1. Given data

• Inductance: $L=1\,\text{H}$
• Resistance: $R=100\,\Omega$
• Battery voltage: $V=6\,\text{V}$

For an $RL$ circuit switched on with a DC source,

$i(t)=I_0\left(1-e^{-t/\tau}\right)$

where

$I_0=\frac{V}{R}, \qquad \tau=\frac{L}{R}$

So,

$I_0=\frac{6}{100}=0.06\,\text{A}$

and

$\tau=\frac{1}{100}=0.01\,\text{s}=10\,\text{ms}$

2. Part (a): Time to reach half the steady-state current

We need

$i(t)=\frac{I_0}{2}$

Thus,

$I_0\left(1-e^{-t/\tau}\right)=\frac{I_0}{2}$

Cancelling $I_0$,

$1-e^{-t/\tau}=\frac{1}{2}$

$e^{-t/\tau}=\frac{1}{2}$

Taking natural log,

$\frac{t}{\tau}=\ln 2$

$t=\tau \ln 2=10\times 0.693\,\text{ms}=6.93\,\text{ms}$

Approximately,

$t\approx 7\,\text{ms}$

3. Part (b): Energy stored after $15\,\text{ms}$

Current at time $t=15\,\text{ms}$:

$i(15\,\text{ms})=I_0\left(1-e^{-t/\tau}\right)$

Here,

$\frac{t}{\tau}=\frac{15}{10}=\frac{3}{2}$

Given $e^{-3/2}=0.25$, so

$i=0.06(1-0.25)=0.06\times 0.75=0.045\,\text{A}$

Energy stored in the magnetic field:

$U=\frac{1}{2}Li^2$

$U=\frac{1}{2}(1)(0.045)^2$

$U=0.5\times 0.002025=0.0010125\,\text{J}$

$U\approx 1.0\times 10^{-3}\,\text{J}=1\,\text{mJ}$

4. Matching with options

We found:

• $t\approx 7\,\text{ms}$
• $U\approx 1\,\text{mJ}$

So the correct option is:

$\boxed{\text{C}}$

5. Comparison with stored correct answer

Stored correct answer: $\text{C}$

Our derived answer also gives $\text{C}$, so they agree.