In a charging RC circuit with a source Vs, the capacitor voltage Vc(t) approaches its final value as t increases. Which expression correctly describes Vc(t)?

In a charging RC circuit with a source Vs, the capacitor voltage Vc(t) approaches its final value as t increases. Which expression correctly describes Vc(t)?

• A. Vc(t) = Vs e^{-t/RC}
• B. Vc(t) = Vs
• C. Vc(t) = Vs (1 + e^{-t/RC})
• D. Vc(t) = Vs (1 - e^{-t/RC})

Answer: (D)

A charging RC circuit drives the capacitor voltage toward the source voltage over time, not instantly. The current is i = C dVc/dt, and the resistor drops the rest of the source voltage, so Vs − Vc = iR = RC dVc/dt. This gives the differential equation RC dVc/dt = Vs − Vc. With the capacitor initially uncharged, Vc(0) = 0, and solving yields Vc(t) = Vs[1 − e^(−t/(RC))]. This starts at zero and asymptotically approaches Vs as time grows, which is exactly the behavior of the charging process. The other forms would describe different situations: for example, a discharging capacitor would follow Vc(t) = Vs e^(−t/(RC)); a constant Vs would imply instantaneous charging; and a form like Vs[1 + e^(−t/(RC))] would start above the final value, which isn’t possible in this setup.

A charging RC circuit drives the capacitor voltage toward the source voltage over time, not instantly. The current is i = C dVc/dt, and the resistor drops the rest of the source voltage, so Vs − Vc = iR = RC dVc/dt. This gives the differential equation RC dVc/dt = Vs − Vc. With the capacitor initially uncharged, Vc(0) = 0, and solving yields Vc(t) = Vs[1 − e^(−t/(RC))]. This starts at zero and asymptotically approaches Vs as time grows, which is exactly the behavior of the charging process. The other forms would describe different situations: for example, a discharging capacitor would follow Vc(t) = Vs e^(−t/(RC)); a constant Vs would imply instantaneous charging; and a form like Vs[1 + e^(−t/(RC))] would start above the final value, which isn’t possible in this setup.