Question

A variable voltage $V=2t$ is applied across an inductor of inductance $L=2\text{H}$ as shown in figure. Then

• A. Current versus time graph is a parabola.
• B. Energy stored in magnetic field at $t=2\text{s}$ is $4\text{J}$.
  
• C. Potential energy at time $t=1\text{s}$ in magnetic field is increasing at a rate of $1\text{J/s}$.
• D. Energy stored in magnetic field is zero in all time.

Answer 1

Answer: (A), (B), (C)

Potential energy at time $t=1\text{s}$ in magnetic field is increasing at a rate of $1\text{J/s}$.

We know that voltage across an inductor is given by,

$V=L\frac{di}{dt}$

$\Rightarrow 2t=2\times \frac{di}{dt}$

$i=\int di=\int tdt=\frac{t^2}{2}$

Therefore, we can say that the graph between $i$ and $t$ is a parabola.

Energy stored in the inductor (or in the magnetic field) is given by

$E=\frac{1}{2}L{i}^2$

$E=\frac{1}{2}\times 2\times {\left(\frac{t^2}{2}\right)}^2=\frac{t^4}{4}$

At, $t=2\text{s}$, $E=\frac{2^4}{4}=4\text{J}$

Rate of change of potential energy is given by,

$\frac{dE}{dt}=\frac{d}{dt}\left(\frac{1}{2}L{i}^2\right)=Li\frac{di}{dt}$

$\Rightarrow \frac{dE}{dt}=2\times \frac{t^2}{2}\times \frac{2t}{2}=t^3$

At, $t=1\text{s}$, $\frac{dE}{dt}=1\text{J/s}$

Hence, options $\left(\text{A}\right)$, $\left(\text{B}\right)$ and $\left(\text{C}\right)$ are the correct answers.