An are lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V(rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

Options

A. 0.08 H
B. 0.044 H
C. 0.065 H
D. 80 H

Answer

To solve the problem, we need to determine the series inductor required for an arc lamp that operates on a direct current of 10A at 80V when connected to a 220V (rms) AC supply.

### Step-by-Step Solution:

1. **Calculate the Resistance of the Lamp:**
\[ R = \frac{V}{I} = \frac{80V}{10A} = 8 \, \Omega \]

2. **Calculate the Required Impedance (Z):**
The impedance required for the lamp to function properly when connected to the AC supply can be calculated using the formula:
\[ Z = \frac{V_{rms}}{I} = \frac{220V}{10A} = 22 \, \Omega \]

3. **Determine the Inductive Reactance (X_L):**
The total impedance in an AC circuit with resistance and inductance is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Rearranging this gives:
\[ X_L = \sqrt{Z^2 - R^2} \]
Substituting the values we calculated:
\[ X_L = \sqrt{22^2 - 8^2} = \sqrt{484 - 64} = \sqrt{420} \, \Omega \]

4. **Relate Inductive Reactance to Inductance:**
The inductive reactance is also given by:
\[ X_L = \omega L = 2 \pi f L \]
where \( f = 50 \, Hz \). Thus:
\[ 2 \pi (50) L = \sqrt{420} \]

5. **Solve for Inductance (L):**
Rearranging gives:
\[ L = \frac{\sqrt{420}}{2 \pi (50)} \]
Calculating \( \sqrt{420} \):
\[ \sqrt{420} \approx 20.49 \]
Now substituting:
\[ L = \frac{20.49}{100 \pi} \approx \frac{20.49}{314.16} \approx 0.065 \, H \]

### Final Answer:
The series inductor needed for the lamp to work is approximately:
\[ L \approx 0.065 \, H \]