An a.c. of frequency f is flowing in a circuit containing only an ideal choke coil of inductance L. If `V_(0) and l_(0)` represent peak values of the voltage and the current respectively. The average power given by the source to the choke coil is equal to

Options

A. `(1)/(2)l_(0)V_(0)`
B. `(1)/(2)l_(0)^(2) (2pifL)`
C. Zero
D. `(1)/(2)V_(0) (2pifL)`

Answer

To solve the problem, we need to determine the average power delivered to an ideal choke coil (inductor) in an AC circuit. The key points to consider are the relationships between voltage, current, and power in an inductive circuit.

### Step-by-Step Solution:

1. **Understand the Circuit**:
The circuit consists of an ideal choke coil (inductor) with inductance \( L \) and an alternating current (AC) of frequency \( f \) flowing through it.

2. **Identify Peak Values**:
Let \( V_0 \) be the peak voltage across the inductor and \( I_0 \) be the peak current through the inductor.

3. **Convert Peak Values to RMS Values**:
The root mean square (RMS) values of voltage and current can be calculated from their peak values:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \quad \text{and} \quad I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]

4. **Determine the Phase Angle**:
In an ideal inductor, the current lags the voltage by \( 90^\circ \) (or \( \frac{\pi}{2} \) radians). Therefore, the power factor \( \cos \phi \) is:
\[ \cos \phi = \cos \left(\frac{\pi}{2}\right) = 0 \]

5. **Calculate Average Power**:
The average power \( P \) delivered to the inductor can be calculated using the formula:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi \]
Substituting the values we found:
\[ P = \left(\frac{V_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot 0 \]
Simplifying this gives:
\[ P = 0 \]

### Conclusion:
The average power given by the source to the choke coil is equal to \( 0 \).