The output of a step-down transformer is measured to be `24 V` when connected to a 12 watt light bulb. The value of the peak current is

Options

A. `1/sqrt(2)A`
B. `sqrt(2)A`
C. 2A
D. `2sqrt(2)A`

Explanation

To find the peak current in the output of a step-down transformer when connected to a 12-watt light bulb, we can follow these steps:

### Step 1: Identify Given Values
- Output Voltage (\( V_s \)) = 24 V
- Output Power (\( P_s \)) = 12 W

### Step 2: Calculate the RMS Current
Using the formula for power in terms of voltage and current:
\[ P_s = V_s \times I_s \]
We can rearrange this to find the RMS current (\( I_s \)):
\[ I_s = \frac{P_s}{V_s} \]
Substituting the given values:
\[ I_s = \frac{12 \, \text{W}}{24 \, \text{V}} = \frac{1}{2} \, \text{A} \]

### Step 3: Convert RMS Current to Peak Current
The relationship between RMS current (\( I_s \)) and peak current (\( I_0 \)) is given by:
\[ I_0 = I_s \times \sqrt{2} \]
Substituting the value of \( I_s \):
\[ I_0 = \left(\frac{1}{2} \, \text{A}\right) \times \sqrt{2} \]
Calculating this gives:
\[ I_0 = \frac{\sqrt{2}}{2} \, \text{A} \]

### Step 4: Final Result
Thus, the peak current (\( I_0 \)) is:
\[ I_0 = \frac{1}{\sqrt{2}} \, \text{A} \]

### Summary
The value of the peak current is \( \frac{1}{\sqrt{2}} \, \text{A} \).

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