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chapter18

# chapter18 - 142 Chapter 18 PROBLEM SOLUTIONS 18.1 From V =...

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142 Chapter 18 PROBLEM SOLUTIONS 18.1 From Δ V I R r = + ( ) , the internal resistance is r V I R = = = Δ Ω Ω 9 00 72 0 4 92 . . . V 0.117 A 18.2 (a) When the three resistors are in series, the equivalent resistance of the circuit is R R R R eq = + + = ( ) = 1 2 3 3 9 0 27 . Ω Ω (b) The terminal potential difference of the battery is applied across the series combination of the three 9 0 . Ω resistors, so the current supplied by the battery and the current through each resistor in the series combination is I V R = = = Δ Ω eq V 27 A 12 0 44 . (c) If the three 9 0 . Ω are now connected in parallel with each other, the equivalent resistance is 1 1 9 0 1 9 0 1 9 0 3 9 0 R eq = + + = . . . . Ω Ω Ω Ω or R eq = = 9 0 3 3 0 . . Ω Ω When this parallel combination is connected to the battery, the potential difference across each resistor in the combination is Δ V = 12 V , so the current through each of the resistors is I V R = = = Δ Ω 12 1 3 V 9.0 A . 18.3 For the bulb in use as intended, R V bulb V W = ( ) = ( ) = Δ Ω 2 2 120 75 0 192 P . Now, assuming the bulb resistance is unchanged, the current in the circuit shown is I V R = = + + = Δ Ω Ω Ω eq V 0.800 A 120 192 0 800 0 620 . . and the actual power dissipated in the bulb is P = = ( ) ( ) = I R 2 2 0 620 192 73 8 bulb A W . . Ω 18.4 (a) When the 8 00 . - Ω resistor is connected across the 9.00-V terminal potential difference of the battery, the current through both the resistor and the battery is I V R = = = Δ Ω 9 00 1 13 . . V 8.00 A (b) The relation between the emf and the terminal potential difference of a battery supplying current I is Δ V Ir = ε , where r is the internal resistance of the battery. Thus, if the battery has r = 0 15 . Ω and maintains a terminal potential difference of Δ V = 9 00 . V while sup- plying the current found above, the emf of this battery must be ε = + = + ( ) ( ) = + ( ) Δ Ω V Ir 9 00 1 13 0 15 9 00 0 17 . . . . . V A Ω Ω = 9 17 .

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Direct-Current Circuits 143 18.5 (a) The equivalent resistance of the two parallel resistors is R p = + = 1 7 00 1 10 0 4 12 1 . . . Ω Ω Ω Thus, R R R R ab p = + + = + + ( ) = 4 9 4 00 4 12 9 00 17 1 . . . . Ω Ω (b) I V R ab ab ab = ( ) = = Δ Ω 34 0 17 1 1 99 . . . V A , so I I 4 9 1 99 = = . A Also, Δ Ω V I R p ab p ( ) = = ( ) ( ) = 1 99 4 12 8 18 . . . A V Then, I V R p 7 7 8 18 1 17 = ( ) = = Δ Ω . . V 7.00 A and I V R p 10 10 8 18 0 818 = ( ) = = Δ Ω . . V 10.0 A 18.6 (a) The parallel combination of the 6 0 . and 12 Ω Ω resistors has an equivalent resistance of 1 1 6 0 1 12 2 1 12 1 R p = + = + . Ω Ω Ω or R p 1 12 3 4 0 = = Ω Ω . Similarly, the equivalent resistance of the 4 0 . and 8.0 Ω Ω parallel combination is 1 1 4 0 1 8 0 2 1 8 0 2 R p = + = + . . . Ω Ω Ω or R p 2 8 3 = Ω The total resistance of the series combination between points a and b is then R R R ab p p = + + = + + = 1 2 5 0 4 0 5 0 8 0 3 35 3 . . . . Ω Ω Ω Ω Ω (b) If Δ V ab = 35 V , the total current from a to b is I V R ab ab ab = = ( ) = Δ Ω 35 3 0 V 35 3 A . and the potential differences across the two parallel combinations are Δ Ω V I R p ab p 1 1 3 0 4 0 12 = = ( ) ( ) = . . A V and Δ Ω V I R p ab p 2 2 3 0 8 0 3 8 0 = = ( ) = .
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chapter18 - 142 Chapter 18 PROBLEM SOLUTIONS 18.1 From V =...

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