The output of a step-down transformer in measured to be12 V when connected to a 6 watt light bulb. The value of the peak current is

Options

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

Answer

To find the peak current in the circuit with a step-down transformer connected to a 6-watt light bulb, we can follow these steps:

### Step 1: Identify the given values
- The output voltage (RMS) of the transformer, \( V_{RMS} = 12 \, \text{V} \)
- The power of the light bulb, \( P = 6 \, \text{W} \)

### Step 2: Calculate the RMS current
Using the formula for power in terms of voltage and current:
\[ P = V_{RMS} \times I_{RMS} \]
We can rearrange this to find the RMS current:
\[ I_{RMS} = \frac{P}{V_{RMS}} = \frac{6 \, \text{W}}{12 \, \text{V}} = 0.5 \, \text{A} \]

### Step 3: Relate RMS current to peak current
The relationship between peak current (\( I_{peak} \)) and RMS current (\( I_{RMS} \)) is given by:
\[ I_{peak} = I_{RMS} \times \sqrt{2} \]

### Step 4: Substitute the value of RMS current
Now, substituting the value of \( I_{RMS} \):
\[ I_{peak} = 0.5 \, \text{A} \times \sqrt{2} \]

### Step 5: Calculate the peak current
Calculating the above expression:
\[ I_{peak} = 0.5 \times 1.414 \approx 0.707 \, \text{A} \]

Thus, the value of the peak current is approximately \( 0.707 \, \text{A} \).

### Summary of the Solution:
- The peak current \( I_{peak} \) is approximately \( 0.707 \, \text{A} \).

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