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 when a step-down transformer outputs 24 V to a 12-watt light bulb, we can follow these steps:

### Step 1: Identify the given values
- Output voltage (Vs) = 24 V
- Power (Ps) = 12 W

### Step 2: Calculate the RMS current (Is)
Using the formula for power:
\[ P_s = V_s \times I_s \]
We can rearrange this to find the RMS current (Is):
\[ I_s = \frac{P_s}{V_s} \]
Substituting the known values:
\[ I_s = \frac{12 \text{ W}}{24 \text{ V}} \]
\[ I_s = 0.5 \text{ A} \]

### Step 3: Convert RMS current to peak current (I0)
The relationship between RMS current and peak current is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
Rearranging this gives:
\[ I_0 = I_{rms} \times \sqrt{2} \]
Substituting the RMS current we found:
\[ I_0 = 0.5 \text{ A} \times \sqrt{2} \]
Calculating this gives:
\[ I_0 = 0.5 \times 1.414 \approx 0.707 \text{ A} \]

### Final Answer
The peak current (I0) is approximately **0.707 A**.

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