EEE 2019 Equivalent Resistance Lesson 8 07-5-21 Basic Laws of Circuits Equivalent Resistance: We know the following for series resistors: R1 R2 . . . Req RN . . . Figure 7.1: Resistors in series. Req = R1 + R2 + . . . + RN 1 Basic Laws of Circuits Equivalent Resistance: We know the following for parallel resistors: . . . Req R1 RN R2 . . . Figure 7.2: Resistors in parallel. 1 1 1 1 . . . Req R1 R2 RN 2 Basic Laws of Circuits Equivalent Resistance: For the special case of two resistors in parallel: R eq R1 R2 Figure 7.3: Two resistors in parallel. 3 R1 R2 Req R1 R2 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. By combination we mean we have a mix of series and Parallel. This is illustrated below. R1 Req R3 R2 R4 R5 Figure 7.4: Resistors In Series – Parallel Combination To find the equivalent resistance we usually start at the output of the circuit and work back to the input. 4 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. R1 R3 R2 Req Rx R4 R5 Rx R4 R5 R1 Req 5 R2 Ry Figure 7.5: Resistance reduction. R y Rx R3 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. R1 RZ Req Req 6 RZ R2 RY R2 RY Req RZ R1 Figure 7.6: Resistance reduction, final steps. Basic Laws of Circuits Equivalent Resistance: Resistors in combination. It is easier to work the previous problem using numbers than to work out a general expression. This is illustrated below. Example 7.1: Given the circuit below. Find Req. 10 Req 8 10 3 Figure 7.7: Circuit for Example 7.1. 7 6 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. Example 7.1: Continued . We start at the right hand side of the circuit and work to the left. 10 Req 10 8 10 2 Req Figure 7.8: Reduction steps for Example 7.1. Ans: 8 Req 15 5 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. Example 7.2: Given the circuit shown below. Find Req. 6 12 c R eq 10 b 4 d Figure 7.9: Diagram for Example 7.2. 9 a Basic Laws of Circuits Equivalent Resistance: Resistors in combination. Example 7.2: Continued. 6 12 c R eq 10 b a 4 d c Fig 7.10: Reduction steps. 10 Req 4 12 b 6 10 d, a Basic Laws of Circuits Equivalent Resistance: Resistors in combination. Example 7.2: Continued. c Req 4 12 b 6 10 resistor shorted out 10 d, a Req Fig 7.11: Reduction steps. 11 4 6 12 Basic Laws of Circuits Equivalent Resistance: Resistors in combination. Example 7.2: Continued. 4 Req Fig 7.12: Reduction steps. 12 Req 6 4 12 4 Basic Electric Circuits Wye to Delta Transformation: You are given the following circuit. Determine Req. I 9 10 V + _ 5 10 Req 8 4 Figure 7.1: Diagram to start wye to delta. 13 Basic Electric Circuits Wye to Delta Transformation: You are given the following circuit. Determine Req. I 9 10 V + _ 5 10 Req 8 4 Figure 7.13: Diagram to start wye to delta. 14 Basic Electric Circuits Wye to Delta Transformation: I 9 10 V + _ 5 10 Req 8 4 We cannot use resistors in parallel. We cannot use resistors in series. If we knew V and I we could solve 15 V Req = I There is another way to solve the problem without solving for I (given, assume, V) and calculating Req for V/I. Basic Electric Circuits Wye to Delta Transformation: Consider the following: a a Ra R1 Rc c R2 Rb b c b R3 (a) wye configuration (b) delta configuration Figure 7.14: Wye to delta circuits. We equate the resistance of Rab, Rac and Rbc of (a) to Rab , Rac and Rbc of (b) respectively. 16 Basic Electric Circuits Wye to Delta Transformation: Consider the following: a a Ra R1 Rc c R2 Rb b c b R3 (a) wye configuration (b) delta configuration R2(R1 + R3) Eq 7.1 Rab = Ra + Rb = R1 + R2 + R3 R1(R2 + R3) Rac = Ra + Rc = Eq 7.2 R1 + R2 + R3 R3(R1 + R2) 17 Rbc = Rb + Rc = R1 + R2 + R3 Eq 7.3 Basic Electric Circuits Wye to Delta Transformation: Consider the following: a a Ra R1 Rc R2 Rb c b c b R3 (a) wye configuration 18 (b) delta configuration Ra R1 R2 R1 R2 R3 R1 Ra Rb Rb Rc Rc Ra Rb Eq 7.4 Rb R2 R3 R1 R2 R3 R2 Ra Rb Rb Rc Rc Ra Rc Eq 7.5 Rc R1 R3 R1 R2 R3 R3 Ra Rb Rb Rc Rc Ra Ra Eq 7.6 Basic Electric Circuits Wye to Delta Transformation: Observe the following: Go to wye Go to delta Ra Rb Rb Rc Rc Ra Rb Eq 7.4 R2 R3 Rb R1 R2 R3 R R Rb Rc Rc Ra R2 a b Rc Eq 7.5 R1 R3 R1 R2 R3 R R Rb Rc Rc Ra R3 a b Ra Eq 7.6 Ra Rc R1 R2 R1 R2 R3 R1 We note that the denominator for Ra, Rb, Rc is the same. We note that the numerator for R1, R2, R3 is the same. We could say “Y” below: “D” 19 Basic Electric Circuits Wye to Delta Transformation: Example 7.3: Return to the circuit of Figure 7.13 and find Req. I 9 a 10 V + _ Req 5 10 c 8 b 4 Convert the delta around a – b – c to a wye. 20 Basic Electric Circuits Wye to Delta Transformation: Example 7.3: continued 9 2 Req 4 2 8 4 Figure 7.15: Example 7.3 diagram. It is easy to see that Req = 15 21 Basic Electric Circuits Wye to Delta Transformation: Example 7.4: Using wye to delta. The circuit of 7.13 may be redrawn as shown in 7.16. 9 a 10 Req c 5 10 8 4 b Figure 7.16: “Stretching” (rearranging) the circuit. Convert the wye of a – b – c to a delta. 22 Basic Electric Circuits Wye to Delta Transformation: Example 7.4: continued 9 9 a 10 7.33 27.5 Req c 8 a 11 Req c 22 11 5.87 b (a) b (b) Figure 7.17: Circuit reduction of Example 7.4. 23 Basic Electric Circuits Wye to Delta Transformation: Example 7.4: continued 9 Req 13.2 11 Figure 7.18: Reduction of Figure 7.17. Req = 15 This answer agrees with the delta to wye solution earlier. 24 READ MORE End of Lesson 7 Equivalent Resistance