In a series resonant circuit, having L,C and R as its element, the resonance current is i. The power dissipated in circuit at resonance is :

Options

A. Zero
B. `i^2R`
C. `i^2omegaL`
D. `(i^2R)/(omegaL-1/(omegaC))`

Answer

To solve the problem of finding the power dissipated in a series resonant circuit with elements L (inductor), C (capacitor), and R (resistor) at resonance, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resonance in an LCR Circuit**:
At resonance, the inductive reactance (Xl) equals the capacitive reactance (Xc). This means that the effects of the inductor and capacitor cancel each other out, leading to a purely resistive circuit.

2. **Impedance at Resonance**:
The impedance (Z) of the circuit at resonance can be expressed as:
\[ Z = \sqrt{(Xl - Xc)^2 + R^2} \]
Since \(Xl = Xc\) at resonance, the equation simplifies to:
\[ Z = R \]

3. **Power Dissipation Formula**:
The power (P) dissipated in a resistive circuit is given by the formula:
\[ P = i^2 \cdot R \]
where \(i\) is the current flowing through the circuit.

4. **Substituting Impedance**:
Since we have established that at resonance, \(Z = R\), we can substitute this into the power formula:
\[ P = i^2 \cdot Z \]
Thus, at resonance:
\[ P = i^2 \cdot R \]

5. **Conclusion**:
Therefore, the power dissipated in the circuit at resonance is:
\[ P = i^2 \cdot R \]

### Final Answer:
The power dissipated in the circuit at resonance is \(P = i^2 R\).

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