# Visualize please refer to figure ex3220 solve from

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Visualize: Please refer to Figure EX32.20. Solve: From Equation 32.21, the potential difference across the battery is ( ) bat bat 9.0 V 1 20 1 1.18 8.5 V R V r R R r V Δ = = = Ω = Ω + Δ E E Assess: 1 Ω is a typical internal resistance for a battery. This causes the battery’s terminal voltage in the circuit to be 0.5 V less than its emf.

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32.20. Visualize: The figure shows a metal wire of resistance R that is cut into two pieces of equal length. This produces two wires each of resistance R /2. Solve: Since these two wires are connected in parallel, eq eq 1 1 1 2 2 4 2 2 4 R R R R R R R R = + = + = =
32.21. Visualize: The three resistors in Figure EX32.21 are equivalent to a resistor of resistance eq 75 . R = Ω Solve: Because the three resistors are in parallel, ( ) eq 1 1 1 1 2 1 400 200 200 200 R R R R R R Ω + = + + = + = Ω Ω Ω ( ) ( ) eq 200 200 75 400 400 1 R R R R Ω Ω = Ω = = Ω Ω + + 400 240 200 1 75 R Ω = = Ω Ω Ω

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32.22. Model: Assume ideal connecting wires. Visualize: Please refer to Figure EX32.22. Solve: The resistance R is given by Ohm’s law, R R R V I = Δ . To determine I R we use Kirchhoff’s junction law. The input current I splits into the three currents I 10 , I 15 , and I R . That is, 2.0 A = I 10 + I 15 + I R R 8 V 8 V 10 15 I = + + Ω Ω I R = 2.0 A – 0.80 A – 0.533 A = 0.667 A Using this value of I R in Ohm’s law, 8 V 12 0.667 A R = = Ω
32.23. Model: The connecting wires are ideal with zero resistance. Solve: For the first step, the 10 Ω and 30 Ω resistors are in series and the equivalent resistance is 40 Ω . For the second step, the 60 Ω and 40 Ω resistors are in parallel and the equivalent resistance is 1 1 1 24 40 60 + = Ω Ω Ω For the third step, the 24 Ω and 10 Ω resistors are in series and the equivalent resistance is 34 Ω .

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32.24. Model: The connecting wires are ideal with zero resistance. Solve: For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance is eq 1 1 1 1 30 45 R = + Ω Ω R eq 1 = 18 Ω For the second step, resistors 42 Ω and R eq 1 = 18 Ω are in series. Therefore, R eq 2 = R eq 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω For the third step, the resistors 40 Ω and R eq 2 = 60 Ω are in parallel. So, eq 3 1 1 1 60 40 R = + Ω Ω R eq 3 = 24 Ω The equivalent resistance of the circuit is 24 Ω .
32.25. Model: The connecting wires are ideal with zero resistance. Solve: In the first step, the resistors 100 Ω , 100 Ω , and 100 Ω in the top branch are in series. Their combined resistance is 300 Ω . In the middle branch, the two resistors, each 100 Ω , are in series. So, their equivalent resistance is 200 Ω . In the second step, the three resistors are in parallel. Their equivalent resistance is eq 1 1 1 1 300 200 100 R = + + Ω Ω Ω R eq = 54.5 Ω The equivalent resistance of the circuit is 54.5 Ω .

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32.26. Model: The connecting wires are ideal with zero resistance. Solve: For the first step, the two resistors in the middle of the circuit are in parallel, so their equivalent resistance is eq 1 1 1 1 100 100 R = + Ω Ω R eq 1 = 50 Ω The three 100 Ω resistors at the end are in parallel. Their equivalent resistance is eq 2 1 1 1 1 100 100 100 R = + + Ω Ω Ω eq 2 100 33.3 3 R = Ω = Ω
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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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