An inductor of inductance 10 mH and a resistance of 5 is connected to a battery of 20 V at t = 0. Find the ratio of current in circuit at t =  to current at t = 40 sec

Options

A. 1.06
B. 1.15
C. 1.46
D. 0.84

Answer

To solve the problem, we need to find the ratio of the current in the circuit at \( t = \infty \) to the current at \( t = 40 \) seconds. We will use the formula for the current in an RL circuit.

### Step-by-Step Solution:

1. **Identify the given values**:
- Inductance \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} = 0.01 \, \text{H} \)
- Resistance \( R = 5 \, \Omega \)
- Voltage \( V = 20 \, \text{V} \)

2. **Calculate the steady-state current (\( I_{\infty} \))**:
The steady-state current in an RL circuit can be calculated using Ohm's law:
\[ I_{\infty} = \frac{V}{R} = \frac{20 \, \text{V}}{5 \, \Omega} = 4 \, \text{A} \]

3. **Determine the time constant (\( \tau \))**:
The time constant \( \tau \) for an RL circuit is given by:
\[ \tau = \frac{L}{R} = \frac{0.01 \, \text{H}}{5 \, \Omega} = 0.002 \, \text{s} \]

4. **Write the expression for the current at time \( t \)**:
The current \( I(t) \) in an RL circuit as a function of time is given by:
\[ I(t) = I_{\infty} \left(1 - e^{-\frac{t}{\tau}}\right) \]
Substituting \( I_{\infty} = 4 \, \text{A} \):
\[ I(t) = 4 \left(1 - e^{-\frac{t}{0.002}}\right) \]

5. **Calculate the current at \( t = 40 \, \text{s} \)**:
Substitute \( t = 40 \, \text{s} \) into the expression:
\[ I(40) = 4 \left(1 - e^{-\frac{40}{0.002}}\right) \]
Since \( \frac{40}{0.002} = 20000 \), we have:
\[ e^{-20000} \approx 0 \quad (\text{very small}) \]
Therefore:
\[ I(40) \approx 4 \left(1 - 0\right) = 4 \, \text{A} \]

6. **Calculate the ratio of currents**:
Now we find the ratio of the current at \( t = \infty \) to the current at \( t = 40 \):
\[ \text{Ratio} = \frac{I_{\infty}}{I(40)} = \frac{4 \, \text{A}}{4 \, \text{A}} = 1 \]

### Final Answer:
The ratio of the current in the circuit at \( t = \infty \) to the current at \( t = 40 \, \text{s} \) is \( 1 \).