An inductor of 0.5 mH, a capacitor of 20 $$\mu$$F and resistance of 20 $$\Omega$$ are connected in series with a 220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $$\sqrt{x}$$ A. The value of $$x$$ is-

Solution :

Given :

$$L = 0.5\text{ mH} = 0.5 \times 10^{-3}\text{ H}$$

$$C = 20\ \mu\text{F} = 20 \times 10^{-6}\text{ F}$$

$$R = 20\ \Omega$$

Source voltage :

$$V_{rms} = 220\text{ V}$$

Current is in phase with emf.

Hence, circuit is in resonance.

At resonance :

$$X_L = X_C$$

and impedance,

$$Z = R$$

Therefore,

$$I_{rms} = \frac{V_{rms}}{R}$$

$$= \frac{220}{20}$$

$$= 11\text{ A}$$

Amplitude of current :

$$I_0 = \sqrt2\ I_{rms}$$

$$= 11\sqrt2\text{ A}$$

Given amplitude is of the form :

$$x^{1/2}$$

Therefore,

$$x = 242$$

Final Answer :

$$242$$