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Difficulty: HardCapacitors

An uncharged capacitor is connected in series with a resistor and a DC voltage source through an open switch. When the switch is suddenly closed, how do the voltage drop across the capacitor and the circuit current behave from the initial instant (t=0t = 0) to steady-state condition (t≥5τt \ge 5\tau)?

  1. A
    Initial capacitor voltage equals the source voltage with zero circuit current; steady-state capacitor voltage is zero with maximum circuit current.
  2. B
    Initial and steady-state capacitor voltage both remain at zero while continuous DC current flows indefinitely through the circuit.
  3. Initial capacitor voltage is zero with maximum circuit current; steady-state capacitor voltage equals the source voltage with zero circuit current.Answer
  4. D
    Initial capacitor voltage equals the source voltage with maximum current; steady-state voltage and current both drop to zero.

Answer

At the initial instant (t=0t = 0), the uncharged capacitor has zero voltage drop and allows maximum circuit current, while at steady state (t≥5τt \ge 5\tau), the capacitor voltage reaches the full source voltage and circuit current drops to zero.
Because a capacitor stores energy in an electrostatic field between its plates, its voltage cannot change instantaneously. When connected to a DC source at t=0t = 0, an uncharged capacitor exhibits zero potential difference (0 V0\text{ V}) across its plates and behaves momentarily like a short circuit, allowing peak current (I=Vsource/RI = V_{\text{source}} / R) to flow through the series circuit. As charge accumulates over five time constants (5τ5\tau), the potential difference across the capacitor rises to match the source voltage (VC=VsourceV_C = V_{\text{source}}), completely balancing the applied EMF. At this steady state, the capacitor acts as an open circuit, reducing direct current flow to zero.

Step-by-Step Solution

1
Analyze the electrical state of an uncharged capacitor at the exact instant the DC switch is closed (t=0t = 0).
Because charge cannot accumulate instantaneously (Q=C×VQ = C \times V), the initial capacitor voltage is VC(0)=0 VV_C(0) = 0\text{ V}. The capacitor acts momentarily as a short circuit, allowing maximum current determined entirely by the series resistor: Imax=VsourceRI_{\text{max}} = \frac{V_{\text{source}}}{R}.
Capacitors oppose instantaneous changes in voltage.
2
Analyze the transient charging phase over time (0<t<5τ0 < t < 5\tau).
As electrons accumulate on the negative plate and depart the positive plate, an opposing electric field develops across the dielectric. The voltage VC(t)V_C(t) increases exponentially according to VC(t)=Vsource(1−e−t/RC)V_C(t) = V_{\text{source}}(1 - e^{-t/RC}), while current decreases exponentially: I(t)=Imaxe−t/RCI(t) = I_{\text{max}} e^{-t/RC}.
The accumulating charge creates a counter-voltage that reduces the net potential driving current through the series resistor.
3
Determine the steady-state circuit values after five time constants (t≥5τt \ge 5\tau).
The capacitor reaches approximately 99.3%99.3\% of full charge. The voltage across the capacitor equals the source voltage (VC=VsourceV_C = V_{\text{source}}), and the net voltage across the resistor becomes 0 V0\text{ V}, causing circuit current to cease entirely (I=0 AI = 0\text{ A}). The fully charged capacitor acts as an open circuit to DC.
Once the opposing electrostatic potential equals the applied DC source potential, no further charge transfer can occur.

Key Concept

DC Transient and Steady-State Behavior of Capacitors in RC Circuits
Estimated Time:1m 30s
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Topics

Capacitors

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