To solve the problem step by step, we need to determine the series inductor required for the arc lamp to operate correctly when connected to an AC supply.
### Step 1: Calculate the Resistance of the Lamp
Using Ohm's Law, we can find the resistance (R) of the lamp. The formula is:
\[
R = \frac{V}{I}
\]
Where:
- \( V = 80 \, \text{V} \) (the voltage required by the lamp)
- \( I = 10 \, \text{A} \) (the current required by the lamp)
Substituting the values:
\[
R = \frac{80 \, \text{V}}{10 \, \text{A}} = 8 \, \Omega
\]
### Step 2: Relate the AC Voltage to the Current and Impedance
When the lamp is connected to an AC supply, the relationship between voltage (V), current (I), and impedance (Z) is given by:
\[
V = I \cdot Z
\]
Where \( Z \) is the total impedance of the circuit. In this case, we have:
\[
Z = \sqrt{R^2 + (ωL)^2}
\]
Where:
- \( ω = 2\pi f \) (angular frequency)
- \( L \) is the inductance
### Step 3: Calculate the Angular Frequency
Given that the frequency \( f = 50 \, \text{Hz} \):
\[
ω = 2\pi f = 2\pi \times 50 \approx 314.16 \, \text{rad/s}
\]
### Step 4: Substitute the Values into the Impedance Equation
We know that the AC supply voltage \( V = 220 \, \text{V} \) (rms) and the current \( I = 10 \, \text{A} \):
\[
220 = 10 \cdot Z
\]
Thus, we can find \( Z \):
\[
Z = \frac{220}{10} = 22 \, \Omega
\]
### Step 5: Set Up the Impedance Equation
Now we can set up the equation using the impedance formula:
\[
22 = \sqrt{8^2 + (ωL)^2}
\]
Squaring both sides gives:
\[
22^2 = 8^2 + (ωL)^2
\]
Calculating \( 22^2 \) and \( 8^2 \):
\[
484 = 64 + (ωL)^2
\]
### Step 6: Solve for \( (ωL)^2 \)
Rearranging the equation:
\[
(ωL)^2 = 484 - 64 = 420
\]
### Step 7: Solve for \( L \)
Taking the square root:
\[
ωL = \sqrt{420}
\]
Now substituting \( ω = 314.16 \):
\[
L = \frac{\sqrt{420}}{ω} = \frac{\sqrt{420}}{314.16}
\]
Calculating \( \sqrt{420} \approx 20.49 \):
\[
L \approx \frac{20.49}{314.16} \approx 0.065 \, \text{H}
\]
### Final Answer
The series inductor needed for the arc lamp to work is approximately:
\[
L \approx 0.065 \, \text{H}
\]
