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. (a)`I_(p)^(2)R cos theta `
B. (b)`1/2 I_(p)^(2)R`
C. (c)`4/(pi)I_(p)^(2)R`
D. (d)`1/piI_(p)^(2)R`

Answer

To solve the problem of finding the power dissipated in a resistor when a sinusoidal AC current flows through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have a sinusoidal AC current flowing through a resistor of resistance \( R \).
- The peak current is given as \( I_p \).

2. **Determine the RMS Current**:
- The root mean square (RMS) value of a sinusoidal current is related to its peak value by the formula:
\[ I_{\text{rms}} = \frac{I_p}{\sqrt{2}} \]
- This formula indicates that the RMS current is the peak current divided by the square root of 2.

3. **Use the Power Formula**:
- The power dissipated in a resistor due to an electric current is given by the formula:
\[ P = I^2 R \]
- Since we are dealing with an AC current, we will use the RMS value of the current in this formula.

4. **Substitute the RMS Current into the Power Formula**:
- Replace \( I \) in the power formula with \( I_{\text{rms}} \):
\[ P = (I_{\text{rms}})^2 R \]
- Substituting the expression for \( I_{\text{rms}} \):
\[ P = \left(\frac{I_p}{\sqrt{2}}\right)^2 R \]

5. **Simplify the Expression**:
- Simplifying the equation gives:
\[ P = \frac{I_p^2}{2} R \]

6. **Final Result**:
- The power dissipated in the resistor is:
\[ P = \frac{1}{2} I_p^2 R \]

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