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. `1.48`
D. `0.84`

Explanation

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.

### Step 1: Identify the parameters
We have:
- Inductance \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \)
- Resistance \( R = 5 \, \Omega \)
- Voltage \( V = 20 \, \text{V} \)

### Step 2: Calculate the time constant \( \tau \)
The time constant \( \tau \) for an RL circuit is given by:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{10 \times 10^{-3}}{5} = 2 \times 10^{-3} \, \text{s} = 0.002 \, \text{s} \]

### Step 3: Determine the maximum current \( I_{\infty} \)
The maximum current \( I_{\infty} \) when \( t \to \infty \) can be calculated using Ohm's law:
\[ I_{\infty} = \frac{V}{R} = \frac{20}{5} = 4 \, \text{A} \]

### Step 4: Write the expression for current \( I(t) \)
The current \( I(t) \) in the circuit as a function of time \( t \) is given by:
\[ I(t) = I_{\infty} \left(1 - e^{-\frac{t}{\tau}}\right) \]
Substituting \( I_{\infty} \) and \( \tau \):
\[ I(t) = 4 \left(1 - e^{-\frac{t}{0.002}}\right) \]

### Step 5: Calculate the current at \( t = 40 \) seconds
Substituting \( t = 40 \) seconds into the equation:
\[ I(40) = 4 \left(1 - e^{-\frac{40}{0.002}}\right) \]
Calculating the exponent:
\[ -\frac{40}{0.002} = -20000 \]
Thus, \( e^{-20000} \) is a very small number, effectively approaching 0:
\[ I(40) \approx 4 \left(1 - 0\right) = 4 \, \text{A} \]

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

### Final Answer
The ratio of the current in the circuit at \( t = \infty \) to the current at \( t = 40 \) seconds is:
\[ \text{Ratio} = 1 \]