An LCR circuit contains resistance of 110 `Omega` and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by `45^(@)`. If on the other hand, only inductor is removed the current leads by `45^(@)` with the applied voltage. The rms current flowing in the circuit will be :

Options

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

Answer

To solve the problem, we need to analyze the LCR circuit and the effects of removing the capacitor and inductor. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Circuit Configuration
We have an LCR series circuit with:
- Resistance (R) = 110 Ω
- Supply Voltage (V) = 220 V
- Angular Frequency (ω) = 300 rad/s

### Step 2: Analyze the First Case (Removing the Capacitor)
When the capacitor is removed, the circuit consists of only the resistor and inductor. The current lags behind the voltage by 45 degrees.

Using the phase relationship:
- The impedance (Z) can be represented as:
\[ Z = R + jX_L \]
- The angle (φ) between the current and voltage is given by:
\[ \tan(φ) = \frac{X_L}{R} \]
- Since φ = 45°, we have:
\[ \tan(45°) = 1 \implies \frac{X_L}{R} = 1 \implies X_L = R = 110 \, \Omega \]

### Step 3: Analyze the Second Case (Removing the Inductor)
When the inductor is removed, the circuit consists of only the resistor and capacitor. The current leads the voltage by 45 degrees.

Using the phase relationship:
- The impedance can be represented as:
\[ Z = R - jX_C \]
- The angle (φ) between the current and voltage is given by:
\[ \tan(φ) = -\frac{X_C}{R} \]
- Since φ = -45°, we have:
\[ \tan(-45°) = -1 \implies -\frac{X_C}{R} = -1 \implies X_C = R = 110 \, \Omega \]

### Step 4: Determine the Total Impedance of the Circuit
Now we have:
- \(X_L = 110 \, \Omega\)
- \(X_C = 110 \, \Omega\)

The total impedance \(Z\) in the circuit when both inductor and capacitor are present is:
\[ Z = R + j(X_L - X_C) = 110 + j(110 - 110) = 110 \, \Omega \]

### Step 5: Calculate the RMS Current
The RMS current (I_rms) can be calculated using Ohm's law:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{220 \, V}{110 \, \Omega} = 2 \, A \]

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