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31.71. Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P31.71 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit to find the current and potential difference of each resistor. Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V 10 Ω = 10 A . Thus, the current through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. In Step 1 of the above diagram, we return the 10 Ω resistor to the 4 Ω, 4 Ω, and 2 Ω resistors in series. These resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2 Ω and the 4 Ω resistors is 10 A. The potential differences are ∆V2 = (10 A)(2 Ω) = 20 V ∆V4 (left) = (10 A)(4 Ω) = 40 V ∆V4 (left) = (10 A)(4 Ω) = 40 V In Step 2, we return the left 4 Ω resistor to the 20 Ω and 5 Ω resistors in parallel. The two resistors must have the same potential difference ∆V = 40 V. From Ohm’s law, I5 = 40 V = 8A 5Ω I20 = 40 V =2A 20 Ω The currents through the various resistors are I2 = I4 = 10 A, I5 = 8 A, and I20 = 2 A. 2 (b) The power dissipated by the 20 Ω resistor is I20 (20 Ω) = (2 A)2 (20 Ω) = 80 W . (c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract the potential differences on the way: 0 V + (20 Ω)I20 + (2 Ω)I2 = (20 Ω)(2 A) + (2 Ω)(10 A) = 60 V = Va