An emf of `15 V` is applied in a circuit containing `5 H` inductance and `10 Omega` resistance. The ratio of the currents at time `t = oo` and `t = 1 s` is

Options

A. `(e^(1//2))/(e^(1//2)-1)`
B. `(e^(2))/(e^(2)-1)`
C. `1-e^(-1)`
D. `e^(-1)`

Explanation

To solve the problem, we need to find the ratio of the currents in an RL circuit at two different times: \( t = \infty \) and \( t = 1 \, \text{s} \).

### Step-by-Step Solution:

1. **Identify the given values:**
- EMF (\( V \)) = 15 V
- Inductance (\( L \)) = 5 H
- Resistance (\( R \)) = 10 Ω

2. **Calculate the time constant (\( \tau \)):**
\[ \tau = \frac{L}{R} = \frac{5 \, \text{H}}{10 \, \Omega} = 0.5 \, \text{s} \]

3. **Determine the maximum current (\( I_0 \)):**
The maximum current in the circuit when it reaches steady state (as \( t \to \infty \)) is given by:
\[ I_0 = \frac{V}{R} = \frac{15 \, \text{V}}{10 \, \Omega} = 1.5 \, \text{A} \]

4. **Write the expression for current \( I(t) \) in the circuit:**
The current as a function of time in an RL circuit is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

5. **Calculate the current at \( t = \infty \):**
As \( t \to \infty \):
\[ I(\infty) = I_0 \left(1 - e^{-\frac{\infty}{\tau}}\right) = I_0 \cdot 1 = 1.5 \, \text{A} \]

6. **Calculate the current at \( t = 1 \, \text{s} \):**
Substitute \( t = 1 \, \text{s} \) into the current equation:
\[ I(1) = 1.5 \left(1 - e^{-\frac{1}{0.5}}\right) = 1.5 \left(1 - e^{-2}\right) \]

Now, calculate \( e^{-2} \):
\[ e^{-2} \approx 0.1353 \quad \text{(using a calculator)} \]
Thus,
\[ I(1) = 1.5 \left(1 - 0.1353\right) = 1.5 \times 0.8647 \approx 1.297 \, \text{A} \]

7. **Find the ratio of the currents:**
Now, we can find the ratio of the currents at \( t = \infty \) and \( t = 1 \, \text{s} \):
\[ \text{Ratio} = \frac{I(\infty)}{I(1)} = \frac{1.5}{1.297} \approx 1.157 \]

### Final Answer:
The ratio of the currents at \( t = \infty \) and \( t = 1 \, \text{s} \) is approximately \( 1.157 \).