Question
Asked Nov 21, 2019
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Chapter 9, Problem 9.017
I )-0.5 cos 2000r A, find the average power absortbed by each element ((a) 150-2 resistor, (b) 40-0 resistor, (c) source, (d) inductor, (e) capacitor) in the circuit in the figure below
ww
40 0
150
60 mH
12.5F
(a)
(ь)
W
(c)
W
(d)
(e)
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Chapter 9, Problem 9.017 I )-0.5 cos 2000r A, find the average power absortbed by each element ((a) 150-2 resistor, (b) 40-0 resistor, (c) source, (d) inductor, (e) capacitor) in the circuit in the figure below ww 40 0 150 60 mH 12.5F (a) (ь) W (c) W (d) (e)

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Expert Answer

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Step 1

The angular frequency of the source is 2000 rad/s.

The reactance offered by the capacitor and the inductor are:-

1
joC
-j
2000 radsx12.5x 10- F
Xj40
And
X j2000 rads1 x 60 x103 H
Xj120
40
0.5L6
T-NoA
120
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1 joC -j 2000 radsx12.5x 10- F Xj40 And X j2000 rads1 x 60 x103 H Xj120 40 0.5L6 T-NoA 120

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Step 2

The current I1 from the given circuit diagram is given by the following expression,

40 j120
- x0.520°
40 j120 +150 j40
40+ j120
-x 0.520
190+ j80
0.306248.73
And
150 j40
40j120 150 j40
150 j40
190 j80
-x 0.520°
-x 0.520°
I 0.376-37.76°
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40 j120 - x0.520° 40 j120 +150 j40 40+ j120 -x 0.520 190+ j80 0.306248.73 And 150 j40 40j120 150 j40 150 j40 190 j80 -x 0.520° -x 0.520° I 0.376-37.76°

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Step 3

The power dissipated by the resistors can be ...

P = 12 R
ms
P50 x150
0.306 A
2
x150Q
P500 7.02 W
And
P0n 1x40
0.376 A
PA0
x400
P0 2.82 W
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P = 12 R ms P50 x150 0.306 A 2 x150Q P500 7.02 W And P0n 1x40 0.376 A PA0 x400 P0 2.82 W

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