# Model the battery and the connecting wires are ideal

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Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.64 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, 100 V 10 10 A. I = Ω = 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 Δ V 2 = (10 A)(2 Ω ) = 20 V Δ V 4 (left) = (10 A)(4 Ω ) = 40 V Δ V 4 (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 Δ V = 40 V. From Ohm’s law, 5 40 V 8 A 5 I = = Ω 20 40 V 2.0 A 20 I = = Ω The currents through the various resistors are I 2 = I 4 = 10 A, I 5 = 8 A, and I 20 = 2.0 A. (b) The power dissipated by the 20 Ω resistor is ( ) ( ) ( ) 2 2 20 20 2.0 A 20 80 W I Ω = Ω = . (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 ) I 20 + (2 Ω ) I 2 = (20 Ω )(2.0 A) + (2 Ω )(10 A) = 60 V = V a

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32.65. Model: The wires and batteries are ideal. Visualize: Solve: Assign currents I 1 , I 2 , and I 3 as shown in the figure. If I 3 turns out to be negative, we’ll know it really flows right to left. Apply Kirchhoff’s loop rule counterclockwise to the top loop from the top right corner: I 1 (5 ) Ω + 12 V – I 3 (10 ) Ω = 0. Apply the loop rule counterclockwise to the bottom loop starting at the lower left corner: I 2 (5 ) Ω + 9 V + I 3 (10 ) Ω = 0. Note that since we went against the current direction through the (10 ) Ω resistor the potential increased. Apply the junction rule to the right middle: I 1 = I 2 + I 3 . These three equations can be solved for the current I 3 by subtracting the second equation from the first then making the substitution I 2 I 1 = – I 3 which was derived from the third equation. The result is I 3 = 3 25 A = 0.12 A, left to right.
32.66. Model: The wires and batteries are ideal. Visualize: Solve: The circuit has been redrawn for clarity. Assign the currents I 1 , I 2 , and I 3 as shown in the figure. To find the power dissipated by the 2 Ω resistor, we must find the current through it. If I 3 turns out to be negative, we’ll know it really flows bottom to top. Apply Kirchhoff’s loop rule clockwise to the left loop from the bottom left corner: + 12 V – I 1 (4 ) Ω I 3 (2 ) Ω = 0.

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