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Current at $t = 1 \: s$ is

 

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

$\begin {array} {1 1} (a)\;1A & \quad (b)\;2A \\ (c)\;4A & \quad (d)\;8A \end {array}$

 

1 Answer

$L(i_f – i_i ) =$ area under $V_L – t$ graph
$ \Rightarrow 2(i_f – V) = \large\frac{1}{2} $$\times 1 \times 8 = 4$
$ \Rightarrow i_f = 2A$
Ans : (b)
answered Mar 1, 2014 by thanvigandhi_1
 

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