A solenoid of inductance 50 mH and resistance `10 Omega` is connected to a battery of 6V. Find the time elapsed before the current acquires half of its steady - state value.

Options

A. `3.5 ms`
B. `2.5 ms`
C. `0.693 ms`
D. `2 ms`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 50 mH = \(50 \times 10^{-3}\) H
- Resistance (R) = 10 Ω
- Voltage (V) = 6 V

### Step 2: Calculate the steady-state current (I₀)
The steady-state current in an LR circuit can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the given values:
\[ I_0 = \frac{6 \, \text{V}}{10 \, \Omega} = 0.6 \, \text{A} \]

### Step 3: Write the equation for the current in the LR circuit
The current (I) in an LR circuit as a function of time (t) is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \(\tau\) is the time constant defined as:
\[ \tau = \frac{L}{R} \]

### Step 4: Calculate the time constant (τ)
Substituting the values of L and R:
\[ \tau = \frac{50 \times 10^{-3} \, \text{H}}{10 \, \Omega} = 5 \times 10^{-3} \, \text{s} = 5 \, \text{ms} \]

### Step 5: Set up the equation to find the time when the current is half of its steady-state value
We need to find the time (t) when the current is half of the steady-state current:
\[ I(t) = \frac{I_0}{2} \]
Substituting into the equation:
\[ \frac{I_0}{2} = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Dividing both sides by \(I_0\):
\[ \frac{1}{2} = 1 - e^{-\frac{t}{\tau}} \]
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = \frac{1}{2} \]

### Step 6: Solve for t
Taking the natural logarithm of both sides:
\[ -\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) \]
This simplifies to:
\[ t = -\tau \ln\left(\frac{1}{2}\right) \]
Using the property \(\ln\left(\frac{1}{2}\right) = -\ln(2)\):
\[ t = \tau \ln(2) \]

### Step 7: Substitute the value of τ
Substituting \(\tau = 5 \times 10^{-3} \, \text{s}\):
\[ t = 5 \times 10^{-3} \ln(2) \]
Using \(\ln(2) \approx 0.693\):
\[ t \approx 5 \times 10^{-3} \times 0.693 \approx 3.465 \times 10^{-3} \, \text{s} \approx 3.5 \, \text{ms} \]

### Final Answer
The time elapsed before the current acquires half of its steady-state value is approximately **3.5 ms**.

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