When 12 w bulb is connected to a step down transformer, the output is 24 V. Peak current is given by:

Options

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

Explanation

To find the peak current when a 12 W bulb is connected to a step-down transformer with an output voltage of 24 V, we can follow these steps:

### Step 1: Understand the relationship between power, voltage, and current.
The power in the secondary circuit of the transformer can be expressed as:
\[ P_s = V_s \times I_s \]
where \( P_s \) is the power in watts, \( V_s \) is the secondary voltage, and \( I_s \) is the secondary current.

### Step 2: Rearrange the formula to find the secondary current.
From the equation above, we can rearrange it to find the secondary current \( I_s \):
\[ I_s = \frac{P_s}{V_s} \]

### Step 3: Substitute the known values.
We know that \( P_s = 12 \, \text{W} \) and \( V_s = 24 \, \text{V} \). Substituting these values into the equation gives:
\[ I_s = \frac{12 \, \text{W}}{24 \, \text{V}} = 0.5 \, \text{A} \]

### Step 4: Convert the secondary current to peak current.
The relationship between the RMS current \( I_s \) and the peak current \( I_m \) is given by:
\[ I_m = I_s \times \sqrt{2} \]
Substituting \( I_s = 0.5 \, \text{A} \) into this equation gives:
\[ I_m = 0.5 \, \text{A} \times \sqrt{2} \]

### Step 5: Calculate the peak current.
Now we can calculate the peak current:
\[ I_m = 0.5 \sqrt{2} \, \text{A} \approx 0.707 \, \text{A} \]

### Final Answer:
The peak current \( I_m \) is approximately \( 0.707 \, \text{A} \).

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