The self inductance of a choke coil is `10mH`. When it is connected to a `10V DC` source, the loss of power is `20 W`. When it is connected to a `10V (AC)` source, the loss of power is `10 W`. The frequency of `(AC)` source (in Hz` ) will be (Answer to be given in nearest integer)

Answer

To solve the problem step by step, we will analyze the information given and apply the relevant formulas.

### Step 1: Determine the Resistance from the DC Source
When the choke coil is connected to a 10V DC source, the power loss is given as 20W. We can use the formula for power in a resistive circuit:

\[ P = \frac{V^2}{R} \]

Where:
- \( P \) = Power (20 W)
- \( V \) = Voltage (10 V)
- \( R \) = Resistance

Rearranging the formula to find \( R \):

\[ R = \frac{V^2}{P} \]

Substituting the known values:

\[ R = \frac{10^2}{20} = \frac{100}{20} = 5 \, \Omega \]

### Step 2: Analyze the AC Circuit
When the choke coil is connected to a 10V AC source, the power loss is 10W. The power in an AC circuit with inductance is given by:

\[ P = \frac{V_{rms}^2}{Z} \cos \phi \]

Where:
- \( Z \) = Impedance
- \( \cos \phi = \frac{R}{Z} \)

### Step 3: Express Power in Terms of Impedance
Using the power formula for AC, we can express it as:

\[ P = \frac{V_{rms}^2}{Z} \cdot \frac{R}{Z} \]

This simplifies to:

\[ P = \frac{V_{rms}^2 R}{Z^2} \]

Substituting the known values:

\[ 10 = \frac{10^2 \cdot 5}{Z^2} \]

This simplifies to:

\[ 10 = \frac{100 \cdot 5}{Z^2} \]

\[ 10Z^2 = 500 \]

\[ Z^2 = 50 \]

### Step 4: Calculate Impedance
Taking the square root:

\[ Z = \sqrt{50} = 5\sqrt{2} \, \Omega \approx 7.07 \, \Omega \]

### Step 5: Relate Impedance to Inductive Reactance
The impedance \( Z \) in an inductive circuit is given by:

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

Where \( X_L = 2\pi f L \) is the inductive reactance. Substituting the values we have:

\[ 7.07 = \sqrt{5^2 + (2\pi f \cdot 10 \times 10^{-3})^2} \]

Squaring both sides:

\[ 50 = 25 + (2\pi f \cdot 10 \times 10^{-3})^2 \]

### Step 6: Solve for Frequency
Rearranging gives:

\[ 25 = (2\pi f \cdot 10 \times 10^{-3})^2 \]

Taking the square root:

\[ 5 = 2\pi f \cdot 10 \times 10^{-3} \]

Solving for \( f \):

\[ f = \frac{5}{2\pi \cdot 10 \times 10^{-3}} \]

Calculating:

\[ f = \frac{5}{0.06283} \approx 79.58 \, \text{Hz} \]

Rounding to the nearest integer:

\[ f \approx 80 \, \text{Hz} \]

### Final Answer:
The frequency of the AC source is approximately **80 Hz**.

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