Chapter 32 - 32.1. Solve: From the circuit in Figure...

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32.1. Solve: From the circuit in Figure EX32.1, we see that 50 Ω and 100 Ω resistors are connected in series across the battery. Another resistor of 75 Ω is also connected across the battery.
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32.2. Solve: In Figure EX32.2, the positive terminal of the battery is connected to a resistor. The other end of that resistor is connected to resistor and a capacitor in parallel.
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32.3. Model: Assume that the connecting wires are ideal. Visualize: Please refer to Figure EX32.3. Solve: The current in the 2 Ω resistor is 1 6 V 2 3 A I = to the left. The current in the 5 Ω resistor is 2 10 V 5 2 A I = downward. Using Kirchhoff’s junction law, we see that I = I 1 + I 2 = 3 A + 2 A = 5 A This current flows toward the junction, that is, downward.
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32.4. Model: The batteries and the connecting wires are ideal. Visualize: Please refer to Figure EX32.4. Solve: (a) Choose the current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit. With the 9 V battery being labeled 1 and the 6 V battery being labeled 2. Kirchhoff’s loop law is bat 1 R bat 2 1 2 0 i VV V V I R Δ= Δ + Δ = +− − = EE 12 9 V 6 V 0.10 A 30 I R −− == = Ω Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I is positive, we can say that I = 0.100 A flows from left to right through the resistor. (b) The graph shows 9 V gained in battery 1, Δ V R = IR = 3 V lost in the resistor, and another 6 V lost in battery 2. The final potential is the same as the initial potential, as required.
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32.5. Model: Assume ideal connecting wires and an ideal battery for which Δ V bat = . E Visualize: Please refer to Figure EX32.5. We will choose a clockwise direction for I . Note that the choice of the current’s direction is arbitrary because, with two batteries, we may not be sure of the actual current direction. The 3 V battery will be labeled 1 and the 6 V battery will be labeled 2. Solve: (a) Kirchhoff’s loop law, going clockwise from the negative terminal of the 3-V battery is ( ) closed loop bat 1 R bat 2 0 i i VV V V V Δ= Δ = Δ + Δ + Δ = + 3 V – (18 Ω ) I + 6 V = 0 9 V 0.5 A 18 I == Ω Thus, the current through the 18 Ω resistor is 0.5 A. Because I is positive, the current is left to right (i.e., clockwise). (b) Assess: The graph shows a 3 V gain in battery 1, a 9 V loss in the resistor, and a gain of 6 V in battery 2. The final potential is the same as the initial potential, as required.
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32.6. Visualize: Please refer to Figure EX32.6. Define the current I as a clockwise flow. Solve: (a) There are no junctions, so conservation of current tells us that the same current flows through each circuit element. From Kirchhoff’s loop law, ΣΔ V i = Δ V bat + Δ V 10 + Δ V 20 = 0 As we go around the circuit in the direction of the current, potential is gained in the battery ( Δ V bat = bat E = + 15 V) and potential is lost in the resistors ( Δ V resistor = IR ). The loop law is 0 = bat E IR 1 IR 2 = bat E I ( R 1 +
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This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 32 - 32.1. Solve: From the circuit in Figure...

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