Question

A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.

(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

(d) What is the Q-factor of the given circuit?

Answer 1

Resistors, Capacitors, AC :D

Symbols Have Usual Meanings :D

Final Answer :
(a) 663.15 Hz, 10√2 A
(b) 663.15 Hz ,2300W
(c) [647.9 Hz ,678.4 Hz] ,10 A
(d) 21.74

Given : $V_ {rms} = 230 V \\ L = 0.12 H, C = 480 * 10^{-9} F, R = 23 \Omega$
Analysis :
(a) At resonance condition, current amplitude is maximum :
$f_r = \frac{1}{2\pi \sqrt{LC}} \\ => \frac{1}{2\pi \sqrt{0.12*480*10^{-9}}}\\ =>663.15Hz$

Maximum Current,
$I_{max} = I_o= \frac{V_o}{R} \\ =>I_{max}= \frac{V_{rms}*\sqrt{2}}{23} \\ = > 10\sqrt{2} A$

(b) At resonace frequency, Average Power absorbed is maximum :
$f_r = 663.15Hz \\ P_{max} = \frac{1}{2}I_o^2 *R = 2300W$

(c)

There are two frequencies for which required condition occurs :
$f= f_r ±\frac{R}{2L} \\ => f = 663.15±\frac{23}{4\pi *0.12} \\ => f = 663.15 ± 15.25 \\ => f_H = 647.9 Hz \\ => f_L = 678.4 Hz$

(d) $Q = \frac{2 \pi f_r *0.12}{23} => Q = 21.74$

Answer 2

plz refer to the attachment for answer

thanks

hope it helps sir.