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 step by step, we can follow these instructions:

### Step 1: Understand the relationship between power, voltage, and current in an AC circuit.
In an AC circuit, the power consumed (P) can be expressed as:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos(\phi) \]
where:
- \( V_{\text{rms}} \) is the root mean square voltage,
- \( I_{\text{rms}} \) is the root mean square current,
- \( \phi \) is the phase difference between the voltage and current.

### Step 2: Define peak power.
The peak power occurs when \( \cos(\phi) \) is maximum, which is 1. Therefore, the peak power \( P_{\text{peak}} \) can be expressed as:
\[ P_{\text{peak}} = V_{\text{rms}} \times I_{\text{rms}} \]

### Step 3: Relate the given power to peak power.
According to the problem, the lamp consumes only 50% of the peak power. Thus, we can write:
\[ P = \frac{1}{2} P_{\text{peak}} \]
Substituting the expression for peak power:
\[ P = \frac{1}{2} (V_{\text{rms}} \times I_{\text{rms}}) \]

### Step 4: Set up the equation using the power formula.
From the power formula, we also know:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos(\phi) \]
Equating the two expressions for power:
\[ \frac{1}{2} (V_{\text{rms}} \times I_{\text{rms}}) = V_{\text{rms}} \times I_{\text{rms}} \times \cos(\phi) \]

### Step 5: Simplify the equation.
We can cancel \( V_{\text{rms}} \) and \( I_{\text{rms}} \) from both sides (assuming they are not zero):
\[ \frac{1}{2} = \cos(\phi) \]

### Step 6: Solve for the phase difference \( \phi \).
To find \( \phi \), we take the inverse cosine:
\[ \phi = \cos^{-1}\left(\frac{1}{2}\right) \]
This gives us:
\[ \phi = 60^\circ \]
or in radians:
\[ \phi = \frac{\pi}{3} \]

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

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