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

Current at t=s is

Options

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

D. Current versus time graph across the inductor will be Fig.

Answer

A, C

Explanation

`V = L(dI)/(dt) rArr underset(0) overset(1) intdI = (1)/(L) underset(0) overset(1)int Vdt`
`rArr I = (1)/(L)` [Area under `nu -t` graph from `t = 0` to `t = t`]
At `t = 2 s`, `I = (1)/(2) [(1)/(2) xx 2xx 10] = 5 A`
For `t = 0` to `2 s, V = 5t`
`I = (1)/(L) underset(0) overset(t) int 5t dt = (1)/(2) [(5t^(2))/(2)]_(0)^(t) = (5t^(2))/(4)`
For `t = 2` to `4 s, V = - 5t + 20`
`I = (1)/(2) underset(0) overset(t)int (-5t + 20)dt rArr I = (-5t^(2))/(4) + 10t`
Hence the correct graph is (c )`.