To solve the problem of determining the value of pure inductance required for a 120 V, 60 W lamp to operate correctly on a 220 V, 50 Hz supply, we will follow these steps:
### Step 1: Calculate the current flowing through the lamp when connected to its rated voltage.
The power \( P \) of the lamp is given as 60 W, and the voltage \( V \) is 120 V. The current \( I \) can be calculated using the formula:
\[
I = \frac{P}{V}
\]
Substituting the values:
\[
I = \frac{60 \, \text{W}}{120 \, \text{V}} = 0.5 \, \text{A}
\]
### Step 2: Calculate the resistance of the lamp.
The resistance \( R \) of the lamp can be calculated using Ohm's law:
\[
R = \frac{V}{I}
\]
Substituting the values:
\[
R = \frac{120 \, \text{V}}{0.5 \, \text{A}} = 240 \, \Omega
\]
### Step 3: Set up the equation for the circuit with the inductor.
When the lamp is connected to a 220 V supply with an inductor in series, the total voltage across the circuit is given by:
\[
V = I \cdot Z
\]
Where \( Z \) is the impedance of the circuit, which can be expressed as:
\[
Z = \sqrt{R^2 + (\omega L)^2}
\]
Where \( \omega = 2\pi f \) and \( f \) is the frequency (50 Hz).
### Step 4: Substitute the known values into the equation.
From the previous steps, we know:
- \( I = 0.5 \, \text{A} \)
- \( V = 220 \, \text{V} \)
- \( R = 240 \, \Omega \)
- \( \omega = 2\pi \times 50 \)
Substituting these into the equation:
\[
0.5 = \frac{220}{\sqrt{240^2 + (2\pi \times 50 \cdot L)^2}}
\]
### Step 5: Cross-multiply and simplify the equation.
Cross-multiplying gives:
\[
0.5 \cdot \sqrt{240^2 + (2\pi \times 50 \cdot L)^2} = 220
\]
Squaring both sides:
\[
0.25 (240^2 + (2\pi \times 50 \cdot L)^2) = 220^2
\]
### Step 6: Solve for \( L \).
Expanding and rearranging gives:
\[
240^2 + (2\pi \times 50 \cdot L)^2 = \frac{220^2}{0.25}
\]
Calculating \( 220^2 \):
\[
220^2 = 48400
\]
Thus,
\[
240^2 + (2\pi \times 50 \cdot L)^2 = 193600
\]
Subtracting \( 240^2 \):
\[
(2\pi \times 50 \cdot L)^2 = 193600 - 57600
\]
Calculating \( 193600 - 57600 \):
\[
(2\pi \times 50 \cdot L)^2 = 136000
\]
Taking the square root:
\[
2\pi \times 50 \cdot L = \sqrt{136000}
\]
Calculating \( \sqrt{136000} \):
\[
\sqrt{136000} \approx 369.83
\]
Now, solving for \( L \):
\[
L = \frac{369.83}{2\pi \times 50}
\]
Calculating \( 2\pi \times 50 \):
\[
2\pi \times 50 \approx 314.16
\]
Finally:
\[
L \approx \frac{369.83}{314.16} \approx 1.175 \, \text{H}
\]
### Final Answer:
The required inductance \( L \) is approximately **1.175 H**.
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