Question

Suppose the circuit in Exercise 7.18 has a resistance of 15 W. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answer 1

Given: The inductance of inductor is 80 mH, the capacitance of the capacitor is 60 μF, the resistance of resistor is 15 Ω and the applied potential difference is 230 V and the frequency of signal is 50 Hz.

The angular frequency is given as,

ω=2πν

Where, the frequency of the signal is ν.

By substituting the given values in the given equation, we get

ω=2π( 50 ) =100π  rad/s

The impedance of the circuit is given as,

Z= R 2 + ( ωL− 1 ωC ) 2

Where, the resistance of the resistor is R, the inductance is L, the capacitance is C and the angular frequency is ω.

By substituting the given values in the above equation, we get

Z= 15 2 + [ ( 100π×80× 10 −3 )− 1 100π×60× 10 −6 ] 2 = 225+ [ 25.12−53.08 ] 2 = 1006.76 =31.728 Ω

The current flowing through the circuit is given as,

I= V Z

Where, the applied potential difference is V.

By substituting the given values in the above equation, we get

I= 230 31.728 =7.25 A

The average power transferred to the resistance is given as,

P R = I 2 R

By substituting the given values in the above equation, we get

P R = ( 7.25 ) 2 ( 15 ) =788.44 W

Since, the average power consumed by the capacitor is P C =0 and the average power consumed by the inductor is P L =0.

The total power absorbed by the circuit is given as,

P= P R + P L + P C

By substituting the given values in the above equation, we get

P=788.44+0+0 =788.44 W

Thus, the total power absorbed by the circuit is 788.44 W.