Question

question 8 (2.5 points) saved immediately after the power is applied (i.e., t = 0, the circuit is not in steady state yet), what is the voltage drop across the capacitor? assume that the capacitor could handle the entire voltage if required. remember: v(t) = q(t) / c r = 1kω vs + - c = 100 μf vc - 0v immediately after the power is applied (i.e., t = 0) 0.006v 9v 0v 4.5v

Explanation:

Step1: Recall Capacitor Voltage at t=0

At \( t = 0^+ \) (immediately after power is applied), a capacitor acts as a short - circuit if it has no initial charge. The voltage across a capacitor is given by \( V(t)=\frac{q(t)}{C} \). Before power is applied (\( t = 0^- \)), the capacitor has \( q(0^ -)=0 \) (since there is no voltage across it initially, as shown in the diagram with \( 0V \) marked). By the principle of continuity of charge for a capacitor (\( q(0^+)=q(0^ -) \)), so \( q(0^+)=0 \).

Step2: Calculate Voltage Across Capacitor

Using the formula \( V_C(0^+)=\frac{q(0^+)}{C} \), substituting \( q(0^+)=0 \) and \( C = 100\mu F \) (the value of capacitance doesn't affect the result here since the charge is zero), we get \( V_C(0^+)=\frac{0}{C}=0V \).

Answer:

\( 0V \) (corresponding to the option with \( 0V \))