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

Explanation

To find the value of the series inductor needed for the arc lamp to function properly when connected to an AC supply, we can follow these steps:

### Step 1: Calculate the Resistance of the Lamp
Using Ohm's Law, we can calculate the resistance (R) of the lamp. The formula is:

\[ R = \frac{V}{I} \]

Where:
- \( V = 80 \, \text{V} \) (voltage required for the lamp)
- \( I = 10 \, \text{A} \) (current required for the lamp)

Substituting the values:

\[ R = \frac{80 \, \text{V}}{10 \, \text{A}} = 8 \, \Omega \]

### Step 2: Determine the Total Voltage in the Circuit
The total voltage (V) from the AC supply is given as:

\[ V = 220 \, \text{V (rms)} \]

### Step 3: Use the Impedance Formula
In an AC circuit with a resistor and inductor in series, the relationship between the total voltage (V), current (I), resistance (R), and inductive reactance (X_L) is given by:

\[ V = I \cdot Z \]

Where \( Z \) (impedance) is given by:

\[ Z = \sqrt{R^2 + X_L^2} \]

And the inductive reactance \( X_L \) is defined as:

\[ X_L = \omega L = 2 \pi f L \]

Where:
- \( f = 50 \, \text{Hz} \) (frequency)

### Step 4: Substitute Known Values
Substituting the known values into the impedance formula:

\[ 220 = 10 \cdot \sqrt{8^2 + (2 \pi \cdot 50 \cdot L)^2} \]

### Step 5: Solve for Impedance
First, calculate the left side:

\[ 220 = 10 \cdot \sqrt{64 + (2 \pi \cdot 50 \cdot L)^2} \]

Dividing both sides by 10:

\[ 22 = \sqrt{64 + (2 \pi \cdot 50 \cdot L)^2} \]

### Step 6: Square Both Sides
Squaring both sides to eliminate the square root:

\[ 22^2 = 64 + (2 \pi \cdot 50 \cdot L)^2 \]

Calculating \( 22^2 \):

\[ 484 = 64 + (2 \pi \cdot 50 \cdot L)^2 \]

### Step 7: Isolate the Inductive Reactance Term
Subtract 64 from both sides:

\[ 484 - 64 = (2 \pi \cdot 50 \cdot L)^2 \]

\[ 420 = (2 \pi \cdot 50 \cdot L)^2 \]

### Step 8: Take the Square Root
Taking the square root of both sides:

\[ \sqrt{420} = 2 \pi \cdot 50 \cdot L \]

### Step 9: Solve for L
Now, solve for \( L \):

\[ L = \frac{\sqrt{420}}{2 \pi \cdot 50} \]

Calculating \( \sqrt{420} \):

\[ \sqrt{420} \approx 20.49 \]

Now substituting back:

\[ L = \frac{20.49}{2 \cdot 3.14 \cdot 50} \]

Calculating the denominator:

\[ L = \frac{20.49}{314} \approx 0.065 \, \text{H} \]

### Conclusion
The value of the series inductor needed for the arc lamp to function properly when connected to the AC supply is approximately:

\[ L \approx 0.065 \, \text{H} \]