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EXAMPLE 10.17 A practical coil is placed between two voltage sources v1 5 cos !1 t and v2 10 cos !2 t 608 , which share the same common reference terminal (see Fig. 9-54). The coil is modeled by a 5-mH inductor in series with a 10-
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resistor. Find the average power in the coil for (a) !2 2!1 4000; b !1 !2 2000; p c !1 2000 and !2 1000 2, all in rad/s. Let v1 by itself produce i1 . Likewise, let v2 produce i2 . Then i i1 i2 . The instantaneous power p and the average power P are
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2 2 p Ri2 R i1 i2 2 Ri1 Ri2 2Ri1 i2 2 2 P hpi Rhi1 i Rhi2 i 2Rhi1 i2 i P1 P2 2Rhi1 i2 i
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where hpi indicates the average of p. Note that in addition to P1 and P2 , we need to take into account a third term hi1 i2 i which, depending on !1 and !2 , may or may not be zero. (a) By applying superposition in the phasor domain we nd the current in the coil (see Example 9.7). I1 P1 V1 5 0:35 Z1 10 j10 458 A; i1 0:35 cos 2000t 458
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2 RI1 10 0:352 0:625 W 2 2 V 10 608 I2 2 0:45 3:48 A; i2 0:45 cos 4000t 3:48 Z2 10 j20 2 RI2 10 0:452 1W 2 2 i i1 i2 0:35 cos 2000t 458 0:45 cos 4000t 3:48
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In this case hi1 i2 i 0 because hcos 2000t 458 cos 4000t 3:48 i 0. applies and P P1 P2 0:625 1 1:625 W. (b) The current in the coil is i 0:61 cos 2000t 1358 (see Example 9.7). coil is P RI 2 =2 5 0:61 2 1:875 W. Note that P > P1 P2 . (c) By applying superposition in the time domain we nd i1 0:35 cos 2000t 458 ; P1 0:625 W p i2 0:41 cos 1000 2t 35:38 ; P2 0:833 W
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Therefore, superposition of power The average power dissipated in the
p i i1 i2 ; P hRi2 =2i P1 P2 1:44hcos 2000t 458 cos 1000 2t 35:38 i p The term hcos 2000t 458 cos 1000 2t 35:38 i is not determined because a common period can t be found. The average power depends on the duration of averaging.
Solved Problems
10.1 The current plotted in Fig. 10-2(a) enters a 0.5-mF capacitor in series with a 1-k
resistor. Find and plot (a) v across the series RC combination and (b) the instantaneous power p entering RC. (c) Compare the results with Examples 10.1 and 10.2.
(a) Referring to Fig. 10-2(a), in one cycle of the current the voltages are  1V 0 < t < 1 ms vR 1 V 1 < t < 2 ms  t 2000t V 0 < t < 1 ms 1 i dt vC C 0 4 2000t V 1 < t < 2 ms  1 2000t V 0 < t < 1 ms v vR vC [See Fig. 10-13 a 3 2000t V 1 < t < 2 ms (b) During one cycle, pR Ri2 1 mW  2000t mW 0 < t < 1 ms pC vC i 2000t 4 mW 1 < t < 2 ms  1 2000t mW 0 < t < 1 ms p vi pR pC 2000t 3 mW 1 < t < 2 ms (c)
[(See Fig. 10-13 b
The average power entering the circuit during one cycle is equal to the average power absorbed by the resistor. It is the same result obtained in Example 10.1. The power exchanged between the source and the circuit during one cycle also agrees with the result obtained in Example 10.2.
A 1-V ac voltage feeds (a) a 1-
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