An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the power dissipated in the `L-C-R` circuit

Options

A. 50 W
B. 100 W
C. 200 W
D. 400 W

Explanation

To solve the problem step by step, we will analyze the L-C-R circuit and calculate the power dissipated in it.

### Step 1: Understand the given information
- Resistance \( R = 100 \, \Omega \)
- Voltage \( V_0 = 200 \, V \)
- Angular frequency \( \omega = 300 \, \text{rad/s} \)
- When capacitance is removed, the current lags behind the voltage by \( 60^\circ \).
- When inductance is removed, the current leads the voltage by \( 60^\circ \).

### Step 2: Analyze the circuit when capacitance is removed
When the capacitance is removed, the circuit behaves inductively. The phase angle \( \phi \) is given as \( 60^\circ \).

Using the tangent of the phase angle:
\[ \tan(60^\circ) = \sqrt{3} \]
The impedance in this case can be expressed as:
\[ \tan(\phi) = \frac{X_L}{R} \implies X_L = R \cdot \tan(60^\circ) = 100 \cdot \sqrt{3} \approx 173.21 \, \Omega \]

### Step 3: Analyze the circuit when inductance is removed
When the inductance is removed, the circuit behaves capacitively, and the phase angle is again \( 60^\circ \).

Using the same approach:
\[ \tan(60^\circ) = \sqrt{3} \]
The impedance in this case can be expressed as:
\[ \tan(\phi) = \frac{X_C}{R} \implies X_C = R \cdot \tan(60^\circ) = 100 \cdot \sqrt{3} \approx 173.21 \, \Omega \]

### Step 4: Calculate the total impedance \( Z \)
Since \( X_L = X_C \), we can find the total impedance \( Z \):
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (173.21 - 173.21)^2} = \sqrt{100^2} = 100 \, \Omega \]

### Step 5: Calculate the maximum current \( I_0 \)
Using Ohm's law:
\[ I_0 = \frac{V_0}{Z} = \frac{200}{100} = 2 \, A \]

### Step 6: Calculate the power dissipated in the circuit
The power dissipated in the circuit can be calculated using the formula:
\[ P = V_0 \cdot I_0 \cdot \cos(\phi) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ P = 200 \cdot 2 \cdot \frac{1}{2} = 200 \, W \]

### Final Answer
The power dissipated in the L-C-R circuit is \( 200 \, W \).