Question

The switch SW shown in the circuit is kept at position '1' for a long duration. At t = $0^{+}$, the switch is moved to position '2'. Assuming $\left|V_{02}\right|>\left|V_{01}\right|$, the voltage $V_c\left(f\right)$ across the capacitor is

• A. $V_c\left(t\right)$ = $V_{02}(1-e^{-t/RC})-V_{01}$
• B. $V_c\left(t\right)$ = $V_{02}(1-e^{-t/RC})+V_{01}$
• C. $V_c\left(t\right)$ = -$(V_{02}+V_{01})(1-e^{-t/2RC})-V_{01}$
• D. $V_c\left(t\right)$ =$(V_{02}-V_{01})(1-e^{-t/2RC})+V_{01}$
• E. None of the above

Answer 1

The correct option is E None of the above

Circuit for t < 0,


Circuit for t = $\infty$

In steady state capacitor becomes open circuit


$\therefore$ $V_c\left(\infty \right)=-V_{02}$

We know that:

$V_c\left(t\right)=V_c\left(\infty \right)-\left[V_c\right(\infty )-V_c(0^{+}\left)\right]e^{-t/\pi }$

$\tau$ = time constant of given circuit

= 2 RC

$\therefore$ $V_c\left(t\right)=-V_{02}-(V_{02}-V_{01})e^{-t/2RC}$

= $(V_{02}+V_{01})-(V_{02}-V_{01})e^{-t/2RC}-V_{01}$

$or,V_c\left(t\right)=(V_{02}+V_{01})(e^{-t/2RC}-1)+V_{01}$