In a series LCR circuit `R= 200(Omega)` and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by `30(@)`. On taking out the inductor from the circuit the current leads the voltage by `30(@)`. The power dissipated in the LCR circuit is

Options

A. 305 W
B. 210 W
C. Zero W
D. 242 W

Explanation

To find the power dissipated in the LCR circuit, we will follow these steps:

### Step 1: Identify the given values
- Resistance, \( R = 200 \, \Omega \)
- Voltage, \( V = 220 \, V \)
- Frequency, \( f = 50 \, Hz \)
- Phase difference when the capacitor is removed (LR circuit), \( \phi = -30^\circ \)
- Phase difference when the inductor is removed (RC circuit), \( \phi = +30^\circ \)

### Step 2: Calculate the inductive reactance \( \omega L \)
In the LR circuit, the phase difference \( \phi \) is given by:
\[ \tan \phi = \frac{\omega L}{R} \]
For \( \phi = -30^\circ \):
\[ \tan(-30^\circ) = -\frac{1}{\sqrt{3}} \]
Thus,
\[ -\frac{1}{\sqrt{3}} = \frac{\omega L}{200} \]
From this, we can express \( \omega L \):
\[ \omega L = -\frac{200}{\sqrt{3}} \]

### Step 3: Calculate the capacitive reactance \( \frac{1}{\omega C} \)
In the RC circuit, the phase difference \( \phi \) is given by:
\[ \tan \phi = \frac{1}{\omega C R} \]
For \( \phi = 30^\circ \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus,
\[ \frac{1}{\sqrt{3}} = \frac{1}{\omega C \cdot 200} \]
From this, we can express \( \frac{1}{\omega C} \):
\[ \frac{1}{\omega C} = \frac{200}{\sqrt{3}} \]

### Step 4: Calculate the impedance \( Z \) of the LCR circuit
The total impedance \( Z \) in the LCR circuit can be calculated as:
\[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]
Substituting \( \omega L \) and \( \frac{1}{\omega C} \):
\[ Z = \sqrt{200^2 + \left(-\frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}}\right)^2} \]
\[ Z = \sqrt{200^2 + \left(-\frac{400}{\sqrt{3}}\right)^2} \]
\[ Z = \sqrt{200^2 + \frac{160000}{3}} \]
Calculating \( Z \):
\[ Z = \sqrt{40000 + \frac{160000}{3}} = \sqrt{\frac{120000}{3}} = \sqrt{40000} = 200 \, \Omega \]

### Step 5: Calculate the power dissipated \( P \)
The power dissipated in the LCR circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
Where \( I_{rms} = \frac{V_{rms}}{Z} \) and \( \cos \phi = \frac{R}{Z} \):
\[ P = V_{rms} \cdot \frac{V_{rms}}{Z} \cdot \frac{R}{Z} \]
Substituting the values:
\[ P = 220 \cdot \frac{220}{200} \cdot \frac{200}{200} \]
\[ P = 220 \cdot 1.1 = 242 \, W \]

### Final Answer
The power dissipated in the LCR circuit is \( 242 \, W \).

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