# 4 a5 ω 4 a3 ω 8 v b using the rules of the series

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(4 A)(5 Ω ) – (4 A)(3 Ω ) = + 8 V (b) Using the rules of the series and the parallel resistors, we have simplified the circuit as shown in figure (b). A battery with emf E = 24 V is connected to an equivalent resistor of 21 8 Ω . The current in this circuit is 21 8 24 V 9.143 A. Ω = Thus, the current that flows through the battery is I bat = 9 A. When the switch is closed, points a and b are connected by an ideal wire and must therefore be at the same potential. Thus V ab = 0 V.

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32.57. Model: The voltage source/battery and the connecting wires are ideal. Visualize: Please refer to Figure P32.57. Solve: Let us first apply Kirchhoff’s loop law starting clockwise from the lower left corner: + V in IR – I (100 Ω ) = 0 V in 100 V I R = + Ω The output voltage is ( ) ( ) in out 100 100 100 V V I R = Ω = Ω + Ω out in 100 100 V V R Ω = + Ω For out in 10 V V = , the above equation can be simplified to obtain R : in in 10 100 100 V V R Ω = + Ω R + 100 Ω = 1000 Ω R = 900 Ω
32.58. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P32.58. Because the ammeter we have shows a full-scale deflection with a current of 500 μ A = 0.500 mA, we must not allow a current more than 0.500 mA to pass through the ammeter. Since we wish to measure a maximum current of 50 mA, we must split the current in such a way that 0.500 mA flows through the ammeter and 49.500 mA flows through the resistor R . Solve: (a) The potential difference across the ammeter and the resistor is the same. Thus, V R = V ammeter (49.500 × 10 3 A) R = (0.500 × 10 3 A)(50.0 Ω ) R = 0.505 Ω (b) Effective resistance is eq eq 1 1 1 0.500 0.505 50.0 R R = + = Ω Ω Ω

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32.59. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P32.59. Because the ammeter we have shows a full-scale deflection with a current of 500 μ A, we must not pass a current greater than this through the ammeter. Solve: The maximum potential difference is 5 V and the maximum current is 500 μ A. Using Ohm’s law, Δ V = I A ( R + R ammeter ) 5.0 V = (500 × 10 6 A)( R + 50.0 Ω ) R = 9.95 k Ω
32.60. Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.60 using the laws of series and parallel resistances. We have labeled the resistors as R 1 = 6 Ω , R 2 = 15 Ω , R 3 = 6 Ω , and R 4 = 4 Ω . Having reduced the circuit to a single equivalent resistance R eq , we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: R 3 and R 4 are combined to get R 34 = 10 Ω , and then R 34 and R 2 are combined to obtain R 234 : 234 2 34 1 1 1 1 1 15 10 R R R = + = + Ω Ω R 234 = 6 Ω Next, R 234 and R 1 are combined to obtain R eq = R 234 + R 1 = 6 Ω + 6 Ω = 12 Ω From the final circuit, eq 24 V 2 A 12 I R = = = Ω E Thus, the current through the battery and R 1 is I R1 = 2 A and the potential difference across R 1 is I ( R 1 ) = (2 A)(6 Ω ) = 12 V 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 Δ V .

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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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