The self-inductance of a choke coil is `10mH`. When it is connected with a `10 V DC` source, then the loss of power is `20 "watt"`. When it is connected with `10 "volt" AC` source loss of power is `10"watt"`. The frequency of `AC` source will be

Options

A. `50 Hz`
B. `60 Hz`
C. `80 Hz`
D. `100 Hz`

Explanation

To solve the problem step-by-step, we will follow the reasoning provided in the video transcript.

### Step 1: Determine the Resistance (R) using the DC source
When the choke coil is connected to a 10 V DC source, the power loss is given as 20 W. The power loss in a resistive circuit can be calculated using the formula:

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

Where:
- \( P \) is the power (20 W)
- \( V \) is the voltage (10 V)

Rearranging the formula to find \( R \):

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

Substituting the values:

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

### Step 2: Determine the Impedance (Z) using the AC source
When the choke coil is connected to a 10 V AC source, the power loss is given as 10 W. The power in an AC circuit can be expressed as:

\[ P = V_{rms} \cdot I_{rms} \cdot \text{power factor} \]

The power factor can be expressed as:

\[ \text{power factor} = \frac{R}{Z} \]

Thus, we can rewrite the power formula as:

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

Substituting the known values:

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

This simplifies to:

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

\[ 10Z^2 = 500 \]

\[ Z^2 = 50 \implies Z = \sqrt{50} = 5\sqrt{2} \, \Omega \]

### Step 3: Relate Impedance (Z) to Resistance (R) and Reactance (X_L)
Using the impedance formula for an RL circuit:

\[ Z^2 = R^2 + X_L^2 \]

Substituting the known values:

\[ 50 = 5^2 + X_L^2 \]

\[ 50 = 25 + X_L^2 \]

\[ X_L^2 = 25 \implies X_L = 5 \, \Omega \]

### Step 4: Calculate the Frequency (f)
The inductive reactance \( X_L \) is given by:

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

Where:
- \( L = 10 \, mH = 10 \times 10^{-3} \, H \)

Substituting \( X_L \) and rearranging for \( f \):

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

\[ f = \frac{5}{2\pi \cdot 10 \times 10^{-3}} = \frac{5}{0.0628} \approx 79.577 \, Hz \]

Rounding to the nearest whole number gives:

\[ f \approx 80 \, Hz \]

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

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