Question

A resistor of \(50~\Omega\), an inductor and a capacitor are connected in series to an AC source of peak voltage \(200\sqrt{2}~\text{V}\).
When the capacitor alone is removed, current lags voltage by \(37^\circ\).
When the inductor alone is removed, current leads voltage by \(37^\circ\).
The power dissipated in the LCR circuit is:

• A. \(400~\text{W}\)
• B. \(800~\text{W}\)
• C. \(200~\text{W}\)
• D. \(100~\text{W}\)

Hint:

At resonance, power is \(P = \frac{V^2}{R}\) as \(X_L = X_C\), net reactance is zero.

Answer 1

The Correct Option is B

Given that when:
- Capacitor is removed, the circuit is RL with current lagging by \(37^\circ\)
- Inductor is removed, the circuit is RC with current leading by \(37^\circ\)
This means in the full LCR circuit, the phase difference is zero — i.e., resonance, so power is maximized.
Peak voltage \(V_0 = 200\sqrt{2}~\text{V}\)
So RMS voltage \(V = \frac{V_0}{\sqrt{2}} = 200~\text{V}\)
Resistance \(R = 50~\Omega\)
Power dissipated: \[ P = \frac{V^2}{R} = \frac{200^2}{50} = \frac{40000}{50} = 800~\text{W} \]

Answer 2

Step 1: Understand the problem and given data
- Resistor, \(R = 50~\Omega\)
- Peak voltage, \(V_0 = 200\sqrt{2}~\text{V}\)
- When capacitor is removed, current lags voltage by \(37^\circ\)
- When inductor is removed, current leads voltage by \(37^\circ\)

Step 2: Interpret the phase angles
- Current lags voltage by \(37^\circ\) means the circuit is inductive with inductive reactance \(X_L\) present.
- Current leads voltage by \(37^\circ\) means the circuit is capacitive with capacitive reactance \(X_C\) present.

Step 3: Use the phase angle formula for R-L and R-C circuits
\[ \tan 37^\circ = \frac{X_L}{R} = \frac{X_C}{R} \]
\(\tan 37^\circ \approx 0.75\)
Thus,
\[ X_L = X_C = 0.75 \times R = 0.75 \times 50 = 37.5~\Omega \]

Step 4: Calculate the net reactance in the full LCR circuit
Since \(X_L = X_C\), the inductive and capacitive reactances cancel each other out:
\[ X = X_L - X_C = 0 \]
So the circuit behaves like a pure resistor of \(50~\Omega\).

Step 5: Calculate the RMS voltage
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200~\text{V} \]

Step 6: Calculate the RMS current
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{200}{50} = 4~\text{A} \]

Step 7: Calculate power dissipated in the circuit
\[ P = I_{\text{rms}}^2 \times R = 4^2 \times 50 = 16 \times 50 = 800~\text{W} \]

Step 8: Final conclusion
The power dissipated in the LCR circuit is \(800~\text{W}\).