Question

For the circuit shown below. The value of $V_c$ at $t=2\mu s$ is

• A. -3.06 V
• B. 12.01 V
• C. -14.02 V
• D. 6.07 V

Answer 1

The correct option is A -3.06 V

At $t<0,$

$V_c\left(0^{-}\right)=3\times \frac{6}{(6+3)}=\frac{18}{9}=2V$
At $t=0$


$v_c\left(\infty \right)=\frac{6}{9}\times (-9)=-6V$
$v_c\left(t\right)=-6+(2+6)e^{-t/\tau }$
$\tau =\frac{18}{9}\times 1=2\mu s$
$V_c\left(t\right)=-6+8e^{\frac{-t}{2}}$
$V_e\left(2\mu s\right)=-6+8e^{-1}=-3.06V$