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Sum of above powers output of battery in part (c), so eq. (26) is satis ed. 18. By eq. (34) the battery sees RT 25 38 = 25 38 15:0794 ohms, and hence battery current I V=RT 15=15:0794 0:9947 amperes; answer: battery power VI 14:9205 watts; answer: 19. By eq. (35), RT 25=16 1:5625 ohms; therefore (a) battery current I V=RT 12=1:5625 7:68 amperes, answer. (b) Ix V=Rx 12=25 0:48 amperes, answer. (c) P VI 12 7:68 92:16 watts, answer. (d) Px VIx 12 0:48 5:76 watts, answer. Note of interest: Let us pause and consider the question At what speed do the charge carriers actually move in a solid conductor such as a copper wire As we know, the speed of propagation of electrical energy along wires is very great, being only slightly less than the speed of light in free space, which is approximately 300 million meters per second (186,300 miles/sec). It should be emphasized, however, that this is the speed at which the electric and magnetic elds are propagated along the line, and is not the speed at which the charge carriers actually move. In a solid conductor, such as a copper wire, the charge carriers move at considerably less than 2.5 centimeters (one inch) per second. To put it somewhat loosely, the wave of electromotive force is propagated along wires at very great speed, but the actual charge carriers, that constitute the current, move quite slowly, with an average speed of less than 1 inch per second. The amount of current owing in a wire depends upon the number of charge carriers in motion, not on the speed of the individual charge carriers. 20. (a) By eq. (34), the parallel combination of the 12- and 6-ohm resistance is equal to a single resistance of (12)(6)/(12 6 4 ohms. Hence Fig. 34 can be redrawn as Fig. 368.
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Fig. 368
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By Ohm s law, I1 36=9 4 amp and I2 36=12 3 amp. Therefore, battery current I I1 I2 4 3 7 amperes, answer. (b) P VI 36 7 252 watts, answer.
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(c) Referring to Fig. 368, we have already found, in part (a), that I2 3 amperes. Hence the voltage drop across the 4-ohm resistor in Fig. 368 is, by eq. (13), equal to 3 4 12 volts, which is the voltage across both the 12-ohm and the 6-ohm resistors in Fig. 34. Therefore the current in the 6-ohm resistor is equal to I 12=6 2 amperes; answer: 21. First, by eq. (34), the parallel 7-ohm and 12-ohm resistors are equivalent to a single resistance of 7 12 =19 4:42105 ohms, approx. Since this is in series with the 3ohm resistor, the gure reduces to the following:
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Note that we now have 4 ohms in series with the parallel combination of 9 ohms and 7.42105 ohms. Hence RT 4 9 7:42105 =16:42105 8:06731 ohms, and therefore, by Ohm s law, eq. (11), we have I battery current 24=8:06731 2:97497 amperes; answer: 22. First let us nd the battery current I, as follows, Let R be the equivalent resistance of 14, 16, 18, and 20 ohms in parallel. To do this we use eq. (32), thus 1=R 1=14 1=16 1=18 1=20 0:239484, hence R 4:17564 ohms. The battery thus sees a resistance RT 3 4:17564 7:17564 ohms, and thus the battery current I is equal to I V=RT 18=7:17564 2:50849 amperes (a) The resistance from point x to ground is Rx 4:17564 ohms, and since the battery current I ows through this resistance we have Vx IRx 10:47455 volts; answer: (b) The current in the 20-ohm branch (7 13 20 is the voltage at point x divided by 20 ohms, that is, 10:47455=20 0:523728 amperes, and this current owing through the 13-ohm resistor gives Vy 0:523728 13 6:80846 volts; answer: R1 RT 23. Let us use eq. (34) as follows. Solving for R2 ; we have R2 , and setting R1 RT R1 36 and RT 20, we have R2 36 20 =16 45 ohms; answer: 24. (a) 7 7 49 elements. (b) Written as a53 or, if you wish, a5;3 .
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