Question

The input voltage, $V_i=5+2\sin \left(\omega t\right)$ is applied to a circuit as shown in the figure below. Assuming all the elements of the circut are ideal, then the minimum value of $V_0$ at steady state would be equal to

• A. -3 V
• B. -5 V
• C. -6 V
• D. -4 V

Answer 1

The correct option is D -4 V

For the positive half cycle, the capacitor will charge with a voltage of $5+2=7V$.

The minimum value of the sine function that could achieve is -2 V, thus the minimum output voltage would be equal to
$V_0=5-2-7=-4V$