The resistance of an R-L circuit is `10 Omega ` . An e.m.f. `E_(0)` applied across the circuit at `Omega ' = 20 rad / s. If the current in the circuit is `(I_(0))/(sqrt(2))` , what is the value of L

Options

A. 1 H
B. 2 H
C. 3 H
D. 0.5 H

Explanation

To solve the problem, we need to find the value of the inductance \( L \) in an R-L circuit given the resistance \( R = 10 \, \Omega \), the angular frequency \( \omega = 20 \, \text{rad/s} \), and the current \( I = \frac{I_0}{\sqrt{2}} \).

### Step-by-Step Solution:

1. **Identify the known values:**
- Resistance, \( R = 10 \, \Omega \)
- Angular frequency, \( \omega = 20 \, \text{rad/s} \)
- Current, \( I = \frac{I_0}{\sqrt{2}} \)

2. **Use the formula for impedance \( Z \) in an R-L circuit:**
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L = \omega L \) is the inductive reactance.

3. **Express the current \( I \) in terms of the applied EMF \( E_0 \) and impedance \( Z \):**
\[ I = \frac{E_0}{Z} \]

4. **Substituting the expression for \( Z \):**
\[ I = \frac{E_0}{\sqrt{R^2 + (\omega L)^2}} \]

5. **Substituting the known values into the equation for current:**
\[ \frac{I_0}{\sqrt{2}} = \frac{E_0}{\sqrt{10^2 + (20L)^2}} \]

6. **Square both sides to eliminate the square root:**
\[ \left(\frac{I_0}{\sqrt{2}}\right)^2 = \frac{E_0^2}{10^2 + (20L)^2} \]
This simplifies to:
\[ \frac{I_0^2}{2} = \frac{E_0^2}{100 + 400L^2} \]

7. **Cross-multiply to isolate the terms involving \( L \):**
\[ I_0^2 (100 + 400L^2) = 2E_0^2 \]

8. **Rearranging gives:**
\[ 100I_0^2 + 400I_0^2 L^2 = 2E_0^2 \]

9. **Isolate the term with \( L^2 \):**
\[ 400I_0^2 L^2 = 2E_0^2 - 100I_0^2 \]

10. **Solve for \( L^2 \):**
\[ L^2 = \frac{2E_0^2 - 100I_0^2}{400I_0^2} \]

11. **Take the square root to find \( L \):**
\[ L = \sqrt{\frac{2E_0^2 - 100I_0^2}{400I_0^2}} \]

12. **Substituting the known values of \( E_0 \) and \( I_0 \) will yield the value of \( L \).** Since we don't have specific values for \( E_0 \) and \( I_0 \), we can use the relationship derived from the earlier steps.

13. **Using the given current \( I = \frac{I_0}{\sqrt{2}} \) and substituting \( R \) and \( \omega \):**
\[ 1 = \frac{1}{1 + \left(\frac{20L}{10}\right)^2} \]
This simplifies to:
\[ 1 + 4L^2 = 2 \implies 4L^2 = 1 \implies L^2 = \frac{1}{4} \implies L = \frac{1}{2} \text{ H} \]

### Final Answer:
Thus, the value of the inductance \( L \) is:
\[ L = 0.5 \, \text{H} \]