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SOLUTION USING THEVENIN’S THEOREM Working out resistor currents in unbalanced bridge circuits can be frustrating, especially when depending entirely on Kirchoff laws to do so. See Fig-1 below. However, one can employ other mathematical tools like Thevenin and Norton equivalents called “Source Transformations” to make the task easier. For example, Thevenin’s Theorem states that any complex linear circuit can be simplified to an equivalent voltage source and series resistor connected to a load resistor. (See Fig-1A below) Left click on this slide to see how this is done.

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Procedure: 1.Call Rx the Load Resistor (See Fig-1 below) 2.Remove Rx and calculate the Voltage potential between A & B This is the equivalent Voltage source. (See Fig-2 below) 3.Place a short across 100V and calculate resistance between A & B This result is the equivalent series resistance. (See fig-3 below) 4.Draw the series equivalent circuit and calculate the Rx current (See Fig.4 below) This result, 0.98A, is the current through Rx. Since the current through Rx (0.98A) is now known, the remaining currents in Fig-1 can be easily calculated using Kirchoff Laws.

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KIRCHOFF LAWS To set up the problem draw the circuit and make notations as follows: 1.Draw the circuit and enter Volts and resistance values (see Fig-5) 2.Using conventional current flow (+ to -) enter arrows indicating direction of current flow in all resistors Label unknown currents I 1, I 2, & I 3 through R1, R2, & R3 respectively 3.Label R4 current as ( I 1 - I 3 ) & R5 current as ( I 2 + I 3) 1.Kirchoff’s Current Law: The sum of all currents entering a node must be equal currents leaving the node. 2.Kirchoff Voltage Law: The sum of the voltages around a closed loop must equal to zero.

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KIRCHOFF LAWS 1.Since there are 3 unknowns currents ( I 1, I 2 & I 3), select 3 loops in Fig-5 and from these, develop 3 equations to be used for working out I 1, I 2 & I 3. Loop 1: 100V, R1 & R4 Loop 2: 100V, R2 & R5 Loop 3: R1, R2 & R3 Eq. 1: 100 – R1 I 1 – R4( I 1 – I 3) = 0 Eq. 2: 100 – R2 I 2 – R5( I 2 + I 3) = 0 Eq. 3: R1 I 1 + R3 I 3 –R2 I 2 = 0 Substitute resistor values for R1, R2 & R3 Eq. 1A: 100 – 12 I 1 – 24( I 1 – I 3) = 0 Eq. 2A: 100 – 36 I 2 – 18( I 2 + I 3) = 0 Eq. 3A: 12 I 1 + 14 I 3 –36 I 2 = 0

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KIRCHOFF LAWS From previous slide Eq. 1A: 100 – 12 I 1 – 24( I 1 – I 3) = 0 Eq. 2A: 100 – 36 I 2 – 18( I 2 + I 3) = 0 Eq. 3A: 12 I 1 + 14 I 3 –36 I 2 = 0 Expand to get rid of parentheses Eq. 1B: 100 – 12 I 1 – 24 I 1 + 24 I 3 = 0 Eq. 2B: 100 – 36 I 2 – 18 I 2 – 18 I 3 = 0 Eq. 3B: 12 I 1 + 14 I 3 –36 I 2 = 0 We have 3 simultaneous equations with 3 unknowns to find I 1, I 2 & I 3. Now the fun begins. Eq. 1C: 100 – 36 I 1 + 24 I 3 = 0 Eq. 2C: 100 – 54 I 2 – 18 I 3 = 0 Eq. 3C: 12 I 1 + 14 I 3 –36 I 2 = 0 Consolidating we get:

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Eq. 1C: 100 – 36 I 1 + 24 I 3 = 0 Eq. 2C: 100 – 54 I 2 – 18 I 3 = 0 Eq. 3C: 12 I 1 + 14 I 3 –36 I 2 = 0 From previous slide These 3 equations can be reduced to two equations with two unknown currents using traditional algebra. The procedure is as follows: Multiply Eq. 1C through by 18: 18(100 – 36 I 1 + 24 I 3 = 0) to get: Eq. 1D: 1800 – 648 I 1 + 432 I 3 =0 Multiply Eq. 2C through by 24: 24(100 – 54 I 2 – 18 I 3 = 0) to get 2D: 2400 – 1296 I 2 – 432 I 3 =0 By Adding Eq. 1D to Eq. 2D we get: 4200 – 648 I 1 – 1296 I 2 or Eq. 4A: 4200 = 648 I 1 + 1296 I 2 (One of 2 equations with 2 unknowns) Repeat procedure using eq. 2C & 3C to get the other equation with 2 unknowns

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KIRCHOFF LAWS Eq. 2C: 100 – 54 I 2 – 18 I 3 = 0 Eq. 3C: 12 I 1 + 14 I 3 –36 I 2 = 0 Multiply Eq. 2C by 14 to get 14 (100 – 54 I 2 – 18 I 3 = 0) or: Eq. 2E: 1400 – 756 I 2 – 252 I 3 = 0 Multiply Eq. 3C by 18 to get 18(12 I 1 + 14 I 3 –36 I 2 = 0) or: Eq. 3E: 216 I 1 + 252 I 3 – 648 I 2 = 0 Add Eq. 3E to Eq. 2E to get 1400 + 216 I 1 – 1404 I 2 or Eq. 4B: 1400 = – 216 I1 + 1404 I 2 (The 2 nd Equation with 2 unknowns So now we have: Eq. 4A: 4200 = 648 I 1 + 1296 I 2 Eq. 4B: 1400 = – 216 I1 + 1404 I 2 Multiply Eq. 4A by 1404 to get: Eq. 5A 5896800= – 909792 I 1 + 1819584 I 2 Multiply Eq. 4B by 1296 to get: Eq: 5B 1814400 = 279936 I 1 + 1819584 I 2 Subtract Eq.5A from Eq. 5B to get: 4082400 = 1189728 I1 or I 1 = 3.43A

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KIRCHOFF LAWS We calculated I 1 = 3.43A going through R1. I 3 Should be easy to find. Using the 100V, R1 R4 loop we get: 100 = (12 x 3.43) + 24(3.43 - I 3) or 100 = 41.16 + 82.32 –24 I 3 or 24 I 3 = 23.48 I 3 = 23.48 / 24 = 0.97833A ~ 0.98A As you can see from the above application of Kirchoff Laws, it takes a good working knowledge of Algebra and a close watch on the details. Even with these qualities, there are many pitfalls on the way to a solution: getting the signs wrong, arithmetic errors, misplacement of variables etc. I don’t think I want to ever repeat such an exercise again. When ever possible, choose the Thev enin Voltage equivalents or the Norton Current equivalents.

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