Question

The switch has been in position 1 for a long time and abruptly changes to position 2 at t = 0.

If time t is in seconds, the capacitor voltage $V_C$ (in volts) for t > 0 is given by

• A. $4(1-6e^{-t/0.5})$
• B. $10-6e^{-t/0.5}$
• C. $4(1-6e^{-t/0.6})$
• D. $10-6e^{-t/0.6}$

Answer 1

The correct option is D $10-6e^{-t/0.6}$

at $t=0^{-},$ Switch is at position 1

where, $V_C\left(0^{-}\right)=\frac{10\times 2}{2+3}=4V$ ...(1)

$\therefore V_C\left(0^{-}\right)=V_C\left(0^{+}\right)=4V$

at $t=\infty$, Switch is at position 2

$V_c\left(\infty \right)=5\times 2=10V$ ...(2)

The time constant of the circuit is
$\tau =R_{eq}{C}_{eq}$

$=(4+2)\times 0.1=0.6sec$

$\therefore V_C\left(t\right)=V_C\left(\infty \right)+\left[V_C\right(0^{+})-V_C(\infty \left)\right]e^{-t/\tau }$

$=10+(4-10)e^{-t/0.6}$

$V_C\left(t\right)=(10-6e^{-t/0.6})V$