A light bulb has the rating 200 W 220 V.
Find the peak value of current.

Answer

To find the peak value of current for a light bulb rated at 200 W and 220 V, we can follow these steps:

### Step 1: Understand the relationship between power, voltage, and current
The power (P) consumed by an electrical device can be expressed using the formula:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \]
where:
- \( P \) is the power in watts (W),
- \( V_{\text{rms}} \) is the root mean square voltage in volts (V),
- \( I_{\text{rms}} \) is the root mean square current in amperes (A).

### Step 2: Substitute the known values into the formula
Given:
- Power \( P = 200 \, \text{W} \)
- Voltage \( V_{\text{rms}} = 220 \, \text{V} \)

We can rearrange the formula to solve for \( I_{\text{rms}} \):
\[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} \]

Substituting the known values:
\[ I_{\text{rms}} = \frac{200 \, \text{W}}{220 \, \text{V}} \]

### Step 3: Calculate \( I_{\text{rms}} \)
Now, we perform the calculation:
\[ I_{\text{rms}} = \frac{200}{220} = \frac{10}{11} \, \text{A} \]

### Step 4: Find the peak current \( I_{\text{peak}} \)
The peak current can be calculated from the root mean square current using the relationship:
\[ I_{\text{peak}} = I_{\text{rms}} \times \sqrt{2} \]

Substituting the value of \( I_{\text{rms}} \):
\[ I_{\text{peak}} = \frac{10}{11} \times \sqrt{2} \]

### Step 5: Simplify the expression
Calculating \( I_{\text{peak}} \):
\[ I_{\text{peak}} = \frac{10 \sqrt{2}}{11} \, \text{A} \]

### Final Answer
The peak value of current is:
\[ I_{\text{peak}} = \frac{10 \sqrt{2}}{11} \, \text{A} \]

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