For a series L C R circuit, the power loss at resonance is

Answer

To find the power loss at resonance in a series LCR circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Resonance Condition**:
In a series LCR circuit, resonance occurs when the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\). Mathematically, this is expressed as:
\[ X_L = X_C \]
where \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\).

2. **Determine the Impedance at Resonance**:
At resonance, since \(X_L = X_C\), the total impedance \(Z\) of the circuit simplifies to:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R \]

3. **Calculate the RMS Current**:
The RMS current \(I_{RMS}\) in the circuit can be calculated using Ohm's law. The voltage across the resistor \(V_R\) is equal to the supply voltage \(V_s\):
\[ I_{RMS} = \frac{V_R}{R} = \frac{V_s}{R} \]

4. **Calculate the Power Loss**:
The power loss \(P\) in the circuit can be calculated using the formula:
\[ P = I_{RMS}^2 \cdot R \]
Substituting the expression for \(I_{RMS}\):
\[ P = \left(\frac{V_s}{R}\right)^2 \cdot R = \frac{V_s^2}{R} \]

5. **Final Result**:
Therefore, the power loss at resonance in a series LCR circuit is given by:
\[ P = \frac{V_s^2}{R} \]