Question

A series LCR circuit with resistance ($R$) $500\ \Omega$ is connected to an a.c. source of $250\ \text{V}$. When only the capacitance is removed, the current lags behind the voltage by $60^\circ$. When only the inductance is removed, the current leads the voltage by $60^\circ$. The impedance of the circuit is ($\tan \frac{\pi}{3} = \sqrt{3}$)

• A. $\frac{500}{\sqrt{3}}\ \Omega$
• B. $500\sqrt{3}\ \Omega$
• C. $250\ \Omega$
• D. $500\ \Omega$

Hint:

Whenever the phase angle of lag (when C is removed) equals the phase angle of lead (when L is removed), it mathematically guarantees $X_L = X_C$. The circuit is resonant, and the impedance is simply equal to the resistance!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the total impedance of a series LCR circuit, given its resistance and the phase angles when either the capacitor or the inductor is removed.

Step 2: Key Formula or Approach:
The phase angle $\phi$ in an RL circuit is given by $\tan \phi = \frac{X_L}{R}$.
The phase angle $\phi$ in an RC circuit is given by $\tan \phi = \frac{X_C}{R}$.
The total impedance of an LCR circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

Step 3: Detailed Explanation:
Given resistance, $R = 500\ \Omega$.
When only capacitance is removed (it becomes an RL circuit), the current lags by $\phi = 60^\circ$:
$$\tan 60^\circ = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = R\sqrt{3}$$
When only inductance is removed (it becomes an RC circuit), the current leads by $\phi = 60^\circ$:
$$\tan 60^\circ = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{R} \implies X_C = R\sqrt{3}$$
From the above, we can clearly see that $X_L = X_C$. This means the LCR circuit is in a state of resonance.
Now, calculate the total impedance $Z$:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
Since $X_L = X_C$, the term $(X_L - X_C)$ becomes $0$:
$$Z = \sqrt{R^2 + 0} = R$$
$$Z = 500\ \Omega$$

Step 4: Final Answer:
The impedance of the circuit is $500\ \Omega$, matching option (D).