A lamp consumes only `50%` of peak power in an `a.c.` circuit. What is the phase difference between the applied voltage and the circuit current

Options

A. `(pi)/6`
B. `(pi)/3`
C. `(pi)/4`
D. `(pi)/2`

Explanation

To solve the problem, we need to determine the phase difference between the applied voltage and the circuit current when a lamp consumes only 50% of peak power in an AC circuit.

### Step-by-Step Solution:

1. **Understanding Power Consumption**:
The problem states that the lamp consumes 50% of the peak power. In AC circuits, the average power (P) consumed can be expressed in terms of the peak power (P_peak) and the power factor (cos φ):
\[ P = P_{\text{peak}} \cdot \cos \phi \]

2. **Setting Up the Equation**:
According to the problem, the power consumed is 50% of the peak power:
\[ P = \frac{1}{2} P_{\text{peak}} \]
Substituting this into the power equation gives:
\[ \frac{1}{2} P_{\text{peak}} = P_{\text{peak}} \cdot \cos \phi \]

3. **Simplifying the Equation**:
We can cancel \(P_{\text{peak}}\) from both sides (assuming \(P_{\text{peak}} \neq 0\)):
\[ \frac{1}{2} = \cos \phi \]

4. **Finding the Phase Difference**:
To find the phase difference (φ), we take the inverse cosine:
\[ \phi = \cos^{-1}\left(\frac{1}{2}\right) \]

5. **Calculating the Angle**:
The value of \(\cos^{-1}\left(\frac{1}{2}\right)\) corresponds to:
\[ \phi = \frac{\pi}{3} \text{ radians} \quad \text{or} \quad 60^\circ \]

### Conclusion:
The phase difference between the applied voltage and the circuit current is \(60^\circ\) or \(\frac{\pi}{3}\) radians.