Question

The potential difference across a $2H$ inductor as a function of time is shown in the figure. At $t=0$, current is zero. Choose the correct statement(s).

• A. Current at $t=2s$ is $5A$
• B. Current at $t=2s$ is $10A$
• C. Current versus time graph across the inductor will be
  
• D. Current versus time graph across the inductor will be
  

Answer 1

Answer: (A), (C)

Current versus time graph across the inductor will be



We know that potential across inductor, $\begin{array}{}V=L\frac{dI}{dt}& \text{(1)}\end{array}$
$\Rightarrow \int dI={\int }_0^2\frac{Vdt}{L}$
Now ${\int }_0^{2}Vdt=$ Area under the V-t graph for $t=0-2sec$
Area, $A=\frac{1}{2}\times 2\times 10=10$
Therefore $\Rightarrow \int dI=I=\frac{A}{L}=\frac{10}{2}=5A$
Hence option A is correct.
For current variation lets consider voltage variation in 2 cases i.e 0-2 sec and 2-4 sec.
Case 1: For $t=0-2$ sec
Equation of straight line passing through origin,
$V=mt+C$, where $m$ is slope and $C$ is y-intercept
From graph, $m=tan\theta =\frac{10}{2}=5$ and $C=0$
Therefore, $V=5t$
Substituting $V$ in equation $\left(1\right)$ and integrating,
$\int dI=\int \frac{Vdt}{L}$
$I=\frac{1}{L}\int 5tdt=\frac{5t^2}{2L}$
Its evident that the variation is not linear and hence option D is incorrect. Also the curve should show increasing slope from 0-2 sec.
Case 2: For $t=2-4$ sec
Variation of $V$ with $t$ can be given using staright line equation as,
$V=-5t+20$
$\int dI=\int \frac{Vdt}{L}$
$I=\frac{1}{L}\int (-5t+20)dt=\frac{1}{L}(\frac{-5t^2}{2}+20t)$
The above equation shows decreasing slope and so does the graph in option C. Hence option C is correct.