Question

A series L-C-R circuit is connected to a 220V variable frequency AC supply. If L = 20 mH, C = (800/$\pi^{2}$) mF and R = 110 $\Omega$ then find:
(i) the frequency of the source, for which average power absorbed by the circuit is maximum.
(ii) the value of maximum current amplitude.

Answer 1

It is given that:

• V = 220 V
• L = 20 mH
• C = $\frac{800}{\pi^2}$ mF
• R = 110 Ohm

i) average power absorbed by the circuit is maximum when the circuit is at resonance.

$\omega_0=\frac{1}{\sqrt{LC}}\\ \omega_0=\frac{1}{\sqrt{20*10^{-3}*800*10^{-3}/\pi^2}}\\\\\omega_0=24.8364\:rad/sec$

$f=\frac{w_0}{2\pi}=\frac{24.8364}{2\pi}=4\:Hz$

ii) $V_{max}=220\sqrt{2}$ V.

For maximum current, the circuit has to be at resonance. Therefore, the impedance "Z" = R = 110 Ohm.

Hence, $I=\frac{220\sqrt{2} }{110}=2.82$ Ampere