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. 80 Hz
B. 100 Hz
C. 120 Hz
D. 220 Hz

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Resistance of the Choke Coil
Given:
- Voltage (V) = 10 V (DC)
- Power (P) = 20 W

Using the formula for power in a DC circuit:
\[ P = \frac{V^2}{R} \]
We can rearrange this to find the resistance (R):
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{10^2}{20} = \frac{100}{20} = 5 \, \Omega \]

### Step 2: Analyze the AC Circuit
For the AC circuit, we know:
- Voltage (V_rms) = 10 V (AC)
- Power (P') = 10 W

The power in an AC circuit can be expressed as:
\[ P' = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
Where:
- \(\cos \phi = \frac{R}{Z}\)
- \(I_{rms} = \frac{V_{rms}}{Z}\)

Substituting these into the power equation:
\[ P' = V_{rms} \cdot \left(\frac{V_{rms}}{Z}\right) \cdot \left(\frac{R}{Z}\right) \]
This simplifies to:
\[ P' = \frac{V_{rms}^2 \cdot R}{Z^2} \]
Rearranging gives:
\[ Z^2 = \frac{V_{rms}^2 \cdot R}{P'} \]

### Step 3: Substitute Known Values
Substituting the known values:
\[ Z^2 = \frac{10^2 \cdot 5}{10} = \frac{100 \cdot 5}{10} = 50 \]

### Step 4: Find the Inductive Reactance
We know:
\[ Z^2 = R^2 + X_L^2 \]
Where \(X_L\) is the inductive reactance. We already calculated \(R = 5 \, \Omega\):
\[ 50 = 5^2 + X_L^2 \]
\[ 50 = 25 + X_L^2 \]
\[ X_L^2 = 50 - 25 = 25 \]
\[ X_L = 5 \, \Omega \]

### Step 5: Relate Inductive Reactance to Frequency
The inductive reactance is given by:
\[ X_L = \omega L = 2 \pi f L \]
Where \(L = 10 \, mH = 10 \times 10^{-3} H\). Substituting:
\[ 5 = 2 \pi f (10 \times 10^{-3}) \]
Solving for frequency \(f\):
\[ f = \frac{5}{2 \pi \cdot 10 \times 10^{-3}} = \frac{5}{0.0628} \approx 79.58 \, Hz \]

### Final Answer
The frequency of the AC source is approximately \(80 \, Hz\).

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