If `V_0` and `I_0` are the peak current and voltge across the resistor in a series `L-C-R` circuit, then the power dissipated in the circuit is `(power factor =costheta)`

Options

A. `(V_0I_0)/2`
B. `(V_0I_0)/sqrt2`
C. `V_0I_0costheta`
D. `(V_0I_0)/2costheta`

Explanation

To solve the problem, we need to derive the expression for the power dissipated in a series L-C-R circuit in terms of the peak current \( I_0 \), peak voltage \( V_0 \), and the power factor \( \cos \theta \).

### Step-by-Step Solution:

1. **Understanding the Power in AC Circuits**:
The power \( P \) dissipated in an AC circuit can be expressed as:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \theta \]
where \( V_{\text{rms}} \) is the root mean square voltage, \( I_{\text{rms}} \) is the root mean square current, and \( \cos \theta \) is the power factor.

2. **Relating Peak and RMS Values**:
The relationship between the peak values and the RMS values is given by:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \quad \text{and} \quad I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]
where \( V_0 \) is the peak voltage and \( I_0 \) is the peak current.

3. **Substituting RMS Values into the Power Formula**:
Substitute the expressions for \( V_{\text{rms}} \) and \( I_{\text{rms}} \) into the power formula:
\[ P = \left(\frac{V_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot \cos \theta \]

4. **Simplifying the Expression**:
Simplifying the expression gives:
\[ P = \frac{V_0 \cdot I_0}{2} \cdot \cos \theta \]

5. **Final Expression for Power**:
Therefore, the power dissipated in the circuit can be expressed as:
\[ P = \frac{V_0 \cdot I_0}{2} \cdot \cos \theta \]

### Conclusion:
The power dissipated in a series L-C-R circuit is given by:
\[ P = \frac{V_0 \cdot I_0}{2} \cdot \cos \theta \]