For an RC circuit with a step input, how does Vc(t) evolve for t ≥ 0?

For an RC circuit with a step input, how does Vc(t) evolve for t ≥ 0?

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

Answer: (B)

When a step input is applied to a series RC circuit, the capacitor voltage rises exponentially from its initial value toward the input voltage. The governing equation is C dVc/dt = (Vs − Vc)/R, which can be written as dVc/dt + (1/RC) Vc = Vs/(RC). With the capacitor initially uncharged (Vc(0) = 0), solving gives Vc(t) = Vs [1 − e^(−t/RC)]. This means the capacitor voltage starts at zero and gradually approaches Vs, with a time constant RC governing the rate of rise. The other forms don’t fit this scenario: an exponential decay from Vs would describe discharge, a linear ramp isn’t produced by an RC with a step input, and a constant Vs would imply an instantaneous charge, which isn’t possible with finite RC.

When a step input is applied to a series RC circuit, the capacitor voltage rises exponentially from its initial value toward the input voltage. The governing equation is C dVc/dt = (Vs − Vc)/R, which can be written as dVc/dt + (1/RC) Vc = Vs/(RC). With the capacitor initially uncharged (Vc(0) = 0), solving gives Vc(t) = Vs [1 − e^(−t/RC)]. This means the capacitor voltage starts at zero and gradually approaches Vs, with a time constant RC governing the rate of rise. The other forms don’t fit this scenario: an exponential decay from Vs would describe discharge, a linear ramp isn’t produced by an RC with a step input, and a constant Vs would imply an instantaneous charge, which isn’t possible with finite RC.