Electric circuits-chapter-2 Basic Laws

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Electric circuits-chapter-2 Basic Laws

  1. 1. 08/01/12 Chapter 2 Basic LawsDKS1113 Electric Circuits
  2. 2. Introduction Fundament laws that govern electric circuits:  Ohm’s Law.  Kirchoff’s Law. These laws form the foundation upon which electric circuit analysis is built. Common techniques in circuit analysis and design:  Combining resistors in series and parallel.  Voltage and current divisions.  Wye to delta and delta to wye transformations. These techniques are restricted to resistive circuits.08/01/12 DKS1113 Electric Circuits 2/43
  3. 3. Ohm’s Law08/01/12 DKS1113 Electric Circuits 3/43
  4. 4. Ohm’s Law Relationship between current and voltage within a circuit element. The voltage across an element is directly proportional to the current flowing through it v α i Thus::v=iR and R=v/i  Where:  R is called resistor.  Has the ability to resist the flow of electric current.  Measured in Ohms (Ω)08/01/12 DKS1113 Electric Circuits 4/43
  5. 5. Ohm’s Law *pay careful attention to current direction v=iR08/01/12 DKS1113 Electric Circuits 5/43
  6. 6. Ohm’s Law Value of R :: varies from 0 to infinity Extreme values == 0 & infinity. Only linear resistors obey Ohm’s Law. Short circuit Open Circuit08/01/12 DKS1113 Electric Circuits 6/43
  7. 7. Ohm’s Law Conductance (G)  Unit mho or Siemens (S).  Reciprocal of resistance R G=1/R  Has the ability to conduct electric current08/01/12 DKS1113 Electric Circuits 7/43
  8. 8. Ohm’s Law Power:  P = iv  i ( i R ) = i2R watts  (v/R) v = v2/R watts  R and G are positive quantities, thus power is always positive.  R absorbs power from the circuit  Passive element.08/01/12 DKS1113 Electric Circuits 8/43
  9. 9. Ohm’s Law Example 1:  Determine voltage (v), conductance (G) and power (p) from the figure below.08/01/12 DKS1113 Electric Circuits 9/43
  10. 10. Ohm’s Law Example 2:  Calculate current i in figure below when the switch is in position 1.  Find the current when the switch is in position 2.08/01/12 DKS1113 Electric Circuits 10/43
  11. 11. Nodes, Branches & Loops Elements of electric circuits can be interconnected in several way. Need to understand some basic concepts of network topology. Branch: Represents a single element (i.e. voltage, resistor & etc) Node: The meeting point between two or more branches. Loop: Any closed path in a circuit.08/01/12 DKS1113 Electric Circuits 11/43
  12. 12. Nodes, Branches & Loops Example 3:  Determine how many branches and nodes for the following circuit.08/01/12 DKS1113 Electric Circuits 12/43
  13. 13. Nodes, Branches & Loops 5 Branches  3 Nodes  1 Voltage Source  a  1 Current Source  b  3 Resistors  c08/01/12 DKS1113 Electric Circuits 13/43
  14. 14. Nodes, Branches & Loops Example 4:  Determine how many branches and nodes for the following circuit.08/01/12 DKS1113 Electric Circuits 14/43
  15. 15. Kirchoff’s Laws08/01/12 DKS1113 Electric Circuits 15/43
  16. 16. Kirchoff’s Laws Kirchoff’s Current Law (KCL)  The algebraic sum of current entering / leaving a node (or closed boundary) is zero.  Current enters = +ve  Current leaves = -ve  ∑ current entering = ∑ current leaving08/01/12 DKS1113 Electric Circuits 16/43
  17. 17. Kirchoff’s Laws Example 5:  Given the following circuit, write the equation for currents.08/01/12 DKS1113 Electric Circuits 17/43
  18. 18. Kirchoff’s Laws Example 6:  Current in a closed boundary08/01/12 DKS1113 Electric Circuits 18/43
  19. 19. Kirchoff’s Laws Example 9:  Use KCL to obtain currents i1, i2, and i3 in the circuit.08/01/12 DKS1113 Electric Circuits 19/43
  20. 20. Kirchoff’s Laws Kirchoff’s Voltage Law (KVL)  Applied to a loop in a circuit.  According to KVL The algebraic sum of voltage (rises and drops) in a loop is zero. + v1 - + + vs V2 - - v3 + -08/01/12 DKS1113 Electric Circuits 20/43
  21. 21. Kirchoff’s Laws Example 10:  Use KVL to obtain v1, v2 and v3.08/01/12 DKS1113 Electric Circuits 21/43
  22. 22. Kirchoff’s Laws Example 11:  Use KVL to obtain v1, and v2.08/01/12 DKS1113 Electric Circuits 22/43
  23. 23. Kirchoff’s Laws Example 12:  Calculate power dissipated in 5Ω resistor. 1008/01/12 DKS1113 Electric Circuits 23/43
  24. 24. Series Resistors & Voltage Division Series resistors  same current flowing through them.  v1= iR1 & v2 = iR2  KVL:  v-v1-v2=0  v= i(R1+R2)  i = v/(R1+R2 ) =v/Req  or v= i(R1+R2 ) =iReq  iReq = R1+R208/01/12 DKS1113 Electric Circuits 24/43
  25. 25. Series Resistors & Voltage Division Voltage Division:  Previously:  v1 = iR1 & v2 = iR2  i = v/(R1+R2 )  Thus:  v1=vR1/(R1+R2)  v2=vR2/(R1+R2)08/01/12 DKS1113 Electric Circuits 25/43
  26. 26. Parallel Resistors & Current Division Parallel resistors  Common voltage across it.  v = i1R1 = i2R2  i = i1+ i2 = v/R1+ v/R2 = v(1/R1+1/R2)  =v/Req  v =iReq  1/Req = 1/R1+1/R2  Req = R1R2 / (R1+R2 )08/01/12 DKS1113 Electric Circuits 26/43
  27. 27. Parallel Resistors & Current Division Current Division:  Previously:  v = i1R1 = i2R2  v=iReq = iR1R2 / (R1+R2 )  and i1 = v /R1 & i2 =v/ R2  Thus:  i1= iR2/(R1+R2)  i2= iR1/(R1+R2 )08/01/12 DKS1113 Electric Circuits 27/43
  28. 28. Conductance (G) Series conductance:  1/Geq = 1/G1 +1/G2+… Parallel conductance:  Geq = G1 +G2+…08/01/12 DKS1113 Electric Circuits 28/43
  29. 29. Voltage and Current Division Example 13:  Calculate v1, i1, v2 and i2.08/01/12 DKS1113 Electric Circuits 29/43
  30. 30. Voltage and Current Division Example 14:  Determine i1 through i4.08/01/12 DKS1113 Electric Circuits 30/43
  31. 31. Voltage and Current Division Example 15:  Determine v and i. Answer v = 3v, I = 6 A.08/01/12 DKS1113 Electric Circuits 31/43
  32. 32. Voltage and Current Division Example 16:  Determine I1 and Vs if the current through 3Ω resistor = 2A.08/01/12 DKS1113 Electric Circuits 32/43
  33. 33. Voltage and Current Division Example 17:  Determine Rab.08/01/12 DKS1113 Electric Circuits 33/43
  34. 34. Voltage and Current Division Example 18:  Determine vx and power absorbed by the 12Ω resistor.  Answer v = 2v, p = 1.92w.08/01/12 DKS1113 Electric Circuits 34/43
  35. 35. Wye-Delta Transformations Given the circuit, how to combine R1 through R6? Resistors are neither in series nor parallel… Use wye-delta transformations08/01/12 DKS1113 Electric Circuits 35/43
  36. 36. Wye-Delta Transformations Y network T network08/01/12 DKS1113 Electric Circuits 36/43
  37. 37. Wye-Delta Transformations Δ network π network08/01/12 DKS1113 Electric Circuits 37/43
  38. 38. Wye-Delta Transformations Delta (Δ) to wye (y) conversion. 08/01/12 DKS1113 Electric Circuits 38/43
  39. 39. Wye-Delta Transformations Thus Δ to y conversion ::  R1 = RbRc/(Ra+Rb+Rc)  R2 = RaRc/(Ra+Rb+Rc)  R3 = RaRb/(Ra+Rb+Rc) # Each resistors in y network is the product of two adjacent branches divide by the 3 Δ resistors08/01/12 DKS1113 Electric Circuits 39/43
  40. 40. Wye-Delta Transformations Y to Δ conversions:  Ra = (R1R2 +R2 R3 +R1R3)/R1  Rb = (R1R2 +R2 R3 +R1R3)/R2  Rc= (R1R2 +R2 R3 +R1R3)/R308/01/12 DKS1113 Electric Circuits 40/43
  41. 41. Wye-Delta Transformations Example 19:  Transform the circuit from Δ to y.  Answer R1=18, R2=6, R3=3.08/01/12 DKS1113 Electric Circuits 41/43
  42. 42. Wye-Delta Transformations Example 20:  Determine Rab.  Answer Rab=142.32.08/01/12 DKS1113 Electric Circuits 42/43
  43. 43. Wye-Delta Transformations Example 21:  Determine Io.08/01/12 DKS1113 Electric Circuits 43/43

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