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

Answer

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

### Step 1: Understand the relationship between peak current and average current
The peak current (Ip) is the maximum value of the sinusoidal current. The average (or effective) current (I_avg) for a sinusoidal waveform is given by:
\[ I_{avg} = \frac{I_p}{\sqrt{2}} \]

### Step 2: Write the formula for power dissipated in a resistor
The power (P) dissipated in a resistor due to current flowing through it can be expressed as:
\[ P = I^2 R \]
where I is the current flowing through the resistor.

### Step 3: Substitute the average current into the power formula
Since we are dealing with AC current, we will use the average current calculated in Step 1:
\[ P = (I_{avg})^2 R = \left(\frac{I_p}{\sqrt{2}}\right)^2 R \]

### Step 4: Simplify the expression
Now, we simplify the expression:
\[ P = \left(\frac{I_p^2}{2}\right) R \]

### Step 5: Final expression for power dissipated
Thus, the power dissipated in the resistor can be written as:
\[ P = \frac{I_p^2 R}{2} \]

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

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