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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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The variation of current versus time is \[\] The potential difference across a 2H inductor as function of time is shown in figure. At time $t=0$, current is zero.


$\begin {array} {1 1} (a)\;\text{ Straight line} & \quad (b)\;\text{ Parabola} \\ (c)\;\text{Hyperbola} & \quad (d)\;\text{Exponential} \end {array}$


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For $t \leq 1s$
$V_L = 8t$
$ \Rightarrow L\: di = V_L dt$
$\Rightarrow \int\: Ldi = \int\: 8 \: tdt$
$\Rightarrow Li = \large\frac{8t^2}{2}$$ = 4t^2$
$ \Rightarrow i = 2t^2$
So, current versus time graph is a parabola.
For $1s \leq t \leq 2s,$
$V_L = 16 – 8t$
$ \Rightarrow \int \: L\: di = \int\: (16 – 8t)\: dt$
$ \Rightarrow Li = \large\frac{16t – 8t^2}{2}$
$ \Rightarrow i = 8t -2t^2$
So, $i$ vs $ t$ graph is a parabola
Ans : (b)


answered Mar 3, 2014 by thanvigandhi_1
edited Oct 8, 2014 by thagee.vedartham

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