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Step-by-Step Solution
Step 1: Express the voltage-current relation for an inductor
For an inductor of inductance L , the voltage across it is given by
L \dfrac{di}{dt} = V(t) .
In this problem, V(t) = 3t and L = 2\,\text{H} . Therefore,
L \dfrac{di}{dt} = 3t .
Step 2: Integrate to find the current as a function of time
Rearrange and integrate both sides with respect to t :
\displaystyle \int \! L\,di = \int \! 3t \, dt \quad \Longrightarrow \quad L \, i = \dfrac{3t^2}{2} + C.
Since the circuit is assumed to start from rest at t = 0 (and thus i(0)=0 ), the constant of integration C = 0 . Hence,
\displaystyle i = \dfrac{3t^2}{2L}.
Substitute L = 2\,\text{H} :
\displaystyle i = \dfrac{3t^2}{2 \times 2} = \dfrac{3t^2}{4}.
Step 3: Write the expression for the energy stored in the inductor
The energy stored in an inductor is given by
\displaystyle E = \dfrac{1}{2} L\, i^2.
Substitute L = 2 H and i = \dfrac{3t^2}{4} :
\displaystyle E = \dfrac{1}{2} \times 2 \times \Bigl(\dfrac{3t^2}{4}\Bigr)^2 = \Bigl(\dfrac{3t^2}{4}\Bigr)^2.
Simplify:
\displaystyle E = \dfrac{9 t^4}{16}.
Step 4: Evaluate the energy at t = 4 s
Plug in t = 4 s:
\displaystyle E = \dfrac{9 \times (4)^4}{16}
= \dfrac{9 \times 256}{16}
= 9 \times 16
= 144 \,\text{J}.
Final Answer
The energy stored in the coil after 4 s is \boxed{144\,\text{J}} .
