A sinusoidal ac current flows through a resistor of resistance R . If the peak current is `I_(p)` , then the power dissipated is

Options

A. `I_(1)^(p)R cos theta`
B. `1/2I_(p)^(2)R`
C. `4/pi I_(p)^(2)R`
D. `1/pi^(2)I_(p)^(2)R`

Explanation

To solve the problem, we need to find the power dissipated in a resistor when a sinusoidal AC current flows through it, given the peak current \( I_p \).

### Step-by-Step Solution:

1. **Understand the relationship between peak current and RMS current**:
The RMS (Root Mean Square) value of the current is related to the peak current \( I_p \) by the formula:
\[ I_{rms} = \frac{I_p}{\sqrt{2}} \]

2. **Calculate the average power dissipated in the resistor**:
The power dissipated in a resistor when an AC current flows through it can be calculated using the formula:
\[ P = I_{rms}^2 \cdot R \]
Substituting the expression for \( I_{rms} \):
\[ P = \left(\frac{I_p}{\sqrt{2}}\right)^2 \cdot R \]

3. **Simplify the expression**:
Now, simplify the equation:
\[ P = \frac{I_p^2}{2} \cdot R \]

4. **Final expression for power**:
Thus, the power dissipated in the resistor is given by:
\[ P = \frac{1}{2} I_p^2 R \]

### Conclusion:
The power dissipated in the resistor when a sinusoidal AC current flows through it with a peak current \( I_p \) is:
\[ P = \frac{1}{2} I_p^2 R \]