To solve the problem step by step, we will analyze the given information and apply the relevant formulas.
### Step 1: Understand the given information
- When a DC voltage of 100 V is applied, the current through the coil is 1 A.
- When an AC voltage of 100 V (50 Hz) is applied, the current is 0.5 A.
### Step 2: Calculate the resistance (R) using DC
For DC, Ohm's law applies:
\[ R = \frac{V}{I} \]
Where:
- \( V = 100 \, \text{V} \)
- \( I = 1 \, \text{A} \)
Substituting the values:
\[ R = \frac{100 \, \text{V}}{1 \, \text{A}} = 100 \, \Omega \]
### Step 3: Analyze the AC circuit
For AC, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where:
- \( X_L = \omega L \)
- \( \omega = 2 \pi f \) (angular frequency)
- \( f = 50 \, \text{Hz} \)
Calculating \( \omega \):
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]
### Step 4: Use the AC current to find impedance (Z)
From the AC circuit:
\[ I = \frac{V_{rms}}{Z} \]
Where:
- \( V_{rms} = 100 \, \text{V} \)
- \( I = 0.5 \, \text{A} \)
Rearranging gives:
\[ Z = \frac{V_{rms}}{I} = \frac{100 \, \text{V}}{0.5 \, \text{A}} = 200 \, \Omega \]
### Step 5: Set up the equation for impedance
Now, we can set up the equation using the values of \( R \) and \( Z \):
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the known values:
\[ 200 = \sqrt{100^2 + (100 \pi L)^2} \]
### Step 6: Square both sides to eliminate the square root
Squaring both sides:
\[ 200^2 = 100^2 + (100 \pi L)^2 \]
\[ 40000 = 10000 + (100 \pi L)^2 \]
### Step 7: Rearrange to solve for \( L \)
Subtract \( 10000 \) from both sides:
\[ 30000 = (100 \pi L)^2 \]
Now, take the square root:
\[ \sqrt{30000} = 100 \pi L \]
\[ L = \frac{\sqrt{30000}}{100 \pi} \]
### Step 8: Simplify \( L \)
Calculating \( \sqrt{30000} \):
\[ \sqrt{30000} = 100 \sqrt{3} \times 10 \]
Thus:
\[ L = \frac{100 \sqrt{3}}{100 \pi} = \frac{\sqrt{3}}{\pi} \]
### Step 9: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ L \approx \frac{1.732}{3.14} \approx 0.55 \, \text{H} \]
### Final Answer
The inductance \( L \) of the coil is approximately \( 0.55 \, \text{H} \).
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