Question

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 -

Hint:

In an AC circuit, when the current is in phase with the emf, the inductive and capacitive reactances are equal. This results in the impedance being purely resistive.

Answer 1

Correct Answer: 242

The correct answer is 242.
$X_L=X_C$
So, $Z=R=20\Omega$
$i_{rms}=\frac{220}{20}=11$
$i_{\max }=11\sqrt{2}=\sqrt{242}$

Answer 2

For an AC circuit, the inductive reactance \( X_L \) and capacitive reactance \( X_C \) are given by: \[ X_L = \omega L = 2 \pi f L \quad \text{and} \quad X_C = \frac{1}{\omega C} \] where: - \( L = 0.5 \, \text{mH} = 0.5 \times 10^{-3} \, \text{H} \), - \( C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F} \), - \( R = 20 \, \Omega \), - \( f \) is the frequency of the AC supply. Since the current is in phase with the emf, we have: \[ X_L = X_C \] Thus, the impedance \( Z \) is given by: \[ Z = R = 20 \, \Omega \] The current amplitude \( I_{\text{max}} \) is then: \[ I_{\text{max}} = \frac{V}{Z} = \frac{220}{20} = 11 \, \text{A} \] Thus, \( x = 242 \).