A 110 V, 60 W lamp is run from a 220 V AC mains using a capacitor in series with the lamp, instead of a resistor then the voltage across the capacitor is about:

Options

A. 110 V
B. 190 V
C. 220 V
D. 311 V

Answer

To solve the problem, we need to find the voltage across the capacitor when a 110 V, 60 W lamp is connected in series with the capacitor to a 220 V AC mains supply.

### Step-by-Step Solution:

1. **Identify the given values:**
- Supply voltage (\( V_S \)) = 220 V
- Lamp voltage (\( V_R \)) = 110 V
- Lamp power (\( P \)) = 60 W

2. **Calculate the current through the lamp:**
- The power of the lamp can be expressed as:
\[ P = V_R \cdot I \]
- Rearranging gives us:
\[ I = \frac{P}{V_R} = \frac{60 \, \text{W}}{110 \, \text{V}} \approx 0.545 \, \text{A} \]

3. **Use the voltage relationship in the series circuit:**
- In a series circuit, the total voltage is the sum of the voltages across the lamp and the capacitor:
\[ V_S = V_R + V_C \]
- Rearranging gives:
\[ V_C = V_S - V_R \]

4. **Substituting the known values:**
- Substitute \( V_S = 220 \, \text{V} \) and \( V_R = 110 \, \text{V} \):
\[ V_C = 220 \, \text{V} - 110 \, \text{V} = 110 \, \text{V} \]

5. **Using the Pythagorean theorem for AC circuits:**
- In an AC circuit with a capacitor and a resistor, the voltages can be related using:
\[ V_S^2 = V_R^2 + V_C^2 \]
- Rearranging to find \( V_C \):
\[ V_C^2 = V_S^2 - V_R^2 \]

6. **Substituting the values into the equation:**
- Substitute \( V_S = 220 \, \text{V} \) and \( V_R = 110 \, \text{V} \):
\[ V_C^2 = (220)^2 - (110)^2 \]
\[ V_C^2 = 48400 - 12100 = 36300 \]
- Taking the square root:
\[ V_C = \sqrt{36300} \approx 190.52 \, \text{V} \]

7. **Final result:**
- The voltage across the capacitor is approximately:
\[ V_C \approx 190 \, \text{V} \]

### Conclusion:
The voltage across the capacitor when the 110 V, 60 W lamp is connected in series with it to a 220 V AC supply is about **190 V**.