Solenoid of self-inductance L is connected in series with a resistor of resistance R with a battery and switch. Switch is closed at t = 0. How much time will be taken by inductor to acquire one-fourth of its maximum energy?

Options

A. `(L)/(R)I n(2)`
B. `(2L)/(R)I n(2)`
C. `(3L)/(R)I n(2)`
D. `(4L)/(R)I n(2)`

Explanation

To solve the problem, we need to find the time taken by the inductor to acquire one-fourth of its maximum energy after the switch is closed. Let's break down the solution step-by-step.

### Step 1: Understand the maximum energy stored in the inductor
The maximum energy \( U_0 \) stored in an inductor is given by the formula:
\[ U_0 = \frac{1}{2} L I_0^2 \]
where \( L \) is the self-inductance and \( I_0 \) is the maximum current flowing through the circuit.

### Step 2: Determine the expression for current \( I \) at time \( t \)
The current \( I \) at any time \( t \) after closing the switch is given by:
\[ I = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant of the circuit.

### Step 3: Substitute the current into the energy formula
The energy \( U \) stored in the inductor at time \( t \) can be expressed as:
\[ U = \frac{1}{2} L I^2 = \frac{1}{2} L \left(I_0 \left(1 - e^{-\frac{t}{\tau}}\right)\right)^2 \]
This simplifies to:
\[ U = \frac{1}{2} L I_0^2 \left(1 - e^{-\frac{t}{\tau}}\right)^2 \]

### Step 4: Set up the equation for one-fourth of maximum energy
We want to find the time \( t \) when the energy \( U \) is one-fourth of the maximum energy \( U_0 \):
\[ U = \frac{1}{4} U_0 \]
Substituting the expressions we have:
\[ \frac{1}{2} L I_0^2 \left(1 - e^{-\frac{t}{\tau}}\right)^2 = \frac{1}{4} \left(\frac{1}{2} L I_0^2\right) \]
This simplifies to:
\[ \left(1 - e^{-\frac{t}{\tau}}\right)^2 = \frac{1}{4} \]

### Step 5: Solve for \( 1 - e^{-\frac{t}{\tau}} \)
Taking the square root of both sides, we get:
\[ 1 - e^{-\frac{t}{\tau}} = \frac{1}{2} \]
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = \frac{1}{2} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ -\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) \]
This can be rewritten as:
\[ \frac{t}{\tau} = -\ln\left(\frac{1}{2}\right) = \ln(2) \]

### Step 7: Solve for \( t \)
Now, substituting \( \tau = \frac{L}{R} \):
\[ t = \tau \ln(2) = \frac{L}{R} \ln(2) \]

### Final Result
The time taken by the inductor to acquire one-fourth of its maximum energy is:
\[ t = \frac{L}{R} \ln(2) \]