# Area under $V_L - t$ graph gives

The potential difference across a 2H inductor as function of time is shown in figure. At time $t=0$, current is zero.

$\begin {array} {1 1} (a)\;\text{Current} & \quad (b)\;\text{Product of current} \\ (c)\;\text{Product of inductance and current} & \quad (d)\;\text{Product of inductance and change in current} \end {array}$

Since $V_L =L \large\frac{di}{dt}$
$\Rightarrow L\: di = V_L\: dt$
Integrating on both sides
$\Rightarrow L(i_f – i_i ) = \: _o\int^t \: V_L\: dt =$ area under $V_L-t$ graph
Ans : (d)

edited Oct 8, 2014