A coil of 10 H inductance and `5 Omega` resistance is connected to 5 volt battery in series . The current in ampere in circuit 2 seconds after switched is on :-

Options

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

Answer

To find the current in the circuit 2 seconds after the switch is turned on, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 10 H
- Resistance (R) = 5 Ω
- Voltage (E) = 5 V
- Time (t) = 2 seconds

### Step 2: Calculate the steady-state current (I0)
The steady-state current (I0) can be calculated using Ohm's Law when the inductor behaves like a short circuit (which occurs after a long time).

\[ I_0 = \frac{E}{R} = \frac{5 \text{ V}}{5 \text{ Ω}} = 1 \text{ A} \]

### Step 3: Calculate the time constant (τ)
The time constant (τ) for an LR circuit is given by the formula:

\[ τ = \frac{L}{R} \]

Substituting the values:

\[ τ = \frac{10 \text{ H}}{5 \text{ Ω}} = 2 \text{ seconds} \]

### Step 4: Use the formula for current at time t
The current (I) in the circuit at any time t can be calculated using the formula:

\[ I(t) = I_0 \left(1 - e^{-\frac{t}{τ}}\right) \]

Substituting the values of I0, t, and τ:

\[ I(2) = 1 \left(1 - e^{-\frac{2}{2}}\right) = 1 \left(1 - e^{-1}\right) \]

### Step 5: Calculate the value of e^{-1}
Using the approximate value of e (Euler's number):

\[ e^{-1} \approx 0.3679 \]

So,

\[ I(2) = 1 \left(1 - 0.3679\right) \approx 1 \times 0.6321 \approx 0.6321 \text{ A} \]

### Final Answer
The current in the circuit 2 seconds after the switch is turned on is approximately **0.632 A**.

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