**Question:**

A series $L C R$ circuit with $R=20 \Omega, L=1.5 \mathrm{H}$ and $C=35 \mu \mathrm{F}$ is connected to a variable-frequency $200 \mathrm{~V}$ ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

**Solution:**

At resonance, the frequency of the supply power equals the natural frequency of the given *LCR* circuit.

Resistance, *R* = 20 Ω

Inductance, *L* = 1.5 H

Capacitance, *C* = 35 μF = 30 × 10−6 F

AC supply voltage to the *LCR* circuit, *V* = 200 V

Impedance of the circuit is given by the relation,

$Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$

At resonance, $\omega L=\frac{1}{\omega C}$

$\therefore Z=R=20 \Omega$

Current in the circuit can be calculated as:

$I=\frac{V}{Z}$

$=\frac{200}{20}=10 \mathrm{~A}$

Hence, the average power transferred to the circuit in one complete cycle= *VI*

= 200 × 10 = 2000 W.

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.