In the given circuit the switch ils closed in the positin 1 at `t=0` and then moved to 2 after `250mus`. Derive and expression for current as a functioin of time for `tgt0`. Also plot the variation of current with time.

Answer

B

Explanation

Time constant of the circuit is
`tau_C=CR=(0.5xx10^-6)(500)=2.5xx10^-4s`
for `tle250musi =i_0e^((-t)/tau_C)`
here e`i_0=20/500=0.04A `
`:. i=(0.04e^(-4000t))amp`
At `t=250mus=2.5xx10^-4s`
`i=0.04e^_1=0.015amp`
At this moment `PD` across the capacitor
`V_C=20(1-e^-1)=12.64V`
So when the switch is shifted to positon 2, the current in the circuit is `0.015` (clockwise) and `PD` across capacitor is `12.64V`

As soon as the switch is shifted to positio `2` current will reverse its direction with minimum current
`i_0^'=(40+12.64)/500`
`=0.11A`
Now, it wil decrease exponentially to zero
For `tge250mus`
`i=-i_0^'e^(-t)/tau_C)=-0.11e^(-4000t)`
The (i-t) graph is as shown in figure