PHY
solutions_chapter19

# R 5 v 2 p 5 1 120 v 2 2 100 w 5 144 v p 5 v 2 r a e 5

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R 5 V 2 P 5 1 120 V 2 2 100 W 5 144 V P 5 V 2 / R ( a ) E 5 48.0 V R 1 5 1.00 V R 3 5 7.00 V R 2 5 3.00 V R 4 5 5.00 V ( b ) E 5 48.0 V R 12 5 4.00 V R 34 5 12.0 V ( c ) E 5 48.0 V R eq 5 3.00 V I 3 5 I 4 5 4.00 A. I 1 5 I 2 5 12.0 A I 12 1 I 34 5 16.0 A. I 34 5 48.0 V 12.0 V 5 4.00 A. I 12 5 48.0 V 4.00 V 5 12.0 A I 5 48.0 V 3.00 V 5 16.0 A. R eq 5 3.00 V . R eq 5 R 1 1 R 2 . R eq 5 R 1 R 2 R 1 1 R 2 . 1 R eq 5 1 R 1 1 1 R 2 9.00 A. I 4 5 V 34 R 4 5 36.0 V 4.00 V 5 I 3 5 V 34 R 3 5 36.0 V 12.0 V 5 3.00 A R 4 , R 3 V 34 4.00 A. I 2 5 V 12 R 2 5 24.0 V 6.00 V 5 I 1 5 V 12 R 1 5 24.0 V 3.00 V 5 8.00 A R 2 , R 1 V 12 60.0 V. V 12 1 V 34 5 V 34 5 IR 34 5 1 12.0 A 21 3.00 V 2 5 36.0 V. V 12 5 IR 12 5 1 12.0 A 21 2.00 V 2 5 24.0 V. ( a ) ( b ) ( c ) R 1 R 2 R 3 R 4 R 12 R 34 R eq E E E I 5 60.0 V 5.00 V 5 12.0 A. R eq 5 5.00 V . Current, Resistance, and Direct-Current Circuits 19-11

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power. Use the rated power to find the resistance of each bulb. In parallel, the full line voltage is applied to each bulb. Solve: The resistance of each bulb is: and (a) In series, The current in each bulb is the current in the equivalent circuit, The power consumed by each bulb is and (b) In parallel, for each bulb, so and (c) P is proportional to and now for each bulb, so they would each consume 4 times their rated power: and The power rating for each bulb would be exceeded and they would quickly burn out. Reflect: In series, the bulb with the greatest resistance dissipates the greatest power. When the bulbs are connected individually across the 120 V outlet, the bulb with the least resistance dissipates the greatest power. 19.56. Set Up: For resistors in series the equivalent resistance is The current is the same in each. so (b) The 20.0 resistor would dissipate more than 2.00 W and would fail. 19.57. Set Up: Assume the unknown currents have the directions shown in Figure 19.57. We have used the junction rule to write the current through the 10.0 V battery as There are two unknowns, and so we will need two equations. Three possible circuit loops are shown in the figure. Figure 19.57 Solve: (a) Apply the loop rule to loop (1), going around the loop in the direction shown: and (b) Apply the loop rule to loop (3): and (c) Reflect: For loop (2) we get so that with the currents we have calculated the loop rule is satisfied for this third loop. 5.00 V 1 5.00 V 2 10.0 V 5 0, 1 5.00 V 1 I 2 1 20.0 V 2 2 I 1 1 30.0 V 2 5 5.00 V 1 1 0.250 A 21 20.0 V 2 2 1 0.333 A 21 30.0 V 2 5 I 1 1 I 2 5 0.333 A 1 0.250 A 5 0.583 A I 2 5 0.250 A. 1 10.0 V 2 1 20.0 V 2 I 2 2 5.00 V 5 0 I 1 5 0.333 A. 0 1 10.0 V 2 1 30.0 V 2 I 1 5 10.0 V 20.0 V 30.0 V 5.00 V 1 1 2 1 2 2 1 3 2 + + I 2 I 1 I 2 , I 1 I 1 1 I 2 . V V 5 IR eq 5 1 0.316 A 21 35.0 V 2 5 11.1 V. R eq 5 5.00 V 1 10.0 V 1 20.0 V 5 35.0 V . I 5 Å P R 20 5 Å 2.00 W 20.0 V 5 0.316 A. P 5 I 2 R R eq 5 R 1 1 R 2 1 R 3 . P 100 5 400 W. P 75 5 300 W P 60 5 240 W, V 5 240 V V 2 P 100 5 100 W. P 75 5 75 W P 60 5 60 W, V 5 120 V P 100 5 1 0.208 A 2 2 1 144 V 2 5 6.23 W. P 75 5 1 0.208 A 2 2 1 192 V 2 5 8.31 W P 60 5 I 2 R 60 5 1 0.208 A 2 2 1 240 V 2 5 10.4 W, I 5 V R eq 5 120 V 576 V 5 0.208 A. R eq 5 R 60 1 R 75 1 R 100 5 576 V . R 100 5 V 2 P 100 5 1 120 V 2 2 100 W 5 144 V . R 75 5 V 2 P 75 5 1 120 V 2 2 75 W 5 192 V R 60 5 V 2 P 60 5 1 120 V 2 2 60 W 5 240 V , P 5 V 2 R 5 I 2 R .
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