A series LCR circuit with an inductor a capacitor and a resistor of 80 `Omega` is connected to an AC source of 180 V and angular frequency of 250 rad `s^(-1)` . Suddenly the inductor is removed from the circuit and it is found that current leads the voltage by `45^(@)` . Similarly when only capacitor is removed from the circuit it is found that the current lags behind the voltage by `45^(@)` . Find the power dissipated in the circuit.

Answer

To solve the problem step by step, we need to analyze the given LCR circuit and the conditions when the inductor and capacitor are removed.

### Step 1: Understanding the Circuit
We have a series LCR circuit with:
- Resistance \( R = 80 \, \Omega \)
- AC source voltage \( V = 180 \, V \)
- Angular frequency \( \omega = 250 \, \text{rad/s} \)

When the inductor is removed, the circuit consists of a capacitor and a resistor, and we know that the current leads the voltage by \( 45^\circ \). When the capacitor is removed, the circuit consists of an inductor and a resistor, and the current lags behind the voltage by \( 45^\circ \).

### Step 2: Analyzing the Capacitive Circuit
In the case where the inductor is removed, the circuit consists of a capacitor and a resistor. The phase difference between the current and voltage is \( 45^\circ \), which means:
\[ \tan(45^\circ) = 1 = \frac{X_C}{R} \]
From this, we can find the capacitive reactance \( X_C \):
\[ X_C = R = 80 \, \Omega \]

### Step 3: Analyzing the Inductive Circuit
In the case where the capacitor is removed, the circuit consists of an inductor and a resistor. The phase difference between the current and voltage is again \( 45^\circ \), meaning:
\[ \tan(45^\circ) = 1 = \frac{R}{X_L} \]
From this, we can find the inductive reactance \( X_L \):
\[ X_L = R = 80 \, \Omega \]

### Step 4: Finding the Impedance of the Original Circuit
In the original LCR circuit, the inductive reactance \( X_L \) and capacitive reactance \( X_C \) cancel each other out since they are equal:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{80^2 + (80 - 80)^2} = \sqrt{80^2} = 80 \, \Omega \]

### Step 5: Calculating the Power Dissipated
The power dissipated in the circuit can be calculated using the formula:
\[ P = I_{\text{rms}} \cdot V_{\text{rms}} \cdot \cos(\phi) \]
where \( \phi \) is the phase angle between the current and voltage. Since \( X_L = X_C \), the circuit is purely resistive, and \( \phi = 0^\circ \) (the current and voltage are in phase). Therefore, \( \cos(0^\circ) = 1 \).

First, we need to find \( I_{\text{rms}} \):
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{180 \, V}{80 \, \Omega} = 2.25 \, A \]

Now substituting the values into the power formula:
\[ P = I_{\text{rms}} \cdot V_{\text{rms}} \cdot \cos(0^\circ) = 2.25 \, A \cdot 180 \, V \cdot 1 = 405 \, W \]

### Final Answer
The power dissipated in the circuit is \( 405 \, W \).

---