The inductance of a solenoid is 5 henery and its resistance id `5 Omega` . If it is connected to a 10 volt battery then time taken by the current to reach `9//10^(th)` of its maximum will be :-

Options

A. 4.0s
B. 2.3 s
C. 1.4 s
D. 1.2 s

Answer

To solve the problem, we need to find the time taken by the current in the solenoid to reach \( \frac{9}{10} \) of its maximum value when connected to a 10-volt battery.

### Step-by-Step Solution:

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

2. **Calculate the maximum current \( I_{\text{max}} \):**
The maximum current in the circuit can be calculated using Ohm's law:
\[ I_{\text{max}} = \frac{V}{R} = \frac{10 \, \text{V}}{5 \, \Omega} = 2 \, \text{A} \]

3. **Determine the time constant \( \tau \):**
The time constant \( \tau \) for an RL circuit is given by:
\[ \tau = \frac{L}{R} = \frac{5 \, \text{H}}{5 \, \Omega} = 1 \, \text{s} \]

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

5. **Set up the equation to find the time when the current reaches \( \frac{9}{10} I_{\text{max}} \):**
We want to find \( t \) when:
\[ I(t) = \frac{9}{10} I_{\text{max}} = \frac{9}{10} \times 2 \, \text{A} = 1.8 \, \text{A} \]
Thus, we set up the equation:
\[ 1.8 = 2 \left(1 - e^{-\frac{t}{1}}\right) \]

6. **Solve for \( t \):**
Rearranging the equation gives:
\[ 1 - e^{-t} = \frac{1.8}{2} = 0.9 \]
\[ e^{-t} = 1 - 0.9 = 0.1 \]
Taking the natural logarithm on both sides:
\[ -t = \ln(0.1) \]
\[ t = -\ln(0.1) \approx 2.3026 \, \text{s} \]

### Final Answer:
The time taken by the current to reach \( \frac{9}{10} \) of its maximum value is approximately \( 2.3026 \, \text{s} \).