A current of 1 A flows through a coil, when a 100 V d.c. is applied to it. But a current of 0.5 A flows through the same coil, when a 100V a.c. of frequency 50 Hz is applied. The resistance and inductance of the coil are given by (take `pi^(2)=10`)

Options

A. `100 Omega , sqrt(0.2) H`
B. `50 Omega , sqrt(0.3) H`
C. `100 Omega , sqrt(0.3) H`
D. `100 Omega , sqrt(0.2) H`

Explanation

To solve the problem, we need to find the resistance (R) and inductance (L) of the coil based on the given conditions for both DC and AC currents.

### Step-by-Step Solution:

1. **Determine Resistance (R) from DC Conditions:**
- Given that a current of 1 A flows through the coil when a 100 V DC is applied, we can use Ohm's Law:
\[ V = I \cdot R \]
- Rearranging gives:
\[ R = \frac{V}{I} = \frac{100 \, \text{V}}{1 \, \text{A}} = 100 \, \Omega \]

2. **Determine Impedance (Z) from AC Conditions:**
- When a 100 V AC is applied, a current of 0.5 A flows through the coil. The relationship between voltage, current, and impedance in AC circuits is:
\[ I = \frac{V}{Z} \]
- Rearranging gives:
\[ Z = \frac{V}{I} = \frac{100 \, \text{V}}{0.5 \, \text{A}} = 200 \, \Omega \]

3. **Relate Impedance to Resistance and Reactance:**
- In an LR circuit, the impedance (Z) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
- We already know \( R = 100 \, \Omega \) and \( Z = 200 \, \Omega \). Plugging in these values:
\[ 200 = \sqrt{100^2 + X_L^2} \]
- Squaring both sides:
\[ 200^2 = 100^2 + X_L^2 \]
\[ 40000 = 10000 + X_L^2 \]
\[ X_L^2 = 40000 - 10000 = 30000 \]
\[ X_L = \sqrt{30000} = 100\sqrt{3} \, \Omega \]

4. **Calculate Inductance (L):**
- The inductive reactance \( X_L \) is related to the inductance (L) and the angular frequency (\( \omega \)):
\[ X_L = \omega L \]
- The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
- Given that the frequency \( f = 50 \, \text{Hz} \):
\[ \omega = 2 \pi \times 50 = 100 \pi \]
- Substituting \( X_L \) into the equation:
\[ 100\sqrt{3} = (100\pi)L \]
- Solving for \( L \):
\[ L = \frac{100\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{\pi} \]
- Using \( \pi^2 = 10 \), we find \( \pi = \sqrt{10} \):
\[ L = \frac{\sqrt{3}}{\sqrt{10}} = \frac{\sqrt{30}}{10} \approx 0.3 \, \text{H} \]

### Final Results:
- **Resistance (R)**: \( 100 \, \Omega \)
- **Inductance (L)**: \( 0.3 \, \text{H} \)