A sinusoidal alternating current of peak value `(I_0)` passes through a heater of resistance R. What is the mean power output of the heater?

Options

A. (A) `I_(0)^(2)R`
B. (B) `(I_(0)^(2)R)/(2)`
C. (C) `2I_(0)^(2)R`
D. (D) `sqrt(2)I_(0)^(2)R`

Explanation

To find the mean power output of a heater through which a sinusoidal alternating current of peak value \( I_0 \) passes, we can follow these steps:

### Step 1: Understand the relationship between peak current and RMS current
The RMS (Root Mean Square) value of a sinusoidal current is related to its peak value by the formula:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]

### Step 2: Write the expression for power in terms of RMS values
The average power \( P \) delivered to a resistive load can be expressed as:
\[ P = I_{\text{rms}}^2 \cdot R \]
where \( R \) is the resistance of the heater.

### Step 3: Substitute the RMS current into the power formula
Substituting the expression for \( I_{\text{rms}} \):
\[ P = \left(\frac{I_0}{\sqrt{2}}\right)^2 \cdot R \]

### Step 4: Simplify the expression
Calculating \( \left(\frac{I_0}{\sqrt{2}}\right)^2 \):
\[ P = \frac{I_0^2}{2} \cdot R \]

### Step 5: Final expression for mean power output
Thus, the mean power output of the heater is given by:
\[ P = \frac{I_0^2 R}{2} \]

### Summary
The mean power output of the heater when a sinusoidal alternating current of peak value \( I_0 \) passes through it is:
\[ P = \frac{I_0^2 R}{2} \]

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