The self inductance of a choke coil is 10mH. When it is connected witrh a 10V dc source loss of power is 20watt. When it is connected witrh a 10V ac source loss of power is 10 watt.The frequency of ac source will be:

Options

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

Answer

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

### Step 1: Determine the resistance of the choke coil using the DC source.

Given:
- Voltage (V) = 10 V
- Power (P) = 20 W (for DC)

Using the formula for power in a resistive 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 to find the impedance (Z).

For the AC source:
- Voltage (V_rms) = 10 V
- Power (P) = 10 W

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

Substituting these into the power formula:
\[ P = V_{rms} \left(\frac{V_{rms}}{Z}\right) \left(\frac{R}{Z}\right) \]
This simplifies to:
\[ P = \frac{V_{rms}^2 R}{Z^2} \]
Now substituting the known values:
\[ 10 = \frac{10^2 \cdot 5}{Z^2} \]
This simplifies to:
\[ 10 = \frac{100 \cdot 5}{Z^2} \]
\[ 10 = \frac{500}{Z^2} \]
Rearranging gives:
\[ Z^2 = \frac{500}{10} = 50 \]

### Step 3: Relate impedance (Z) to inductive reactance (X_L) and resistance (R).

We know:
\[ Z^2 = R^2 + X_L^2 \]
Substituting the known values:
\[ 50 = 5^2 + X_L^2 \]
\[ 50 = 25 + X_L^2 \]
\[ X_L^2 = 50 - 25 = 25 \]
\[ X_L = 5 \, \Omega \]

### Step 4: Relate inductive reactance (X_L) to frequency (f).

Inductive reactance is given by:
\[ X_L = \omega L \]
Where \(\omega = 2 \pi f\) and \(L\) is the self-inductance.

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

Substituting:
\[ 5 = 2 \pi f (10 \times 10^{-3}) \]
Solving for frequency \(f\):
\[ f = \frac{5}{2 \pi (10 \times 10^{-3})} \]
\[ f = \frac{5}{0.02 \pi} \]
Calculating this gives:
\[ f \approx \frac{5}{0.06283} \approx 79.58 \, \text{Hz} \approx 80 \, \text{Hz} \]

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

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