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, we can follow these steps:

### Step 1: Identify the given values
- Output voltage (\( V_s \)) = 24 V
- Power (\( P \)) = 12 W

### Step 2: Calculate the RMS current (\( I_{rms} \))
Using the formula for power in terms of voltage and current:
\[ P = V_s \times I_{rms} \]
We can rearrange this to find the RMS current:
\[ I_{rms} = \frac{P}{V_s} \]
Substituting the values we have:
\[ I_{rms} = \frac{12 \, \text{W}}{24 \, \text{V}} = 0.5 \, \text{A} \]

### Step 3: Calculate the peak current (\( I_{peak} \))
The relationship between the RMS current and the peak current is given by:
\[ I_{peak} = I_{rms} \times \sqrt{2} \]
Substituting the value of \( I_{rms} \):
\[ I_{peak} = 0.5 \, \text{A} \times \sqrt{2} \approx 0.5 \, \text{A} \times 1.414 \approx 0.707 \, \text{A} \]

### Step 4: Final answer
Thus, the peak current is approximately:
\[ I_{peak} \approx 0.707 \, \text{A} \quad \text{or} \quad \frac{1}{\sqrt{2}} \, \text{A} \]