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Electric circuits.pptx
- 1. E18302 ELECTRIC CIRCUITS A.MOHAMED NAZEER LECTURER –EEE PSG POLYTECHNIC COLLEGE
- 2. Syllabus DC CIRCUITS: Electrical Quantities -Definition and Units of Charge, Voltage, Current, Resistance, Power, Energy- Ohm’s law, Kirchoff’s laws -Series, Parallel and Series Parallel Circuits - Simple Problems. (12) DC NETWORK THEOREM: Thevenin’s Theorem – Norton’s Theorem – Super Position Theorem –Maximum Power Transfer Theorem – Simple Problems (12) AC CIRCUITS: Single Phase Circuits -Alternating Current -Sinusoidal waveform, Frequency, Peak value, RMS and Average values – Form factor –Peak factor –simple problems-power in AC circuits - power and power factor- Inductance, Capacitance - series circuits - RL,RC,RLC - simple problems-Introduction to Resonance (No Derivations) (12) THREE PHASE CIRCUITS: Three phase circuits (balanced supply and load only ) Concept of 3 phase system – star/delta connections – phasor diagram – power in Three phase circuits –Measurement of power in 3 phase circuits using two wattmeter method. (12) WIRING: Different types of wires – types of wiring – comparison of different types of wiring –rules for wiring – testing procedures – Importance of earthing – plate and pipe earthing. (10) TOTAL HOURS - 58 HOURS THEOREY - 4 HOURS/WEEK TUTORIAL - 1 HOURS/WEEK NAZEER-L-EEE-PSGPTC 2
- 3. Books for Study • TEXT BOOK: 1. B.L. Theraja A.K. Theraja S.G. Tarnekar, ”Electrical Technology”- Volume I, S. CHAND Publishing & Company Ltd, 2014. • REFERENCES: 1. R.Muthusubramanian S.Salivahanan “ Basic Electrical , Electronics and Engineering” Tata McGraw Publishing Company Limited ,2015. 2. John Bird “ Electrical Circuit Theory and Technology” Published by Elsevier, a division of Reed Elsevier India Private Limited, 2012. 3. S Salivahanan, S Pravin Kumar “Circuit Theory “ Vikas publishing house Pvt. Ltd. 2014. NAZEER-L-EEE-PSGPTC 3
- 4. Electricity • Motion of electrical charges or particles. • Gives current to people to light, heat, drive the motor etc. • Creates electric field which has positive negative charges. • Creates magnetic field using the principle of electromagnetism. NAZEER-L-EEE-PSGPTC 4
- 5. Comes to Mind on Electricity • Voltage • Current • AC • DC • Power • Energy • Conductor • Insulator • Frequency • Resistor • Inductor • Capacitor • Battery • Magnet • Motor • Generator • Switch • Plug • Socket • Cable • Wires • Light • Speaker • Heater NAZEER-L-EEE-PSGPTC 5
- 6. Electric Charge • Every atom in metal has electrical charges. • Electric Charge is an inbuilt physical, electrical property of a particle to react in electric field. • It is an internal energy levels to move particle inside atom. • Every atom has positive charge(protons) and negative charge(electrons). • Denotes as “Q” • Unit – coulomb. NAZEER-L-EEE-PSGPTC 6
- 7. Voltage • EMF(Electro Motive Force) • Potential Difference • It is an electrical force to move charged particle from one level(lower potential) to another level(higher potential). • Since it is the electrical force, called as electro motive force. • It is always between two points. So called as potential difference. • Denotes as “V” • Unit – Volt(V). NAZEER-L-EEE-PSGPTC 7
- 8. Current • It is the flow of electrons in a circuit. • It is the movement of electrical charge in a circuit. • It is the rate of flow of charge. • It is continuous. • Denotes as “I” • Current =charge/ time ( I=Q/t). • Direction of current is opposite to direction of flow of electrons. • Current flows from lower potential to upper potential. • Unit – Ampere (A) NAZEER-L-EEE-PSGPTC 8
- 13. DC • Direct Current. • Electrical parameter is unidirectional and constant with respect to time. AC • Alternating Current. • Electrical parameter is bidirectional and varying with respect to time. NAZEER-L-EEE-PSGPTC 13
- 14. Ripples • Pulsated dc. • These waveforms are unidirectional. • Magnitude is varying with respect to time. • Pure dc + unwanted ac. • unwanted ac mixed with pure dc is called “ripples”. NAZEER-L-EEE-PSGPTC 14
- 15. Harmonics • These waveforms are bidirectional and varying with time. • Sine waveform is pure form of ac. • Other waveforms are mixed with unwanted ac components. • Unwanted ac component mixed with pure ac component is harmonics NAZEER-L-EEE-PSGPTC 15
- 16. Circuit Open Circuit • Current does not flow in the circuit. • Disconnection in the circuit. Closed Circuit • Current flows in the circuit. • The circuit connection is continuous. NAZEER-L-EEE-PSGPTC 16
- 17. Electrical Material Classification Conductor Insulator Semiconductor Allows the current to flow Blocks the current to flow At certain conditions it will allow the current to flow otherwise blocks. Copper, Aluminum Paper, Mica Bakelite Silicon, Germanium Low Resistance Very High Resistance Low resistance when conducting High Resistance when not conducting NAZEER-L-EEE-PSGPTC 17
- 18. Resistance • Resistance oppose the flow of current in an electric circuit. • Resistances dissipate electrical energy as heat. • Symbol : • Denotes as “R” in circuit. • Unit – ohm (Ω) • Resistance value changes with temperature. • Depends on material – positive or negative temperature cd. NAZEER-L-EEE-PSGPTC 18
- 19. Resistance(contd..) • Resistance is reciprocal of conductance. R= 1/σ ( σ - conductance) • Resistance of wire is calculated as R= ρ l /A (ρ –Resistivity or self resistance of material l- length of conductor A- area of cross section of conductor) NAZEER-L-EEE-PSGPTC 19
- 23. Ohm’s Law • When temperature remain constant, voltage across a conductor is directly proportional to the current flow through it. V α I V = I R (V- Voltage ; I- Current; R- Resistance) I = V / R R = V/ I NAZEER-L-EEE-PSGPTC 23
- 25. NAZEER-L-EEE-PSGPTC 25 1. A small light bulb is connected to a 6 V battery and draws 2 A of current. What is the net resistance of the bulb? voltage: V = 6 V current: I = 2 A resistance: R = ? R = V/I R = 6/2 R = 3 Ω 2. A motor with an operating resistance of 32 Ω is connected to a voltage source. The current in the circuit is 1.5 A. What is the voltage of the source? resistance: R = 32 Ω; current: I = 1.5 A voltage: V = ? V = I x R V = 1.5 x 32 V = 48 V 3. Determine the amount of current going through a 50 Ω resistor with a voltage of 120 V. resistance: R = 50 Ω voltage: V = 120 V current: I = ? I = V/R I = 120/50 I = 2.4 A
- 26. Power • It is the capacity to do the work. • Denotes as “P” • Unit – watt P = V x I (P- power;V- voltage ;I- current) NAZEER-L-EEE-PSGPTC 26
- 30. Energy • Power utilised over the period. • Denoted as “E” • Unit – Wh(watt-hour) • 1 KWh = 1000 Wh = 1 unit E = P x t (E-Energy; P- Power; T-time) NAZEER-L-EEE-PSGPTC 30
- 31. NAZEER-L-EEE-PSGPTC 31 A 220 V – 5 A electric lamp is used for 30 minutes. How much energy does it require? Solution : Voltage (V) = 220 V Electric current (I) = 5 A Time (t) = 30 minutes = 30 / 60 hour = 0.5 hour Electric power (P) : P = V I = (220)(5 ) P = 1100 Watt Electric energy E = P t= 1100 x 0.5 Electric energy E = 550 Wh = 550/1000 KWh E= 0.55 KWh=0.55 unit
- 32. NAZEER-L-EEE-PSGPTC 32 Someone watches TV on average 6 hours each day. The TV is connected to a 220 Volt voltage so that the electric current flows through the TV is 0.5 Amperes. If the electric company charges Rs 2 per unit, then find the cost of using electric energy for TV for 1 month (30 days). Given : Time t = 6 hours x 30days = 180 hours Voltage (V) = 220 V Current (I) = 0.5 A To find: The cost per month Solution : Power of TV : P = V I = 220 x 0.5 = 110 W Electric energy E for 30 days = P x t for 30days Electric energy of TV for 30 days= 110 x 180 = 19800 Wh = 19800/1000 kWh = 19.8 kWh = 19.8 units The cost of using electric energy for TV during 1 month = 19.8 units x Rs 2 per unit = Rs.39.60 = Rs 40
- 34. Kirchoff’s Laws • Kirchoff’s Current Law (KCL): In a junction or node, sum of the currents flowing towards the junction is equal to sum of the currents flowing away from the junction. • Kirchoff’s Voltage Law (KVL): In a closed loop, sum of the voltage raises is equal to sum of the voltage drops. NAZEER-L-EEE-PSGPTC 34
- 35. Kirchoff’s Current Law (KCL) • Yellow dot is a junction or node. • Currents flowing towards the junction = i1+i2 +i6 • Currents flowing away from the junction = i3+i4+i5 • As per KCL i1+i2+i6 = i3+i4+i5 NAZEER-L-EEE-PSGPTC 35
- 39. Kirchoff’s Voltage Law (KVL) • In the closed loop ABCDA, • Voltage raises = V4 • Voltage drops = V1 + V2 +V3 • As per KVL V4 = V1 + V2 +V3 NAZEER-L-EEE-PSGPTC 39
- 40. Circuit Terminologies • Node :A point at which two or more elements are joints together. • Junction : A point where three or more branches meet together. • Branch: The portion of the circuit between two nodes. • Loop: Starts from one node and after going through a set of nodes returns to same starting node. NAZEER-L-EEE-PSGPTC 40
- 48. Series Circuit • Equivalent Resistance Req = R1+R2+R3… • Current I = I1=I2=I3 • Voltage V = V1+V2+V3 • Power P = P1+ P2+P3 = V x I • Voltage V = I x Req • V1 = I1 x R1; V2 = I2 x R2; V3 = I3 x R3 • P1 = V1x I1 ; P2 = V2 x I2 ; P3 = V3 x I3 NAZEER-L-EEE-PSGPTC 48
- 49. Series Circuit (Characteristics) • If no junction in between two resistances, they are connected in series. • Same current flows through each series connected resistors. • Voltage across each series connected resistor is different. • Power dissipated in each series connected resistor is different. • If one resistor in series connection is disconnected, circuit will open. NAZEER-L-EEE-PSGPTC 49
- 54. Parallel Circuit • Equivalent Resistance 1/ Req = 1/R1 + 1/R2 + 1/R3… • Current I = I1+I2+I3 • Voltage V = V1=V2=V3 • Power P = P1+ P2+P3 = V x I • Voltage V = I x Req • V1 = I1 x R1; V2 = I2 x R2; V3 = I3 x R3 • P1 = V1x I1 ; P2 = V2 x I2 ; P3 = V3 x I3 NAZEER-L-EEE-PSGPTC 54
- 55. Parallel Circuit (Characteristics) • If there is a junction in between two resistances, they are connected in parallel. • Different current flows through each parallel connected resistors. • Voltage across each parallel connected resistor is same. • Power dissipated in each parallel connected resistor is different. • If one resistor in parallel connection is disconnected, current flows in other branches. NAZEER-L-EEE-PSGPTC 55
- 62. NAZEER-L-EEE-PSGPTC 62 Req1 = R2+R3 = 8+4 =12 Req2 = Req1 //l with R4 1 / Req2 = 1/Req1 + 1/R4 1/Req2 = 1/12 + 1/12 = 0.166 Req2 = 1/0.166 = 6Ω Req = Req2 + R1 = 6 +6 = 12 Ω From circuit 4 I= V / Req = 12/12 = 1A From circuit 2 Apply KVL V = V1 + Vreq1 12 = I x R1 + I1 x Req1 = 1 x 6 + I1 x 12 12 = 6 +12 I1 12 – 6 = 12 I1 6 = 12 I1 I1 = 6/12 = 0.5 A I = I1 +I2 1 = 0.5 + I2 I2 = 1-0.5 = 0.5 A V1 = 1 x 6 = 6V V2 = 0.5 x 4 = 2 V V3 = 0.5 x 8 = 4 V V4 = 0.5 x 12 = 6V
- 63. • Find the equivalent Resistance, current through each resistor and voltage across each resistor NAZEER-L-EEE-PSGPTC 63
- 71. Linear Bilateral Active Network • Linear means a network in which the relation between voltage and current is straight line. • Bilateral means the relation between voltage and current does not change in both the directions in the network. • An active network contains at least one voltage source or current source that can supply energy to the network indefinitely. NAZEER-L-EEE-PSGPTC 71
- 72. Superposition Theorem It states that in any linear, active, bilateral network having more than one source, the current in any branch or element is the sum of the currents obtained from each source considered separately and all other sources are replaced by their internal resistance. NAZEER-L-EEE-PSGPTC 72
- 74. NAZEER-L-EEE-PSGPTC 74 Step 1 – Take only one independent source of voltage or current and deactivate the other sources. Step 2 – In the circuit diagram B shown above, consider the source E1 and replace the other source E2 by its internal resistance. If its internal resistance is not given, then it is taken as zero and the source is short-circuited. Step 3 – If there is a voltage source than short circuit it and if there is a current source then just open circuit it. Step 4 – Thus, by activating one source and deactivating the other source find the current in each branch of the network. Taking the above example find the current I1’, I2’and I3’. Step 5 – Now consider the other source E2 and replace the source E1 by its internal resistance r1 as shown in the circuit diagram C. Step 6 – Determine the current in various sections, I1’’, I2’’ and I3’’. Step 7 – Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch. Step 8 – If the current obtained by each branch is in the same direction then add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.
- 76. Thevenin’s Theorem • It states that any linear active network consisting of independent or dependent voltage and current source and the network elements can be replaced by an equivalent circuit having a voltage source in series with a resistance. NAZEER-L-EEE-PSGPTC 76
- 78. NAZEER-L-EEE-PSGPTC 78 Step 1 – First of all remove the load resistance rL of the given circuit. Step 2 – Replace all the sources by their internal resistance. Step 3 – If sources are ideal then short circuit the voltage source and open circuit the current source. Step 4 – Now find the equivalent resistance at the load terminals, known as Thevenin’s Resistance (RTH). Step 5 – Draw the Thevenin’s equivalent circuit by connecting the load resistance and after that determine the desired response.
- 80. Norton’s Theorem It states that in a linear active network consisting of the independent or dependent voltage source and current sources and the various circuit elements can be substituted by an equivalent circuit consisting of a current source in parallel with a resistance. NAZEER-L-EEE-PSGPTC 80
- 82. NAZEER-L-EEE-PSGPTC 82 Step 1 – Remove the load resistance of the circuit. Step 2 – Find the internal resistance Rint of the source network by deactivating the constant sources. Step 3 – Now short the load terminals and find the short circuit current ISC flowing through the shorted load terminals using conventional network analysis methods. Step 4 – Norton’s equivalent circuit is drawn by keeping the internal resistance Rint in parallel with the short circuit current ISC. Step 5 – Reconnect the load resistance RL of the circuit across the load terminals and find the current through it known as load current IL.
- 84. Maximum Power Transfer Theorem It states that in a resistive load, being connected to a DC network, receives maximum power when the load resistance is equal to the internal resistance known as (Thevenin’s equivalent resistance) of the source network as seen from the load terminals. NAZEER-L-EEE-PSGPTC 84
- 86. NAZEER-L-EEE-PSGPTC 86 Step 1 – Remove the load resistance of the circuit. Step 2 – Find the Thevenin’s resistance (RTH) of the source network looking through the open- circuited load terminals. Step 3 – As per the maximum power transfer theorem, this RTH is the load resistance of the network, i.e., RL = RTH that allows maximum power transfer. Step 4 – Maximum Power Transfer is calculated.