current decreases 17. Which of the following is not a valid expression of Ohms Law a. R = PI b. E = IR c. I = E/R d. R = E/I 18. Using ohms Law, what happens to the circuit current if the applied voltage increases? a. Current doubles b. Current increases c. Current remians constant d. Current decreases
  37. 37. REVIEW QUESTIONSREVIEW QUESTIONS 19. According to ohms law, what happen to the circuit current if the circuit resistance increases? a. Current doubles b. Current decreases c. Current increases d. Current remains constant 20. If the resistance of a circuit is doubled and the applied voltage is kept constant, the current will be _______ . a. Be quaddrupled b. Remains c. Be cut in half d. Be doubled 19. According to ohms law, what happen to the circuit current if the circuit resistance increases? a. Current doubles b. Current decreases c. Current increases d. Current remains constant 20. If the resistance of a circuit is doubled and the applied voltage is kept constant, the current will be _______ . a. Be quaddrupled b. Remains c. Be cut in half d. Be doubled
  38. 38. REVIEW QUESTIONSREVIEW QUESTIONS 21. It is an electrical current that flows in one direction only? a. Normal current b. Alternating current c. Direct current d. Eddy current 22. In Ohms Law, what is E/R? a. Amperage b. Voltage c. Resistance d. Power 21. It is an electrical current that flows in one direction only? a. Normal current b. Alternating current c. Direct current d. Eddy current 22. In Ohms Law, what is E/R? a. Amperage b. Voltage c. Resistance d. Power
  39. 39. REVIEW QUESTIONSREVIEW QUESTIONS 23. A 33-Kohm resistor is connected in series with a parallel combination made up of 56-Kohm resistor and a 7.8-kohm resistor. What is the total combined reistance of the three resistors? a. 390667 ohms b. 49069 ohms c. 63769 ohms d. 95000 ohms 24. Which of the following cannot be included in a loop of Kirchoff’s Volatage Law a. Current source b. Voltage source c. Resistance d. Reactance 23. A 33-Kohm resistor is connected in series with a parallel combination made up of 56-Kohm resistor and a 7.8-kohm resistor. What is the total combined reistance of the three resistors? a. 390667 ohms b. 49069 ohms c. 63769 ohms d. 95000 ohms 24. Which of the following cannot be included in a loop of Kirchoff’s Volatage Law a. Current source b. Voltage source c. Resistance d. Reactance
  40. 40. REVIEW QUESTIONSREVIEW QUESTIONS 25. A series connected circuit consists of 3 loads and consume a total power of 50 Watts. It was reconfigured such that 2 are in parallel and the other load is in series with a combination. What is the applied expected powers to be consumed them? a. 50 watts b. 25 watts c. 75 watts d. 45 watts 26. If the number of valence electrons of an atom is less than 4, the substance is usually a. Semiconductor b. An insulator c. A conductor d. None of the above 25. A series connected circuit consists of 3 loads and consume a total power of 50 Watts. It was reconfigured such that 2 are in parallel and the other load is in series with a combination. What is the applied expected powers to be consumed them? a. 50 watts b. 25 watts c. 75 watts d. 45 watts 26. If the number of valence electrons of an atom is less than 4, the substance is usually a. Semiconductor b. An insulator c. A conductor d. None of the above
  41. 41. REVIEW QUESTIONSREVIEW QUESTIONS 27. Electric current in a wire is the flow of a. Free electrons b. Valence electrons c. Bound electrons d. Atoms 28. EMF in a circuit is a form a. Power b. Energy c. Charge d. none 27. Electric current in a wire is the flow of a. Free electrons b. Valence electrons c. Bound electrons d. Atoms 28. EMF in a circuit is a form a. Power b. Energy c. Charge d. none
  42. 42. REVIEW QUESTIONSREVIEW QUESTIONS 29. The SI unit of specific-resistance is a. Mho b. Ohm-n c. Ohm-sq.-m d. Ohm-cm 30. The resistance of a material is ___ its area of cross-section a. Directly proportional b. Inversely proportional c. Independent d. None of these 29. The SI unit of specific-resistance is a. Mho b. Ohm-n c. Ohm-sq.-m d. Ohm-cm 30. The resistance of a material is ___ its area of cross-section a. Directly proportional b. Inversely proportional c. Independent d. None of these
  43. 43. REVIEW QUESTIONSREVIEW QUESTIONS 31. The value of α, i.e., the temperarure coefficient of resistance depends upon the ____ of the material. a. Length b. Volume c. X-sectional area d. Nature and temperature 32. The value of α of a conductor is 1/236 C. The value of a α is a. 1/218 C b. 1/272 C c. 1/254 C d. 1/265 C 31. The value of α, i.e., the temperarure coefficient of resistance depends upon the ____ of the material. a. Length b. Volume c. X-sectional area d. Nature and temperature 32. The value of α of a conductor is 1/236 C. The value of a α is a. 1/218 C b. 1/272 C c. 1/254 C d. 1/265 C
  44. 44. REVIEW QUESTIONSREVIEW QUESTIONS 33. Electrical appliances are not connected in series because a. Series circuit is complicated b. Power loss is greater c. Appliances have different current rating d. None of these 34. Electrical appliances are connected in parallel because it a. Is a simple circuit b. Results in reduced power loss c. Draw less current d. Makes the operation of the appliances independent from each other 33. Electrical appliances are not connected in series because a. Series circuit is complicated b. Power loss is greater c. Appliances have different current rating d. None of these 34. Electrical appliances are connected in parallel because it a. Is a simple circuit b. Results in reduced power loss c. Draw less current d. Makes the operation of the appliances independent from each other
  45. 45. REVIEW QUESTIONSREVIEW QUESTIONS 35. The hot resistance of a 100W, 250V incandecent lamp is a. 2.5 ohms b. 625 ohms c. 25 ohms d. None of these 36. When a number of resistances are connecte in parallel, the total resistance is a. Less than the smallest resistance b. Greater than the smallest resistance c. Between the smallest and greater resistance d. None of these 35. The hot resistance of a 100W, 250V incandecent lamp is a. 2.5 ohms b. 625 ohms c. 25 ohms d. None of these 36. When a number of resistances are connecte in parallel, the total resistance is a. Less than the smallest resistance b. Greater than the smallest resistance c. Between the smallest and greater resistance d. None of these
  46. 46. REVIEW QUESTIONSREVIEW QUESTIONS 37. If the resistances, each of value 36 ohms are connected in parallel, the total resistance is a. 2 ohms b. 54 ohms c. 36 ohms d. None of these 38. Two incandecent lamps of 100 W, 200V are in parallel across the 200 V. The total resistance will be a. 800 ohms b. 200 ohms c. 400 ohms d. 600 ohms 37. If the resistances, each of value 36 ohms are connected in parallel, the total resistance is a. 2 ohms b. 54 ohms c. 36 ohms d. None of these 38. Two incandecent lamps of 100 W, 200V are in parallel across the 200 V. The total resistance will be a. 800 ohms b. 200 ohms c. 400 ohms d. 600 ohms
  47. 47. REVIEW QUESTIONSREVIEW QUESTIONS 39. Three resistors are connected in parallel and draws 1A, 2.5A, and 3.5A, rspectively. If the applied voltage is 210V, what is the total resistance of the circuit? a. 5 ohms b. 147 ohms c. 3 ohms d. 73.5 ohms 40. An ordinary dry cell can deliver about ____ continuously. a. 3 A b. 2 A c. 1/8 A d. None of these 39. Three resistors are connected in parallel and draws 1A, 2.5A, and 3.5A, rspectively. If the applied voltage is 210V, what is the total resistance of the circuit? a. 5 ohms b. 147 ohms c. 3 ohms d. 73.5 ohms 40. An ordinary dry cell can deliver about ____ continuously. a. 3 A b. 2 A c. 1/8 A d. None of these
  48. 48. REVIEW QUESTIONSREVIEW QUESTIONS 41. Four cells of internal resistance 1 ohms, are connected in parallel. The battery resistance will be a. 4 ohms b. 0.25 ohms c. 2 ohms d. 1 ohms 42. Of the following combination of units, the one that is not equal to the watt is a. Joule/sec b. Ampere-volt c. Ampere-ohm d. Ohm/volt 41. Four cells of internal resistance 1 ohms, are connected in parallel. The battery resistance will be a. 4 ohms b. 0.25 ohms c. 2 ohms d. 1 ohms 42. Of the following combination of units, the one that is not equal to the watt is a. Joule/sec b. Ampere-volt c. Ampere-ohm d. Ohm/volt
  49. 49. REVIEW QUESTIONSREVIEW QUESTIONS 43. The power dissipated in a circuit is not equal to a. VI b. IR c. V/R d. IR/V 44. An electric iron draws a current of 15A when connected to 120V power source. Its resistance is a. 0.125 ohms b. 8 ohms c. 16 ohms d. 1,800 ohms 43. The power dissipated in a circuit is not equal to a. VI b. IR c. V/R d. IR/V 44. An electric iron draws a current of 15A when connected to 120V power source. Its resistance is a. 0.125 ohms b. 8 ohms c. 16 ohms d. 1,800 ohms
  50. 50. REVIEW QUESTIONSREVIEW QUESTIONS 45. The power rating of an electric motor which draws a current of 3 A when operated at 120 V is a. 40 W b. 360 W c. 540 W d. 1,080 W 46. When a 100W, 240V, light bulb is operated at 200V, the current that flows in it is a. 0.35 A b. 0.42 A c. 0.5 A d. 0.58 A 45. The power rating of an electric motor which draws a current of 3 A when operated at 120 V is a. 40 W b. 360 W c. 540 W d. 1,080 W 46. When a 100W, 240V, light bulb is operated at 200V, the current that flows in it is a. 0.35 A b. 0.42 A c. 0.5 A d. 0.58 A
  51. 51. REVIEW QUESTIONSREVIEW QUESTIONS 47. The equivalent resistance of a network of three 2 ohm resistors cannot be a. 0.67 ohms b. 1.5 ohms c. 3 ohms d. 6 ohms 48. A 12V potential difference is applied across a series combination of four six-ohms resistors. The current in each six-ohm resistor will be a. 0.5 A b. 2 A c. 8 A d. 6 A 47. The equivalent resistance of a network of three 2 ohm resistors cannot be a. 0.67 ohms b. 1.5 ohms c. 3 ohms d. 6 ohms 48. A 12V potential difference is applied across a series combination of four six-ohms resistors. The current in each six-ohm resistor will be a. 0.5 A b. 2 A c. 8 A d. 6 A
  52. 52. REVIEW QUESTIONSREVIEW QUESTIONS 49. A 12V potential difference is applied across a parallel combination of four six-ohms resistors. The current in each six-ohm resistor will be a. 0.5 A b. 2 A c. 8 A d. 6 A 50. The dissipation of energy can cause burns because it proceduces a. Heat b. Fire c. Friction d. Overload 49. A 12V potential difference is applied across a parallel combination of four six-ohms resistors. The current in each six-ohm resistor will be a. 0.5 A b. 2 A c. 8 A d. 6 A 50. The dissipation of energy can cause burns because it proceduces a. Heat b. Fire c. Friction d. Overload
  53. 53. REVIEW QUESTIONSREVIEW QUESTIONS 51. The rate of expenditure of energy is a. Voltage b. Power c. Current d. Energy 52. In a simple DC power line, the wire that carries the current from the generator to the load is called a. Return wire b. Feeder c. Outgoing wire d. Conductor 51. The rate of expenditure of energy is a. Voltage b. Power c. Current d. Energy 52. In a simple DC power line, the wire that carries the current from the generator to the load is called a. Return wire b. Feeder c. Outgoing wire d. Conductor
  54. 54. REVIEW QUESTIONSREVIEW QUESTIONS 53. A circuit in which the resistance are connected in a continuous run, i.e., end-to-end is a _____ circuit. a. Saries b. Parallel c. Series-parallel d. None of these 54. A battery is connected to an external circuit. The potential drop with the battery is proportional to a. The EMF of the battery b. The current of the circuit c. The equivqlent circuit resistance d. Power dissipated in the circuit 53. A circuit in which the resistance are connected in a continuous run, i.e., end-to-end is a _____ circuit. a. Saries b. Parallel c. Series-parallel d. None of these 54. A battery is connected to an external circuit. The potential drop with the battery is proportional to a. The EMF of the battery b. The current of the circuit c. The equivqlent circuit resistance d. Power dissipated in the circuit
  55. 55. REVIEW QUESTIONSREVIEW QUESTIONS 55. Two wires A and B have the same cross-sectional area and are made of the same material. Ra = 600 ohms and Rb = 100 ohms. The number of times A is longer tahn B is a. 6 b. 2 c. 4 d. 5 56. A coil has a resistance of 100 ohms at 90 C. At 100 C, its resistance is 101 ohms. The temperature coefficient of the wire is a. 0.01 b. 0.1 c. 0.0001 d. 0.001 55. Two wires A and B have the same cross-sectional area and are made of the same material. Ra = 600 ohms and Rb = 100 ohms. The number of times A is longer tahn B is a. 6 b. 2 c. 4 d. 5 56. A coil has a resistance of 100 ohms at 90 C. At 100 C, its resistance is 101 ohms. The temperature coefficient of the wire is a. 0.01 b. 0.1 c. 0.0001 d. 0.001
  56. 56. REVIEW QUESTIONSREVIEW QUESTIONS 57. The resistance of a conductor does not depend on its a. Resistivity b. Length c. Cross-section d. Mass 58. A material which has a negative temperature coefficient of resistance is usually a/an a. Insulator b. Conductor c. Semi-conductor d. All of these 57. The resistance of a conductor does not depend on its a. Resistivity b. Length c. Cross-section d. Mass 58. A material which has a negative temperature coefficient of resistance is usually a/an a. Insulator b. Conductor c. Semi-conductor d. All of these
  57. 57. REVIEW QUESTIONSREVIEW QUESTIONS 59. Which of the following statements is true both for a series and a parallel dc circuit? a. Power additive b. Current are additive c. Voltage are additive d. All of these 60. Two resistors are said to be in series when a. Both carry the same value of current b. Total current equals the sum of the branch current c. Sum of IR drops equal to EMF d. Same current phases through both 59. Which of the following statements is true both for a series and a parallel dc circuit? a. Power additive b. Current are additive c. Voltage are additive d. All of these 60. Two resistors are said to be in series when a. Both carry the same value of current b. Total current equals the sum of the branch current c. Sum of IR drops equal to EMF d. Same current phases through both
  58. 58. REVIEW QUESTIONSREVIEW QUESTIONS 61. According to KCL as applied to a juction in a network of conductors. a. Total sum of currents meeting at the juction is Zero b. No current can leave the juction without same current passing throuhg it c. Net current flow at he juction is positive d. Algebraic sum of the currents meeting at the juction is zero 62. Kirchoff’s Current Law is applicable only to a. Closed-loop circuit b. Electronic circuits c. Juctions in a network d. Electric circuit 61. According to KCL as applied to a juction in a network of conductors. a. Total sum of currents meeting at the juction is Zero b. No current can leave the juction without same current passing throuhg it c. Net current flow at he juction is positive d. Algebraic sum of the currents meeting at the juction is zero 62. Kirchoff’s Current Law is applicable only to a. Closed-loop circuit b. Electronic circuits c. Juctions in a network d. Electric circuit
  59. 59. REVIEW QUESTIONSREVIEW QUESTIONS 63. Kirchoff’s Voltage Lasw is concerned with a. IR drops b. Battery EMF’s c. Juction voltages d. A and b 64. According to KVL, the algebraic sum of all IR drops and EMF’s in any closed loop of a network is always a. Zero b. Negative c. Positive d. Determined by battery EMF 63. Kirchoff’s Voltage Lasw is concerned with a. IR drops b. Battery EMF’s c. Juction voltages d. A and b 64. According to KVL, the algebraic sum of all IR drops and EMF’s in any closed loop of a network is always a. Zero b. Negative c. Positive d. Determined by battery EMF
  60. 60. REVIEW QUESTIONSREVIEW QUESTIONS 65. The algebraic sign of an IR drops primarily dependent upon a. The amount of current flowing through it b. Direction of current c. The value of the resistance d. The battery connection 66. Choose the wrong statement. In the node voltage technique of solvingnetwork parameters, the choice of the reference node does not a. Affect the operation of the circuit b. Change the voltage across the element c. Alter the potential difference between any pair of nodes d. Affect the volatge of various nodes 65. The algebraic sign of an IR drops primarily dependent upon a. The amount of current flowing through it b. Direction of current c. The value of the resistance d. The battery connection 66. Choose the wrong statement. In the node voltage technique of solvingnetwork parameters, the choice of the reference node does not a. Affect the operation of the circuit b. Change the voltage across the element c. Alter the potential difference between any pair of nodes d. Affect the volatge of various nodes
  61. 61. REVIEW QUESTIONSREVIEW QUESTIONS 67. The nodal analysis is primarily based on the application of a. KVL b. KCL. c. Ohms Law d. b and c 68. Superposition theorem can be applied only to circuits having ____ elements a. Non-linear b. Passive c. Linear bilateral d. Resistive 67. The nodal analysis is primarily based on the application of a. KVL b. KCL. c. Ohms Law d. b and c 68. Superposition theorem can be applied only to circuits having ____ elements a. Non-linear b. Passive c. Linear bilateral d. Resistive
  62. 62. REVIEW QUESTIONSREVIEW QUESTIONS 69. The superposition theorem is essentially based on the concept of a. Reciprocity b. Linearity c. non-linearity d. Duality 70. What are the electrons in motion called? a. Current variation b. Electric current c. Electron velocity d. Dynamic electricity 69. The superposition theorem is essentially based on the concept of a. Reciprocity b. Linearity c. non-linearity d. Duality 70. What are the electrons in motion called? a. Current variation b. Electric current c. Electron velocity d. Dynamic electricity
  63. 63. REVIEW QUESTIONSREVIEW QUESTIONS 71. An active element in a circuit is one which _____ . a. Receives energy b. Supplies energy c. a or b d. None of these 72. The siperposition theorem is used when the circuit contains a. A single voltage source b. A number of voltage source c. Passive element only d. None of these 71. An active element in a circuit is one which _____ . a. Receives energy b. Supplies energy c. a or b d. None of these 72. The siperposition theorem is used when the circuit contains a. A single voltage source b. A number of voltage source c. Passive element only d. None of these
  64. 64. REVIEW QUESTIONSREVIEW QUESTIONS 73. Thevenin’s theorem is ____ form of equivalent circuit. a. Voltage b. Current c. Both a and b d. None of these 74. Norton’s theorem is ____ form of an equivalent circuit. a. Voltage b. Current c. Both a and b d. None of these 73. Thevenin’s theorem is ____ form of equivalent circuit. a. Voltage b. Current c. Both a and b d. None of these 74. Norton’s theorem is ____ form of an equivalent circuit. a. Voltage b. Current c. Both a and b d. None of these
  65. 65. REVIEW QUESTIONSREVIEW QUESTIONS 75. In the analysis of vacuum tube circuit, we can generally use ____ theorem. a. Norton’s b. Thevenin’s c. Superposition d. Reciprocity 76. In the analysis of transistor circuits, we generally use _____ theorem. a. Voltage b. Current c. Both a and b d. None of these 75. In the analysis of vacuum tube circuit, we can generally use ____ theorem. a. Norton’s b. Thevenin’s c. Superposition d. Reciprocity 76. In the analysis of transistor circuits, we generally use _____ theorem. a. Voltage b. Current c. Both a and b d. None of these
  66. 66. REVIEW QUESTIONSREVIEW QUESTIONS 77. Under the condition of Maximum Power Transfer, the efficiency is a. 75 % b. 100 % c. 50 % d. 25 % 78. The maximum power transfer theorem is used in a. Electronic circuits b. Home lightning c. Power sytem d. None of the above 77. Under the condition of Maximum Power Transfer, the efficiency is a. 75 % b. 100 % c. 50 % d. 25 % 78. The maximum power transfer theorem is used in a. Electronic circuits b. Home lightning c. Power sytem d. None of the above
  67. 67. REVIEW QUESTIONSREVIEW QUESTIONS 79. delta/star or star/delta transformation technique is applied to a. One terminal b. Two terminal c. Three terminal d. None of these 80. _____ will be used under elctrostatics. a. Incandecent lamp b. Electric motor c. Electric iron d. Lightning rod 79. delta/star or star/delta transformation technique is applied to a. One terminal b. Two terminal c. Three terminal d. None of these 80. _____ will be used under elctrostatics. a. Incandecent lamp b. Electric motor c. Electric iron d. Lightning rod
  68. 68. REVIEW QUESTIONSREVIEW QUESTIONS 81. The value of the absulute permitivity of air is ______ F/m. a. 9 x 10 b. 8,854 x 10 ^ -12 c. 5 x 10 d. 9 X 10 82. When two similar charges eaach of 1 coulumb each are placed 1m apart in air, then the force of repusion is a. 8 x10 N b. 10 N c. 9 x 10^9 N d. 5 x 10 N 81. The value of the absulute permitivity of air is ______ F/m. a. 9 x 10 b. 8,854 x 10 ^ -12 c. 5 x 10 d. 9 X 10 82. When two similar charges eaach of 1 coulumb each are placed 1m apart in air, then the force of repusion is a. 8 x10 N b. 10 N c. 9 x 10^9 N d. 5 x 10 N
  69. 69. REVIEW QUESTIONSREVIEW QUESTIONS 83. Another name for dielectric strength is a. Potetial gradient b. Breakdown voltage c. Dielectric constant d. Electric intensity 84. A heater connected to a 100 V supply, generates 10,000 J of heat energy is 10 sec. How much time is needed to generate the same amount of heat when it is used on 220V line? a. 5 sec b. 2.5 sec c. 7.5 sec d. 4 sec. 83. Another name for dielectric strength is a. Potetial gradient b. Breakdown voltage c. Dielectric constant d. Electric intensity 84. A heater connected to a 100 V supply, generates 10,000 J of heat energy is 10 sec. How much time is needed to generate the same amount of heat when it is used on 220V line? a. 5 sec b. 2.5 sec c. 7.5 sec d. 4 sec.
  70. 70. REVIEW QUESTIONSREVIEW QUESTIONS 85. Kirchoff’s current law is applied in what type of circuit analysis? a. Mesh b. Thevenin’s c. Superposition d. Nodal 86. Inductance and capacitance are not relevant in a dc circuit because a. Frequency of DC is zero b. It is a simple circuit c. They do not exist in dc circuit d. None of the above 85. Kirchoff’s current law is applied in what type of circuit analysis? a. Mesh b. Thevenin’s c. Superposition d. Nodal 86. Inductance and capacitance are not relevant in a dc circuit because a. Frequency of DC is zero b. It is a simple circuit c. They do not exist in dc circuit d. None of the above
  71. 71. REVIEW QUESTIONSREVIEW QUESTIONS 87. Three resistors of 3 ohms resistance each are connected in delta, the equivalent wye-connected resistors will be a. 1 ohm b. 3 ohms c. 9 ohms d. 0.111 ohm 88. Cells are conneted in series when _____ is required a. High current b. High voltage c. High power d. All of these 87. Three resistors of 3 ohms resistance each are connected in delta, the equivalent wye-connected resistors will be a. 1 ohm b. 3 ohms c. 9 ohms d. 0.111 ohm 88. Cells are conneted in series when _____ is required a. High current b. High voltage c. High power d. All of these
  72. 72. ALTERNATING CURRENTALTERNATING CURRENT ALTERNATING CURRENT  A current that is constantly changing in amplitude and direction. ALTERNATING CURRENT  A current that is constantly changing in amplitude and direction. Advantages of AC:  Magnitude can easily be changed (stepped-up or stepped down) with the use of a transformer  Can be produced either single phase for light loads, two phase for control motors, three phase for power distribution and large motor loads or six phase for large scale AC to DC conversion. Advantages of AC:  Magnitude can easily be changed (stepped-up or stepped down) with the use of a transformer  Can be produced either single phase for light loads, two phase for control motors, three phase for power distribution and large motor loads or six phase for large scale AC to DC conversion. BASIC AC THEORY
  73. 73. AC WAVEFORMSAC WAVEFORMS BASIC AC THEORY
  74. 74. AC WAVEFORMSAC WAVEFORMS  Period (T) – the time of one complete cycle in seconds  Frequency (f) – the number of cycles per second (Hertz) a. 1 cycle/second (cps) = 1 Hertz (Hz) b. Proper operation of electrical equipmnent requires specific frequency c. Frequencies lower than 60 Hz would cause flicker when used in lightning  Wavelength (λ) – the length of one complete cycle  Propagation Velocity (v) – the speed of the signal  Phase (Φ) – an angilar measurement that specifies the position of a sine wave relative to reference  Period (T) – the time of one complete cycle in seconds  Frequency (f) – the number of cycles per second (Hertz) a. 1 cycle/second (cps) = 1 Hertz (Hz) b. Proper operation of electrical equipmnent requires specific frequency c. Frequencies lower than 60 Hz would cause flicker when used in lightning  Wavelength (λ) – the length of one complete cycle  Propagation Velocity (v) – the speed of the signal  Phase (Φ) – an angilar measurement that specifies the position of a sine wave relative to reference f = 1 T λ = v f Parameters of Alternating Signal BASIC AC THEORY
  75. 75. AC WAVEFORMSAC WAVEFORMS THE SINUSOIDAL WAVE  Is the most common AC waveform that is practically generated by generators used in household and industries  General equation for sine wave: THE SINUSOIDAL WAVE  Is the most common AC waveform that is practically generated by generators used in household and industries  General equation for sine wave: Where: a(t) – instantaneous amplitude of voltage or current at a given time (t) Am – maximum voltage or current amplitude of the signal ω – angular velocity in rad/sec; ω = 2πf t – time (sec) Φ – phase shift ( + or – in degrees) Where: a(t) – instantaneous amplitude of voltage or current at a given time (t) Am – maximum voltage or current amplitude of the signal ω – angular velocity in rad/sec; ω = 2πf t – time (sec) Φ – phase shift ( + or – in degrees) A(t) = Am sin (ωt + Φ) BASIC AC THEORY
  76. 76. AC WAVEFORMSAC WAVEFORMS  PEAK AMPLITUDE – the height of an AC waveform as measured from the zero mark to the highest positive or lowest negative point on a graph. Also known as the crest amplitude of a wave.  PEAK AMPLITUDE – the height of an AC waveform as measured from the zero mark to the highest positive or lowest negative point on a graph. Also known as the crest amplitude of a wave. AMPLITUDE  It is the height of an AC waveform as depicted on a graph over time (peak, peak-to-peak, average, or RMS quantity) AMPLITUDE  It is the height of an AC waveform as depicted on a graph over time (peak, peak-to-peak, average, or RMS quantity) Measurements of AC Magnitude BASIC AC THEORY
  77. 77. AC WAVEFORMSAC WAVEFORMS  PEAK-TO-PEAK AMPLITUDE – the total height of an AC waveform as measured from maximum positive to maximum negative peaks on a graph. Often abbreviated as “P-P”  PEAK-TO-PEAK AMPLITUDE – the total height of an AC waveform as measured from maximum positive to maximum negative peaks on a graph. Often abbreviated as “P-P” BASIC AC THEORY
  78. 78. AC WAVEFORMSAC WAVEFORMS  AVERAGE AMPLITUDE – the mathematical “mean” of all a waveform’s points over the period of one cycle. Technically, the average amplitude of any waveform with equal-area portions above and below the “zero” line on a graph is zero.  For a sine wave, the average value so calculated is approximately 0.637 of its peak value.  AVERAGE AMPLITUDE – the mathematical “mean” of all a waveform’s points over the period of one cycle. Technically, the average amplitude of any waveform with equal-area portions above and below the “zero” line on a graph is zero.  For a sine wave, the average value so calculated is approximately 0.637 of its peak value. BASIC AC THEORY
  79. 79. AC WAVEFORMSAC WAVEFORMS  RMS AMPLITUDE - “RMS” stands for Root Mean Square, and is a way of expressing an AC quantity of voltage or current in terms functionally equivalent to DC. Also known as the “equivalent” or “DC equivalent” value of an AC voltage or current.  For a sine wave, the RMS value is approximately 0.707 of its peak value.  RMS AMPLITUDE - “RMS” stands for Root Mean Square, and is a way of expressing an AC quantity of voltage or current in terms functionally equivalent to DC. Also known as the “equivalent” or “DC equivalent” value of an AC voltage or current.  For a sine wave, the RMS value is approximately 0.707 of its peak value. BASIC AC THEORY
  80. 80. AC WAVEFORMSAC WAVEFORMS  The crest factor of an AC waveform is the ratio of its peak (crest) to its RMS value.  The form factor of an AC waveform is the ratio of its RMS value to its average value.  The crest factor of an AC waveform is the ratio of its peak (crest) to its RMS value.  The form factor of an AC waveform is the ratio of its RMS value to its average value. BASIC AC THEORY
  81. 81. AC QUANTITIESAC QUANTITIES BASIC AC THEORY
  82. 82. AC QUANTITIESAC QUANTITIES Inductive Reactance (XL) • The property of the inductor to oppose the alternating current Inductive Susceptance (BL) • Reciprocal of inductive reactance Capacitive Reactance (XC) • The property of a capacitor to oppose alternating current Capacitive Susceptance (BC) • Reciprocal of capacitive reactance d Inductive Reactance (XL) • The property of the inductor to oppose the alternating current Inductive Susceptance (BL) • Reciprocal of inductive reactance Capacitive Reactance (XC) • The property of a capacitor to oppose alternating current Capacitive Susceptance (BC) • Reciprocal of capacitive reactance d XL = 2πfLXL = 2πfL BL = 1 BL = 1 XL 2πfL BL = 1 BL = 1 XL 2πfL XC = 1 2πfC XC = 1 2πfC BL = 1 BL = 2πfC XC BL = 1 BL = 2πfC XC BASIC AC THEORY
  83. 83. AC QUANTITIESAC QUANTITIES IMPEDANCE (Z)  Total opposition to the flow of Alternating current  Combination of the resistance in a circuit and the reactances involved IMPEDANCE (Z)  Total opposition to the flow of Alternating current  Combination of the resistance in a circuit and the reactances involved Z = R + jXeq Z = |Z| ∠φ Where: |Z| = √ R2 + X2 φ = Arctan Xeq R Where: |Z| = √ R2 + X2 φ = Arctan Xeq R BASIC AC THEORY
  84. 84. AC QUANTITIESAC QUANTITIES  If I = Im ∠β is the resulting current drawn by a passive, linear RLC circuit from a source voltage V = Vm ∠θ, then Where: Z = Vm = √ R2 + X2 = magnitude of the impedance Im φ = θ – β = tan-1 X = phase angle of the impedance R R = Zcos φ = active or real component of the impedance X = Zsin φ = reactive or quadrature component of impedance  If I = Im ∠β is the resulting current drawn by a passive, linear RLC circuit from a source voltage V = Vm ∠θ, then Where: Z = Vm = √ R2 + X2 = magnitude of the impedance Im φ = θ – β = tan-1 X = phase angle of the impedance R R = Zcos φ = active or real component of the impedance X = Zsin φ = reactive or quadrature component of impedance Z = V = Vm ∠θ = Z ∠φ I Im ∠β Z = V = Vm ∠θ = Z ∠φ I Im ∠β Z cosφ + jZsin φ = R + jX = √ R2 + X2 ∠ tan-1 X R Z cosφ + jZsin φ = R + jX = √ R2 + X2 ∠ tan-1 X R BASIC AC THEORY
  85. 85. AC QUANTITIESAC QUANTITIES ADMITTANCE (Y)  The reciprocal of impedance  Expressed in siemens or mho (S) ADMITTANCE (Y)  The reciprocal of impedance  Expressed in siemens or mho (S) Where: Y = Im = √ G2 + B2 = 1 = magnitude of the admittance Z φy = β – θ = φ = tan-1 B = phase angle of the admittance G G = Ycos φy = conductive/conductance component B = Ysin φy = susceptive/susceptance component Where: Y = Im = √ G2 + B2 = 1 = magnitude of the admittance Z φy = β – θ = φ = tan-1 B = phase angle of the admittance G G = Ycos φy = conductive/conductance component B = Ysin φy = susceptive/susceptance component Y = Im ∠ β – θ = Y = Ycos φy + jYsin φy = G + jB Vm Y = Im ∠ β – θ = Y = Ycos φy + jYsin φy = G + jB Vm Y = √ G2 + B2 ∠ tan-1 B G Y = √ G2 + B2 ∠ tan-1 B G BASIC AC THEORY
  86. 86. AC RESISTOR CIRCUITAC RESISTOR CIRCUIT With an AC circuit like this which is purely resistive, the relationship of the voltage and current is as shown:  Voltage (e) is in phase with the current (i)  Power is never a negative value. When the current is positive (above the line), the voltage is also positive, resulting in a power (p=ie) of a positve value  This consistent “polarity” of a power tell us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy. Whether the current is negative or positive, a resistor still dissipated energy. With an AC circuit like this which is purely resistive, the relationship of the voltage and current is as shown:  Voltage (e) is in phase with the current (i)  Power is never a negative value. When the current is positive (above the line), the voltage is also positive, resulting in a power (p=ie) of a positve value  This consistent “polarity” of a power tell us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy. Whether the current is negative or positive, a resistor still dissipated energy. AC CIRCUITS Impedance (Z) = RImpedance (Z) = R
  87. 87. AC INDUCTOR CIRCUITAC INDUCTOR CIRCUIT  The most distinguishing electrical characteristics of an L circuit is that current lags voltage by 90 electrical degrees  Because the current and voltage waves arae 90o out of phase, there sre times when one is positive while the other is negative, resulting in equally frequent occurences of negative instantaneous power.  Negative power means that the inductor is releasing power back to the circuit, while a positive power means that it is absorbing power from the circuit.  The inductor releases just as much power back to the circuit as it absorbs over the span of a complete cycle.  The most distinguishing electrical characteristics of an L circuit is that current lags voltage by 90 electrical degrees  Because the current and voltage waves arae 90o out of phase, there sre times when one is positive while the other is negative, resulting in equally frequent occurences of negative instantaneous power.  Negative power means that the inductor is releasing power back to the circuit, while a positive power means that it is absorbing power from the circuit.  The inductor releases just as much power back to the circuit as it absorbs over the span of a complete cycle. Impedance (Z) = jXL Impedance (Z) = jXL AC CIRCUITS
  88. 88. AC INDUCTOR CIRCUITAC INDUCTOR CIRCUIT o Inductive reactance is the opposition that an inductor offers to alternating current due to its phase-shifted storage and release of energy in its magnetic field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). o Inductive reactance can be calculated using this formula: XL = 2πfL o The angular velocity of an AC circuit is another way of expressing its frequency, in units of electrical radians per second instead of cycles per second. It is symbolized by the lowercase Greek letter “omega,” or ω. o Inductive reactance increases with increasing frequency. In other words, the higher the frequency, the more it opposes the AC flow of electrons. o Inductive reactance is the opposition that an inductor offers to alternating current due to its phase-shifted storage and release of energy in its magnetic field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). o Inductive reactance can be calculated using this formula: XL = 2πfL o The angular velocity of an AC circuit is another way of expressing its frequency, in units of electrical radians per second instead of cycles per second. It is symbolized by the lowercase Greek letter “omega,” or ω. o Inductive reactance increases with increasing frequency. In other words, the higher the frequency, the more it opposes the AC flow of electrons. AC CIRCUITS
  89. 89. AC CAPACITOR CIRCUITAC CAPACITOR CIRCUIT  The most distinguishing electrical characteristics of an C circuit is that leads the voltage by 90 electrical degrees  The current through a capacitor is a reaction against the change in voltage across it  A capacitor’s opposition to change in voltage translates to an opposition to alternating voltage in general, which is by definition always changing in instantaneous magnitude and direction. For any given magnitude of AC voltage at a given frequency, a capacitor of given size will “conduct” a certain magnitude of AC current.  The phase angle of a capacitor’s opposition to current is -90o ,meaning that a capacitor’s opposition to current is a negative imaginary quantity  The most distinguishing electrical characteristics of an C circuit is that leads the voltage by 90 electrical degrees  The current through a capacitor is a reaction against the change in voltage across it  A capacitor’s opposition to change in voltage translates to an opposition to alternating voltage in general, which is by definition always changing in instantaneous magnitude and direction. For any given magnitude of AC voltage at a given frequency, a capacitor of given size will “conduct” a certain magnitude of AC current.  The phase angle of a capacitor’s opposition to current is -90o ,meaning that a capacitor’s opposition to current is a negative imaginary quantity Impedance (Z) = -jXC Impedance (Z) = -jXC AC CIRCUITS
  90. 90. AC CAPACITOR CIRCUITAC CAPACITOR CIRCUIT o Capacitive reactance is the opposition that a capacitor offers to alternating current due to its phase-shifted storage and release of energy in its electric field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). o Capacitive reactance can be calculated using this formula: XC = 1/(2πfC) o Capacitive reactance decreases with increasing frequency. In other words, the higher the frequency, the less it opposes (the more it “conducts”) the AC flow of electrons. o Capacitive reactance is the opposition that a capacitor offers to alternating current due to its phase-shifted storage and release of energy in its electric field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). o Capacitive reactance can be calculated using this formula: XC = 1/(2πfC) o Capacitive reactance decreases with increasing frequency. In other words, the higher the frequency, the less it opposes (the more it “conducts”) the AC flow of electrons. AC CIRCUITS
  91. 91. SERIES RESITOR-INDCUTOR CIRCUITSERIES RESITOR-INDCUTOR CIRCUIT  For a series resistor-inductor circuit, the voltage and current relation is determined in its phase shift. Thus, current lags voltage by a phase shift (θ)  For a series resistor-inductor circuit, the voltage and current relation is determined in its phase shift. Thus, current lags voltage by a phase shift (θ) Impedance (Z) = R + jXL Impedance (Z) = R + jXL Admittance (Y) = 1 = R – jXL R + jXL R2 + jXL 2 Admittance (Y) = 1 = R – jXL R + jXL R2 + jXL 2 AC CIRCUITS
  92. 92. SERIES RESITOR-INDCUTOR CIRCUITSERIES RESITOR-INDCUTOR CIRCUIT o When resistors and inductors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0o and +90o . The circuit current will have a phase angle somewhere between 0o and -90o . Series AC circuits exhibit the same fundamental properties as series DC circuits: current is uniform throughout. o When resistors and inductors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0o and +90o . The circuit current will have a phase angle somewhere between 0o and -90o . Series AC circuits exhibit the same fundamental properties as series DC circuits: current is uniform throughout. Phase shift (θ) = Arctan ( XL ) |Z| = √ R2 + jXL 2 = e R i Phase shift (θ) = Arctan ( XL ) |Z| = √ R2 + jXL 2 = e R i AC CIRCUITS
  93. 93. SERIES RESISTOR-CAPACITOR CIRCUITSERIES RESISTOR-CAPACITOR CIRCUIT  For a series resistor – capacitor circuit, the voltage and current relation is determined by the phase shift. Thus the current leads the voltage by an angle less than 90 degrees but greater than 0 degrees.  For a series resistor – capacitor circuit, the voltage and current relation is determined by the phase shift. Thus the current leads the voltage by an angle less than 90 degrees but greater than 0 degrees. Impedance (Z) = R – jXC Impedance (Z) = R – jXC Admittance (Y) = 1 = R + jXC R – jXC R2 + jXC 2 Admittance (Y) = 1 = R + jXC R – jXC R2 + jXC 2 AC CIRCUITS
  94. 94. SERIES RESISTOR-CAPACITOR CIRCUITSERIES RESISTOR-CAPACITOR CIRCUIT Phase shift (θ) = Arctan ( XC ) |Z| = √ R2 + jXC 2 = e R i Phase shift (θ) = Arctan ( XC ) |Z| = √ R2 + jXC 2 = e R i AC CIRCUITS
  95. 95. PARALLEL RESISTOR-INDUCTORPARALLEL RESISTOR-INDUCTOR Y = G – jβL where: G – conductance = 1/R βL – susceptance = 1/XL Z = E , by Ohm’s Law I  The basic approachwith regarda to parallel circuits is using admittance because it is additive Y = G – jβL where: G – conductance = 1/R βL – susceptance = 1/XL Z = E , by Ohm’s Law I  The basic approachwith regarda to parallel circuits is using admittance because it is additive AC CIRCUITS
  96. 96. PARALLEL RESISTOR-INDUCTORPARALLEL RESISTOR-INDUCTOR o When resistors and inductors are mixed together in parallel circuits (just like in series cicuits), the total impedance will have a phase angle somewhere between 0o and +90o . The circuit current will have a phase angle somewhere between 0o and -90o . o Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughour the circuit, brach currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance. o When resistors and inductors are mixed together in parallel circuits (just like in series cicuits), the total impedance will have a phase angle somewhere between 0o and +90o . The circuit current will have a phase angle somewhere between 0o and -90o . o Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughour the circuit, brach currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance. AC CIRCUITS
  97. 97. PARALLEL RESISTOR-CAPACITORPARALLEL RESISTOR-CAPACITOR Y = G + jβC where: G – conductance = 1/R βC – susceptance = 1/XC o When resistors and capacitors are mixxed together in circuits, the total impedance will have a phase angle somewhere between 0o and -90o . Y = G + jβC where: G – conductance = 1/R βC – susceptance = 1/XC o When resistors and capacitors are mixxed together in circuits, the total impedance will have a phase angle somewhere between 0o and -90o . AC CIRCUITS
  98. 98. APPARENT POWER (S)APPARENT POWER (S) APPARENT POWER  Represents the rate at which the total energy is supplied to the system  Measured in volt-amperes (VA)  It has two components, the Real Power and the Capacitive or Inductive Reactive Power APPARENT POWER  Represents the rate at which the total energy is supplied to the system  Measured in volt-amperes (VA)  It has two components, the Real Power and the Capacitive or Inductive Reactive Power POWER IN AC CIRCUITS S = Vrms Irms = Irms 2 |Z|
  99. 99. APPARENT POWER (S)APPARENT POWER (S) Power Triangle Complex Power S = P ± jQ POWER IN AC CIRCUITS
  100. 100. REAL POWER (R)REAL POWER (R) REAL POWER  The power consumed by the resistive component  Also called True Power, Useful Power and Productive Power  Measured in Watts (W)  It is equal to the product of the apparent power and the power factor REAL POWER  The power consumed by the resistive component  Also called True Power, Useful Power and Productive Power  Measured in Watts (W)  It is equal to the product of the apparent power and the power factor  Cosine of the power factor angle (θ)  Measure of the power that is dissipated by the cicuit in relation to the apparent power and is usually given as a decimal or percentage  Cosine of the power factor angle (θ)  Measure of the power that is dissipated by the cicuit in relation to the apparent power and is usually given as a decimal or percentage POWER IN AC CIRCUITS Pf = cos θ P = Scos θ Power Factor
  101. 101. REAL POWER (R)REAL POWER (R)  Ratio of the Real Power to the Apparent Power ( P ) S when: Pf = 1.0 I is in phase with V; resistive system Pf = lagging I lags V by θ; inductive system Pf = leading I leads V by θ; capacitive system Pf = 0.0 lag I lags V by 90o; purely inductive Pf = 0.0 lead I leads V by 90o; purely capacitive  The angle between the apparent power and the real poweer in the power triangle Let v(t) = Vm cos(ωt + θv) volts V = Vrms ∠θv i(t) = Im cos(ωt + θi) A I = Irms ∠θi  Ratio of the Real Power to the Apparent Power ( P ) S when: Pf = 1.0 I is in phase with V; resistive system Pf = lagging I lags V by θ; inductive system Pf = leading I leads V by θ; capacitive system Pf = 0.0 lag I lags V by 90o; purely inductive Pf = 0.0 lead I leads V by 90o; purely capacitive  The angle between the apparent power and the real poweer in the power triangle Let v(t) = Vm cos(ωt + θv) volts V = Vrms ∠θv i(t) = Im cos(ωt + θi) A I = Irms ∠θi POWER IN AC CIRCUITS Power factor Angle (θ)
  102. 102. REAL POWER (R)REAL POWER (R) Where: θ = phase shfit between v(t) and i(t) or the phase angle of the equivalent impedance Where: θ = phase shfit between v(t) and i(t) or the phase angle of the equivalent impedance POWER IN AC CIRCUITS Instantaneous Power (watts) Average Power (watts) P(t) = v(t) i(t)P(t) = v(t) i(t) P(t) = ½ VmIm cos (θv – θi) + ½ VmIm cos (2ωt + θv + θi)P(t) = ½ VmIm cos (θv – θi) + ½ VmIm cos (2ωt + θv + θi) P(t) = ½ VmIm cos (θv – θi) = VmIm cos θP(t) = ½ VmIm cos (θv – θi) = VmIm cos θ
  103. 103. REACTIVE POWER (QL or QC)REACTIVE POWER (QL or QC) REACTIVE POWER  Represents the rate at which energy is stored or released in any of the energy storing elements (the inductor or the capacitor)  Also called the imaginary power, non-productive or wattless power  Measured in volt-ampere reactive (Var)  When the capacitor and inductor are both present, the reactive power associated with them take opposite signs since they do not store or release energy at the same time  It is positive for inductive power (QL) and negative for capacitive power (QC) REACTIVE POWER  Represents the rate at which energy is stored or released in any of the energy storing elements (the inductor or the capacitor)  Also called the imaginary power, non-productive or wattless power  Measured in volt-ampere reactive (Var)  When the capacitor and inductor are both present, the reactive power associated with them take opposite signs since they do not store or release energy at the same time  It is positive for inductive power (QL) and negative for capacitive power (QC)  Ratio of the Reactive Power to the Apparent Power  Sine of the power factor angle (θ)  Ratio of the Reactive Power to the Apparent Power  Sine of the power factor angle (θ) POWER IN AC CIRCUITS Q = VmIm sin θQ = VmIm sin θ Reactive factor Rf = sin θRf = sin θ
  104. 104. BALANCED THREE PHASE SYSTEMSBALANCED THREE PHASE SYSTEMS BALANCED 3-PHASE SYSTEM  Comprises of three identical single-phase systems operating at a 120o phase displacement from one another. This means that a balance three-phase system provides three voltages(and currents) that are equal in magnitude and separated by 120o from each other BALANCED 3-PHASE SYSTEM  Comprises of three identical single-phase systems operating at a 120o phase displacement from one another. This means that a balance three-phase system provides three voltages(and currents) that are equal in magnitude and separated by 120o from each other Three-Phase, 3-wire systems  Provide only one type of voltage(line to line) both single phase and three phase loads Three-Phase, 4-wire systems  Provide two types of voltages(line to line and line to neutral) to both single phase and three phase loads Three-Phase, 3-wire systems  Provide only one type of voltage(line to line) both single phase and three phase loads Three-Phase, 4-wire systems  Provide two types of voltages(line to line and line to neutral) to both single phase and three phase loads BALANCED THREE PHASE SYSTEM CLASSIFICATION
  105. 105. BALANCED THREE PHASE SYSTEMSBALANCED THREE PHASE SYSTEMS and VLL and VLN are out of phase by 30o and VLL and VLN are out of phase by 30o BALANCED Y-system VLL = √3 VLN VLL = √3 VLN IL = IP IL = IP and IL and IP are out of phase by 30o Where: VLL or VL - line to line or line voltage VLN or VP - line to neutral or phase voltage IL - line current IP - phase current and IL and IP are out of phase by 30o Where: VLL or VL - line to line or line voltage VLN or VP - line to neutral or phase voltage IL - line current IP - phase current BALANCED ∆-system IL = √3 IP IL = √3 IP VLL = VLN VLL = VLN BALANCED THREE PHASE SYSTEM
  106. 106. ALTERNATING CURRENTALTERNATING CURRENT VJH watts vars va VJH watts vars va BALANCED THREE PHASE SYSTEM Note: for balanced 3-phase systems:Note: for balanced 3-phase systems: IA + IB + IC = 0IA + IB + IC = 0 VAN + VBN + VCN = 0VAN + VBN + VCN = 0 VAB + VBC + VCA = 0VAB + VBC + VCA = 0 P = 3VPIPcos θ = √3 VLIL cos θP = 3VPIPcos θ = √3 VLIL cos θ Q = 3VPIPsin θ = √3 VLIL sin θQ = 3VPIPsin θ = √3 VLIL sin θ S = 3VPIP = √3 VLIL S = 3VPIP = √3 VLIL THREE-PHASE POWER
  107. 107. REVIEW QUESTIONSREVIEW QUESTIONS 1. The description of two sine waves that are in step with each other going through their maximum and minimum points ate the same time and in the same direction. a. Sine waves in phase b. Stepped sine waves c. Phased sine waves d. Sine waves in coordination 2. Term used for the out of phase, non-productive power associated with inductors and capacitors. a. Effective power b. True power c. Reactive power d. Peak envelope power 1. The description of two sine waves that are in step with each other going through their maximum and minimum points ate the same time and in the same direction. a. Sine waves in phase b. Stepped sine waves c. Phased sine waves d. Sine waves in coordination 2. Term used for the out of phase, non-productive power associated with inductors and capacitors. a. Effective power b. True power c. Reactive power d. Peak envelope power
  108. 108. 3. Refers to a reactive power. a. Wattles, non productive power b. Power consumed in circuit Q c. Power loss because of capacitor leakage d. Power consumed in wire resistance in an inductor 4. Term used for an out-of-phase, non-productive power associated with inductors and capacitors. a. Effective power b. Reactive power c. Peak envelope power d. True power 3. Refers to a reactive power. a. Wattles, non productive power b. Power consumed in circuit Q c. Power loss because of capacitor leakage d. Power consumed in wire resistance in an inductor 4. Term used for an out-of-phase, non-productive power associated with inductors and capacitors. a. Effective power b. Reactive power c. Peak envelope power d. True power REVIEW QUESTIONSREVIEW QUESTIONS
  109. 109. 5. The product of current and voltage in an AC circuit refers to the a. Real power b. Useful power c. Apparent power d. DC power 6. The distance covered or traveled by a waveform during the time interval of one complete cycle. a. Frequency b. Wavelength c. Time slot d. Wave time 5. The product of current and voltage in an AC circuit refers to the a. Real power b. Useful power c. Apparent power d. DC power 6. The distance covered or traveled by a waveform during the time interval of one complete cycle. a. Frequency b. Wavelength c. Time slot d. Wave time REVIEW QUESTIONSREVIEW QUESTIONS
  110. 110. 7. The power dissipated accross the resistance in an AC circuit. a. Real power b. Reactive power c. Apparent power d. True power 8. It is the number of complete cycles of alternating voltage or current complete each second a. Period b. Frequency c. Amplitude d. Phase 7. The power dissipated accross the resistance in an AC circuit. a. Real power b. Reactive power c. Apparent power d. True power 8. It is the number of complete cycles of alternating voltage or current complete each second a. Period b. Frequency c. Amplitude d. Phase REVIEW QUESTIONSREVIEW QUESTIONS
  111. 111. 9. How many degrees are there in one complete cycle? a. 720 deg b. 360 deg c. 180 deg d. 90 deg 10. The impedance in the study of electronics is represented by resistance and ___ . a. Reactance b. Inductance and capacitance c. Inductance d. capacitance 9. How many degrees are there in one complete cycle? a. 720 deg b. 360 deg c. 180 deg d. 90 deg 10. The impedance in the study of electronics is represented by resistance and ___ . a. Reactance b. Inductance and capacitance c. Inductance d. capacitance REVIEW QUESTIONSREVIEW QUESTIONS
  112. 112. 11. It is the current that is eliminated by a synchro capacitor? a. Magnetizing stator b. Loss c. Stator d. Rotor 12. It is a rotaing sector that represent either current or voltage in an AC circuit. a. Resistance b. Phasor c. Solar diagram d. velocity 11. It is the current that is eliminated by a synchro capacitor? a. Magnetizing stator b. Loss c. Stator d. Rotor 12. It is a rotaing sector that represent either current or voltage in an AC circuit. a. Resistance b. Phasor c. Solar diagram d. velocity REVIEW QUESTIONSREVIEW QUESTIONS
  113. 113. 13. The relationship of the voltage accros an inductor to its current is described as a. Leading the current by 90 deg b. Lagging the current by 90 deg c. Leading the current by 180 deg d. In phase with the current 14. Find the phase angle between the voltage across through the cicuit when Xc is 25ohms, R is 100 ohms and XL is 50 ohms. a. 76 deg with the voltage leading the current b. 24 deg with the voltage lagging the current c. 14 deg with the voltage lagging the current d. 76 deg with the voltage lagging the current 13. The relationship of the voltage accros an inductor to its current is described as a. Leading the current by 90 deg b. Lagging the current by 90 deg c. Leading the current by 180 deg d. In phase with the current 14. Find the phase angle between the voltage across through the cicuit when Xc is 25ohms, R is 100 ohms and XL is 50 ohms. a. 76 deg with the voltage leading the current b. 24 deg with the voltage lagging the current c. 14 deg with the voltage lagging the current d. 76 deg with the voltage lagging the current REVIEW QUESTIONSREVIEW QUESTIONS
  114. 114. 15. Calculate the period of an alternating current having a equation of I=20sin 120πt a. 4.167 ms b. 8.33 ms c. 16.67 ms d. 33.33 ms 16. What do you mean by root-mean-square (rms) value? a. It is the average value b. It is the effective value c. It is the value that causes the same heating effect as the DC voltage d. b or c 15. Calculate the period of an alternating current having a equation of I=20sin 120πt a. 4.167 ms b. 8.33 ms c. 16.67 ms d. 33.33 ms 16. What do you mean by root-mean-square (rms) value? a. It is the average value b. It is the effective value c. It is the value that causes the same heating effect as the DC voltage d. b or c REVIEW QUESTIONSREVIEW QUESTIONS
  115. 115. 17. The maximum instances value of a vrying current, voltage or power equal to 1.414 times the effective value of a sine wave. a. RMS value b. Peak value c. Effective value d. Peak to Peak value 18. If an AC signal has a peak voltage of 55V, what is the average value? a. 34.98 V b. 61.05V c. 86.34 V d. 38.89 V 17. The maximum instances value of a vrying current, voltage or power equal to 1.414 times the effective value of a sine wave. a. RMS value b. Peak value c. Effective value d. Peak to Peak value 18. If an AC signal has a peak voltage of 55V, what is the average value? a. 34.98 V b. 61.05V c. 86.34 V d. 38.89 V REVIEW QUESTIONSREVIEW QUESTIONS
  116. 116. 19. If an AC signal has an average voltage of 18V, what is the rms voltage? a. 12.726 V b. 19.980 V c. 25.380 V d. 16.213 V 20. A 220-V, 60Hz is driving a series RL circuit. Determine the current if R=100 ohms and 20 mH inductance a. 2.2 A (lagging) b. 2.0 A (lagging) c. 2.2 A (leading) d. 2.0 A (leading) 19. If an AC signal has an average voltage of 18V, what is the rms voltage? a. 12.726 V b. 19.980 V c. 25.380 V d. 16.213 V 20. A 220-V, 60Hz is driving a series RL circuit. Determine the current if R=100 ohms and 20 mH inductance a. 2.2 A (lagging) b. 2.0 A (lagging) c. 2.2 A (leading) d. 2.0 A (leading) REVIEW QUESTIONSREVIEW QUESTIONS
  117. 117. 21. Ignoring any inductive effects, what is the impedance of RC series capacitor made up of a 56K ohm resistor and a 0.33uF capacitor at a signal frequency of 4650 Hz. a. 66730 ohms b. 57019 ohms c. 45270 ohms d. 10730 ohms 22.What is the time constant of a 500mH coil and a 3300 ohm resistor in series? a. 0.00015 sec b. 6.6 sec c. 0.0015 sec d. 0.0000015 sec 21. Ignoring any inductive effects, what is the impedance of RC series capacitor made up of a 56K ohm resistor and a 0.33uF capacitor at a signal frequency of 4650 Hz. a. 66730 ohms b. 57019 ohms c. 45270 ohms d. 10730 ohms 22.What is the time constant of a 500mH coil and a 3300 ohm resistor in series? a. 0.00015 sec b. 6.6 sec c. 0.0015 sec d. 0.0000015 sec REVIEW QUESTIONSREVIEW QUESTIONS
  118. 118. 23. What is the realtionship between frequency and the value of XC ? a. Frequency has no effect b. XC varies inversely with frequency c. XC varies indirectly with frequency d. XC varies directly with frequency 24. The reactance of a 25mH coil at 5000Hz which of the following? a. 785 ohms b. 785000 ohms c. 13 ohms d. 0.0012 ohms 23. What is the realtionship between frequency and the value of XC ? a. Frequency has no effect b. XC varies inversely with frequency c. XC varies indirectly with frequency d. XC varies directly with frequency 24. The reactance of a 25mH coil at 5000Hz which of the following? a. 785 ohms b. 785000 ohms c. 13 ohms d. 0.0012 ohms REVIEW QUESTIONSREVIEW QUESTIONS
  119. 119. 25. There are no transients in pure resistive circuites becaus they a. Offer high resistance b. Obey ohm’s Law c. Are linear circuits d. Have no stored energy 26. The reciprocal of capacitance is called a. Elastance b. Conductance c. Permitivity d. permeability 25. There are no transients in pure resistive circuites becaus they a. Offer high resistance b. Obey ohm’s Law c. Are linear circuits d. Have no stored energy 26. The reciprocal of capacitance is called a. Elastance b. Conductance c. Permitivity d. permeability REVIEW QUESTIONSREVIEW QUESTIONS
  120. 120. REVIEW QUESTIONSREVIEW QUESTIONS 27. The AC system is prefered over DC system because a. Ac voltages can easily changed in amgnitude b. Dc motors do not have fine speed control c. High voltage AC transmission is less efficient d. DC voltage cannot be used for domestic aplliences 28. An altenating voltage is given by v = 20 sin 157 t. The frequency of the alternating voltage is a. 50 Hz b. 25 HZ c. 100 Hz d. 75 Hz 27. The AC system is prefered over DC system because a. Ac voltages can easily changed in amgnitude b. Dc motors do not have fine speed control c. High voltage AC transmission is less efficient d. DC voltage cannot be used for domestic aplliences 28. An altenating voltage is given by v = 20 sin 157 t. The frequency of the alternating voltage is a. 50 Hz b. 25 HZ c. 100 Hz d. 75 Hz
  121. 121. REVIEW QUESTIONSREVIEW QUESTIONS 29. An alternating current given by i = 10 sin 314 t. The time taken to generate two cycles of current is a. 20 ms b. 10 ms c. 40 ms d. 50 ms 30. In a pure resistive circuit, the instantaneous voltage and are current are given by: v=250 sin 314t i=10sin314t The peak power in the circuit is a.1250 W b. 25 W c. 2500 W d. 250 w 29. An alternating current given by i = 10 sin 314 t. The time taken to generate two cycles of current is a. 20 ms b. 10 ms c. 40 ms d. 50 ms 30. In a pure resistive circuit, the instantaneous voltage and are current are given by: v=250 sin 314t i=10sin314t The peak power in the circuit is a.1250 W b. 25 W c. 2500 W d. 250 w
  122. 122. REVIEW QUESTIONSREVIEW QUESTIONS 31. An average value of 6.63 A is _____ the effective value of 7.07 A. a. The same area b. Less than c. Greater than d. Any of these 32. In an R-L series circuit, the resistance is 10 ohms and the inductive reactance is 10 ohms. The phase angle between the applied voltage and circuit current will be a. 45 deg b. 30 deg c. 60 deg d. 36.8 deg 31. An average value of 6.63 A is _____ the effective value of 7.07 A. a. The same area b. Less than c. Greater than d. Any of these 32. In an R-L series circuit, the resistance is 10 ohms and the inductive reactance is 10 ohms. The phase angle between the applied voltage and circuit current will be a. 45 deg b. 30 deg c. 60 deg d. 36.8 deg
  123. 123. REVIEW QUESTIONSREVIEW QUESTIONS 33.An R-L series ac circuit has 15V across the resistor and 20V across the inductor. The supply volatge is a. 35 V b. 5 V c. 25 V d. 175 V 34. The active and reactive powers of an inductive circuit are equal. The power factor of the circuit is a. 0.8 lagging b. 0.707 lagging c. 0.6 lagging d. 0.5 lagging 33.An R-L series ac circuit has 15V across the resistor and 20V across the inductor. The supply volatge is a. 35 V b. 5 V c. 25 V d. 175 V 34. The active and reactive powers of an inductive circuit are equal. The power factor of the circuit is a. 0.8 lagging b. 0.707 lagging c. 0.6 lagging d. 0.5 lagging
  124. 124. REVIEW QUESTIONSREVIEW QUESTIONS 35. A circuit when connected to 200V mains takes a current of 20 A, leading the voltage by one-twelfth of time period. The circuit resistance is a. 10 ohms b. 8.66 ohms c. 20 ohms d. 17.32 ohms 36. An AC circuit has a resistance of 6 ohms, inductive reactance of 20 ohms, and capacitive reactance of 12 ohms. The circuit power will be a. 0.8 lagging b. 0.8 leading c. 0.6 lagging d. 0.6 leading 35. A circuit when connected to 200V mains takes a current of 20 A, leading the voltage by one-twelfth of time period. The circuit resistance is a. 10 ohms b. 8.66 ohms c. 20 ohms d. 17.32 ohms 36. An AC circuit has a resistance of 6 ohms, inductive reactance of 20 ohms, and capacitive reactance of 12 ohms. The circuit power will be a. 0.8 lagging b. 0.8 leading c. 0.6 lagging d. 0.6 leading
  125. 125. REVIEW QUESTIONSREVIEW QUESTIONS 37. An alternating voltage of 80 + j60 V is applied to a circuit and the current flowing is -40 + j10 A. Find the phase angle. a. 25 deg b. 50 deg c. 75 deg d. 100 deg 38. A current wave is represented by the equation i = 10 sin 251t. The average and RMS value of current are a. 7.07 A; 6.63A b. 6.36A; 7.07A c. 10A; 7.07A d. 6.36A; 10A 37. An alternating voltage of 80 + j60 V is applied to a circuit and the current flowing is -40 + j10 A. Find the phase angle. a. 25 deg b. 50 deg c. 75 deg d. 100 deg 38. A current wave is represented by the equation i = 10 sin 251t. The average and RMS value of current are a. 7.07 A; 6.63A b. 6.36A; 7.07A c. 10A; 7.07A d. 6.36A; 10A
  126. 126. REVIEW QUESTIONSREVIEW QUESTIONS 39. Calculate the susceptance in mho of a circuit consisting of resistor of 10 ohms in series with a conductor of 0.1H, when the frequency is 50Hz. a. 0.0303 b. 0.0092 c. -0.029 d. 32.95 40. An inductive circuit of resistance 16.5 ohms and inductive of 0.14H takes a current of 25 A. If the frequency is 50Hz, the supply voltage is a. 117.4 V b. 1174 V c. 1714 V d. 1471 V 39. Calculate the susceptance in mho of a circuit consisting of resistor of 10 ohms in series with a conductor of 0.1H, when the frequency is 50Hz. a. 0.0303 b. 0.0092 c. -0.029 d. 32.95 40. An inductive circuit of resistance 16.5 ohms and inductive of 0.14H takes a current of 25 A. If the frequency is 50Hz, the supply voltage is a. 117.4 V b. 1174 V c. 1714 V d. 1471 V
  127. 127. REVIEW QUESTIONSREVIEW QUESTIONS 41. The current taken by a circuit is 1.2 A when the applied potential difference is 250 V and the power taken is 135W. The power factor is a. 0.35 b. 0.45 c. 0.55 d. 0.65 42. A capacitor has a capacitance of 20uF. The current supplied if it is placed across a 1100 V, 25 Hz supply. a. 3.554 A b. 6.91 A c. 3.45 A d. 9.61 A 41. The current taken by a circuit is 1.2 A when the applied potential difference is 250 V and the power taken is 135W. The power factor is a. 0.35 b. 0.45 c. 0.55 d. 0.65 42. A capacitor has a capacitance of 20uF. The current supplied if it is placed across a 1100 V, 25 Hz supply. a. 3.554 A b. 6.91 A c. 3.45 A d. 9.61 A
  128. 128. REVIEW QUESTIONSREVIEW QUESTIONS 43. A non-inductive load takes a 100A at 100V. Calculate the inductance of the inductor to be connected in series in order that the same current is supplied from 220 V, 50 Hz mains. a. 1.96 ohms b. 6.91 ohms c. 19.6 ohms d. 9.61 ohms 44. An inductor having negligible resistance and an inductance of 0.07H is connected in series with a resiostor of 20 ohms resitance across a 200, 50 Hz supply. The maximum energy stored in the coil is a. 3.175 J b. 1.585 J c. 0.236 J d. 0.33 J 43. A non-inductive load takes a 100A at 100V. Calculate the inductance of the inductor to be connected in series in order that the same current is supplied from 220 V, 50 Hz mains. a. 1.96 ohms b. 6.91 ohms c. 19.6 ohms d. 9.61 ohms 44. An inductor having negligible resistance and an inductance of 0.07H is connected in series with a resiostor of 20 ohms resitance across a 200, 50 Hz supply. The maximum energy stored in the coil is a. 3.175 J b. 1.585 J c. 0.236 J d. 0.33 J
  129. 129. REVIEW QUESTIONSREVIEW QUESTIONS 45. A coil has 1200 turns and procedures 100 uWb mwhen the current flowing is 1A. The inductance of the coil is a. 0.21 H b. 0.12 H c. 0.31 H d. 0.41 H 46. A capacitor connected to a 115 V, 25 Hz supply takes 5 A. What current will it take when the capacitance and frequency are doubled? a. 2 A b. 5 S c. 10 A d. 20 A 45. A coil has 1200 turns and procedures 100 uWb mwhen the current flowing is 1A. The inductance of the coil is a. 0.21 H b. 0.12 H c. 0.31 H d. 0.41 H 46. A capacitor connected to a 115 V, 25 Hz supply takes 5 A. What current will it take when the capacitance and frequency are doubled? a. 2 A b. 5 S c. 10 A d. 20 A
  130. 130. REVIEW QUESTIONSREVIEW QUESTIONS 47. A resistor of 20 ohms is connected in parallel with a capacitor across a 110 V, 40 Hz supply. If the current taken is 6A, what is the capacitance? a. 88.6 uF b. 68.8 uF c. 86.8 uF d. 76.8 uF 48. What capacitance must be placed in series with an inductance of 0.05H, so that when the frequency is 100 Hz, the impedance becomes equal to the ohmic resitance? a. 70.5 uF b. 50.7 uF c. 5.7 uF d. 7.05 uF 47. A resistor of 20 ohms is connected in parallel with a capacitor across a 110 V, 40 Hz supply. If the current taken is 6A, what is the capacitance? a. 88.6 uF b. 68.8 uF c. 86.8 uF d. 76.8 uF 48. What capacitance must be placed in series with an inductance of 0.05H, so that when the frequency is 100 Hz, the impedance becomes equal to the ohmic resitance? a. 70.5 uF b. 50.7 uF c. 5.7 uF d. 7.05 uF
  131. 131. REVIEW QUESTIONSREVIEW QUESTIONS 49. A reactance of 20 ohms and inductance of 0.1 H is connected in parallel with a capacitor. The capacitance of the capacitor required to produce a resonance when connected to a 100V, 50 Hz is a. 71.2 uF b. 1.277 uF c. 17.2 uF d. 72.1 uF 50. What is the resonant frequency of a circuit when an inductance of 1uH and capacitance of 10 picofarad are in series? a. 15.9 MHz b. 50.3 MHz c. 15.9 kHz d. 50.3 KHz 49. A reactance of 20 ohms and inductance of 0.1 H is connected in parallel with a capacitor. The capacitance of the capacitor required to produce a resonance when connected to a 100V, 50 Hz is a. 71.2 uF b. 1.277 uF c. 17.2 uF d. 72.1 uF 50. What is the resonant frequency of a circuit when an inductance of 1uH and capacitance of 10 picofarad are in series? a. 15.9 MHz b. 50.3 MHz c. 15.9 kHz d. 50.3 KHz
  132. 132. REVIEW QUESTIONSREVIEW QUESTIONS 51. The _____ the Q of a circuit, the narrower the bandwidth. a. Lower b. Higher c. Broader d. Selective 52. Find the half power bandwidth of a parallel resonant circuit which has a resonant frequency of 3.6MHz and Q of 218. a. 606 kHz b. 58.7 kHz c. 16.5 kHz d. 47.3 kHz 51. The _____ the Q of a circuit, the narrower the bandwidth. a. Lower b. Higher c. Broader d. Selective 52. Find the half power bandwidth of a parallel resonant circuit which has a resonant frequency of 3.6MHz and Q of 218. a. 606 kHz b. 58.7 kHz c. 16.5 kHz d. 47.3 kHz
  133. 133. REVIEW QUESTIONSREVIEW QUESTIONS 53. At series resonance _____ . a. Circuit impedance is very large b. Cicuit power factor is minimum c. Voltage across L or C is zero d. Circuit power factor is unity 54. At series resonance, the voltage across the inductor is a. Equal to the applied voltage b. Much more than the apllied voltage c. Equal to voltage across R d. Less than the applied voltage 53. At series resonance _____ . a. Circuit impedance is very large b. Cicuit power factor is minimum c. Voltage across L or C is zero d. Circuit power factor is unity 54. At series resonance, the voltage across the inductor is a. Equal to the applied voltage b. Much more than the apllied voltage c. Equal to voltage across R d. Less than the applied voltage
  134. 134. REVIEW QUESTIONSREVIEW QUESTIONS 55. The Q factor of the coil is _____ the resistance of the coil. a. Inversely proportional to b. Directly proportional to c. Independent of d. None of these 56. An RLC circuit is connected is connected to 200V AC source. If Q factor of the coil is 10, then the voltage across the capacitor at resonance is a. 200 V b. 2000 V c. 20 V d. 210 V 55. The Q factor of the coil is _____ the resistance of the coil. a. Inversely proportional to b. Directly proportional to c. Independent of d. None of these 56. An RLC circuit is connected is connected to 200V AC source. If Q factor of the coil is 10, then the voltage across the capacitor at resonance is a. 200 V b. 2000 V c. 20 V d. 210 V
  135. 135. REVIEW QUESTIONSREVIEW QUESTIONS 57. At parallel resonance a. Circuit impedance is minimum b. Power factor is zero c. Line current is maximum d. Power factor is unity 58. The dynamic impedance of parallel resonant circuit is 1 Mohms. If the capacitance is 1uF, and the resistance is 1ohm, then the value of the inductance a. 1 H b. 10 H c. 10 pH d. 10 uH 57. At parallel resonance a. Circuit impedance is minimum b. Power factor is zero c. Line current is maximum d. Power factor is unity 58. The dynamic impedance of parallel resonant circuit is 1 Mohms. If the capacitance is 1uF, and the resistance is 1ohm, then the value of the inductance a. 1 H b. 10 H c. 10 pH d. 10 uH
  136. 136. REVIEW QUESTIONSREVIEW QUESTIONS 59. When supply frequency is less than the resonant frequency in a parallel ac circuit, then the circuit is a. Resistive b. Capacitive c. Inductive d. None of these 60. At parallel resonance, the circuit drwas a current of 2mA. If the Q of the circuit is 100, then the current through the capacitor is a. 2 mA b. 1 mA c. 200 mA d. None of these 59. When supply frequency is less than the resonant frequency in a parallel ac circuit, then the circuit is a. Resistive b. Capacitive c. Inductive d. None of these 60. At parallel resonance, the circuit drwas a current of 2mA. If the Q of the circuit is 100, then the current through the capacitor is a. 2 mA b. 1 mA c. 200 mA d. None of these
  137. 137. REVIEW QUESTIONSREVIEW QUESTIONS 61. A circuit has an impedance of (1-j2) ohms. The susceptance of the circuit in mho is a. 0.1 b. 0.2 c. 0.4 d. None of these 62. If the admittance of a parallel ac circuit increased, the circuit current a. Remains constant b. Is increased c. Is decreased d. None of these 61. A circuit has an impedance of (1-j2) ohms. The susceptance of the circuit in mho is a. 0.1 b. 0.2 c. 0.4 d. None of these 62. If the admittance of a parallel ac circuit increased, the circuit current a. Remains constant b. Is increased c. Is decreased d. None of these
  138. 138. REVIEW QUESTIONSREVIEW QUESTIONS 63. The resistance between any pair two terminals of a balanced wye- connected load is 12 ohms. a. 6 ohms b. 18 ohms c. 24 ohms d. None of these 64. If an AC circuit contains three nodes, the number of each mesh equations that can be formulated is a. 1 b. 2 c. 3 d. 4 63. The resistance between any pair two terminals of a balanced wye- connected load is 12 ohms. a. 6 ohms b. 18 ohms c. 24 ohms d. None of these 64. If an AC circuit contains three nodes, the number of each mesh equations that can be formulated is a. 1 b. 2 c. 3 d. 4
  139. 139. REVIEW QUESTIONSREVIEW QUESTIONS 65. The relation of the voltage across an inductor to its current is describe as a. Leading the current by 90 deg b. Lagging the current by 90 deg c. Leading the current by 180 deg d. In phase with the current 65. The relation of the voltage across an inductor to its current is describe as a. Leading the current by 90 deg b. Lagging the current by 90 deg c. Leading the current by 180 deg d. In phase with the current
  140. 140. THANk YOUTHANk YOU

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