If the frequency of the source `e.m.f.` in an `AC` circuit in `n`,the power varies with a frequency:

Options

A. `n`
B. `2 n`
C. `n//2`
D. zero

Explanation

To solve the problem of how power varies with frequency in an AC circuit where the frequency of the source e.m.f. is \( n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Voltage and Power**:
The power \( P \) in an AC circuit can be expressed in terms of voltage \( V \) and resistance \( R \) using the formula:
\[ P = \frac{V^2}{R} \]
This indicates that power is proportional to the square of the voltage.

2. **Identify the Frequency of Voltage**:
Given that the frequency of the source e.m.f. is \( n \), we can denote the voltage as a function of time, which can be expressed in a sinusoidal form:
\[ V(t) = V_0 \sin(2\pi nt) \]
where \( V_0 \) is the maximum voltage.

3. **Calculate the Instantaneous Power**:
The instantaneous power can be calculated by substituting the voltage function into the power formula:
\[ P(t) = \frac{(V_0 \sin(2\pi nt))^2}{R} \]
Simplifying this gives:
\[ P(t) = \frac{V_0^2 \sin^2(2\pi nt)}{R} \]

4. **Determine the Frequency of Power**:
The term \( \sin^2(2\pi nt) \) can be rewritten using the double angle identity:
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]
Applying this identity:
\[ P(t) = \frac{V_0^2}{2R} (1 - \cos(4\pi nt)) \]
From this expression, we can see that the frequency of the power \( P(t) \) is \( 2n \) because the cosine term \( \cos(4\pi nt) \) oscillates with a frequency of \( 2n \).

5. **Conclusion**:
Therefore, if the frequency of the source e.m.f. in the AC circuit is \( n \), the frequency at which the power varies is \( 2n \).

### Final Answer:
The power varies with a frequency of \( 2n \).

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