Question

Initially the switch is in position 1 for a long time. At $t=0$, the switch is moved from 1 to 2. Obtain expressions for $V_C$ and $V_R$ for t > 0

Answer 1

$q_i=C{V}_i=100\mu C,q_f=C{V}_f=-50\mu C$

Therefore charge will vary from $100\mu C$ to $-50\mu C$ exponentially.

$\therefore q=-50+150e^{-t/{\tau }_C}$ Here q is in $\mu C$

${\tau }_C=CR=\left({10}^{-6}\right)(5\times {10}^3)=5\times {10}^{-3}$

$\therefore q=-50+150e^{-t/{\tau }_C}$

$V_C=\frac{q}{C}=(-50+150e^{-t/{\tau }_C})V$

or $V_C=50(3e^{-200t}-1)$

$i=-\frac{dq}{dt}=\frac{150\times {10}^{-6}}{{\tau }_C}{e}^{-200t}$

$=\frac{150\times {10}^{-6}}{5\times {10}^{-3}}{e}^{-200t}$

$=30\times {10}^{-3}{e}^{-200t}$

$V_R=iR=150e^{-200t}$