In the circuit shown, the switch `S` has been close for a long time. At time `t=0` the switch is opened. It remains open for a long time `T`, after which it is closed again.

Options

A. voltage drop across `100 k Omega` resister is `10e^(-t//1.5)V` for `tltT`
B. voltage drop cross `100kOmega` resister is `(20e^(-t//1.5))/3V` for `tltT`
C. voltage drop acros `100OkOmega` resister is `100(e^(T-t)-e^((T-3t)//3))` for `tgtT`
D. The time constant of the circuit is `1.5` sec for `tgtT`

Answer

B, C

Explanation

When the which `'S'` is opened,
`tau=RC=1.5`sec
Voltage across the capacitor
`V_(c)=10(1-e^(-t/1.5))` for `tltT`
The current `i=(dQ)/(dt)=C(dV_(c))/(dt)`
`=200/3xx10^(-6)(e^(-t//1.5))` for `tltT`
Voltage drop across `100k mu` resister,
`V=iR=20/3e^(-t//1.5)` for `t lt T`
After `t=T`,
`tau=100xxkOmegaxx10 mu F=1` sec
Voltage across capacitor
`V_(c)=V_(@)t^(-[(t-T)//tau])=10(1-e^(-T//1.5)),e^(-[(t-|T)//1])`
The charge on capaclitor, `Q=V_(c).C`
The current thorugh `100 k mu` resistor `i=-C(dV_(c))/(dt)=10^(-3)(1-e^(-T//1.5))(1-e^(-(t/T)//1))`