An LCR circuit contains resistance of `100 Omega` and a supply of `200V` at `300 rads^(-1)` angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by `60^(@)` . If, on the other hand , only inductor is taken out,the current leads the voltage by `60^(o)`. The current flowing in the circuit is

Options

A. 1.5 Amp
B. 2 Amp
C. 2.5 Amp
D. 1 Amp

Answer

To solve the problem step by step, we will analyze the given LCR circuit and apply the relevant formulas.

### Step 1: Understand the Circuit Configuration
We have an LCR circuit with:
- Resistance (R) = 100 Ω
- Supply Voltage (V) = 200 V
- Angular Frequency (ω) = 300 rad/s

### Step 2: Analyze the Case When Capacitance is Removed
When the capacitance is removed, the circuit behaves as an RL circuit. The problem states that the current lags behind the voltage by 60 degrees. This indicates that the circuit is inductive.

From the phase relationship in an RL circuit:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \(X_L\) is the inductive reactance and \(\phi\) is the phase angle (60 degrees in this case).

### Step 3: Calculate Inductive Reactance
Using the tangent of the angle:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can write:
\[ \sqrt{3} = \frac{X_L}{100} \]
From this, we can solve for \(X_L\):
\[ X_L = 100\sqrt{3} \, \Omega \]

### Step 4: Analyze the Case When Inductance is Removed
When the inductance is removed, the circuit behaves as a capacitive circuit. The current leads the voltage by 60 degrees, indicating that the circuit is capacitive.

From the phase relationship in a capacitive circuit:
\[ \tan(\phi) = \frac{X_C}{R} \]
where \(X_C\) is the capacitive reactance.

### Step 5: Calculate Capacitive Reactance
Using the same tangent value:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can write:
\[ \sqrt{3} = \frac{X_C}{100} \]
From this, we can solve for \(X_C\):
\[ X_C = 100\sqrt{3} \, \Omega \]

### Step 6: Find the Total Impedance
In the original LCR circuit, we have:
\[ X_L = X_C \]
Thus, the total impedance \(Z\) can be calculated as:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \(X_L = X_C\), we have:
\[ Z = R = 100 \, \Omega \]

### Step 7: Calculate the Current
Using Ohm's Law, the current \(I\) in the circuit can be calculated as:
\[ I = \frac{V}{Z} \]
Substituting the values:
\[ I = \frac{200}{100} = 2 \, A \]

### Final Answer
The current flowing in the circuit is **2 A**.