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Does a Capacitor Pass DC Current?

Published By the 9to5Science editorial team

At a glance

  1. 1. Use the current relation

    A capacitor’s current depends on how quickly the voltage across it changes.

  2. 2. Check the circuit state

    A DC source can produce current while the capacitor charges, but not after an ideal capacitor reaches steady state.

  3. 3. Calculate the transient

    For a series resistor and capacitor, use the RC time constant to find how current decays.

  4. 4. Allow for real components

    Real capacitors can have small leakage current, unlike the ideal model.

An ideal capacitor does not carry a steady current from a DC source after it is fully charged. It does carry current while its voltage is changing: connecting a DC supply causes a charging transient, usually limited by resistance in the circuit. Real capacitors also have a small leakage current, so their steady current is not exactly zero.

An ideal capacitor blocks steady DC current, but it does not block current at every moment when a DC source is connected. During charging, charge accumulates on the capacitor’s plates and the voltage across it changes. That changing voltage produces a current in the circuit. Once the capacitor is fully charged and its voltage stops changing, the ideal current falls to zero.

The key relation: current depends on changing voltage

A capacitor stores charge according to

Q=CV,Q=CV,

where QQ is the charge on one plate, CC is the capacitance, and VV is the voltage across the capacitor. Current is the rate at which charge changes, so differentiating gives

i=CdVdt.i=C\frac{dV}{dt}.

This relation explains the DC case directly:

  • If the voltage across the capacitor is changing, then dV/dtdV/dt is nonzero and current flows.
  • If the voltage is constant, then dV/dt=0dV/dt=0 and the ideal capacitor current is zero.

A DC supply has constant voltage, but that does not mean the capacitor voltage is instantly constant. The capacitor must first charge, and that changing voltage creates a temporary current.

What happens in a DC charging circuit?

A common circuit connects a DC source, a resistor, and a capacitor in series. The resistor limits the charging current. For a source voltage VsV_s, resistance RR, capacitance CC, and an initially uncharged capacitor, the capacitor voltage and current are

VC(t)=Vs(1−e−t/(RC)),i(t)=VsRe−t/(RC).\begin{aligned} V_C(t)&=V_s\left(1-e^{-t/(RC)}\right),\\ i(t)&=\frac{V_s}{R}e^{-t/(RC)}. \end{aligned}

The time constant is τ=RC\tau=RC. At first, the capacitor voltage is low, so the current is at its maximum, Vs/RV_s/R. As charge builds up, the capacitor voltage rises and the current decreases. After several time constants, the capacitor is close to its final voltage and the current is very small. In the ideal model, it reaches zero only in the limit as time goes to infinity.

Worked example: a 12 V source, resistor, and capacitor

Suppose a 12 V12\,\text{V} DC supply charges an initially uncharged 100 μF100\,\mu\text{F} capacitor through a 10 kΩ10\,\text{k}\Omega resistor.

First find the time constant:

τ=RC=(10,000 Ω)(100×10−6 F)=1 s.\tau=RC=(10{,}000\,\Omega)(100\times10^{-6}\,\text{F})=1\,\text{s}.

The initial current is

i(0)=VsR=12 V10,000 Ω=1.2 mA.i(0)=\frac{V_s}{R}=\frac{12\,\text{V}}{10{,}000\,\Omega}=1.2\,\text{mA}.

After one time constant, t=1 st=1\,\text{s}, the current is

i(1 s)=1.2 mA e−1≈0.44 mA.i(1\,\text{s})=1.2\,\text{mA}\,e^{-1}\approx0.44\,\text{mA}.

The capacitor voltage at that time is

VC(1 s)=12(1−e−1) V≈7.6 V.V_C(1\,\text{s})=12(1-e^{-1})\,\text{V}\approx7.6\,\text{V}.

The capacitor is still charging, so current still flows. After a long time, its voltage approaches 12 V12\,\text{V} and the ideal current approaches zero.

Does current pass through the capacitor itself?

In a basic circuit model, charge carriers do not travel through the insulating dielectric between the plates. Instead, charge flows along the wires onto one plate and away from the other while the capacitor charges. The changing electric field between the plates is associated with what is called displacement current in electromagnetic theory; this makes current consistent through the circuit during the transient, without requiring conduction electrons to cross the dielectric.

What if a capacitor is connected directly to a DC source?

The ideal equation i=C dV/dti=C\,dV/dt says that an instantaneous voltage step would require an impulse of current. That is a mathematical idealization, not a realistic finite current. In a physical circuit, the source has finite rise time and internal resistance, and wires and components have parasitic resistance and inductance; these factors determine the actual current transient. The capacitor may also have a maximum permissible rate of voltage change (a dV/dtdV/dt rating) or a pulse-current rating, which must not be exceeded. A resistor is normally included to limit the current.

Common mistakes

  • Saying a capacitor never carries current: It carries current during charging or discharging, whenever its voltage changes.
  • Saying current continues indefinitely with a DC supply: In an ideal circuit with a constant DC voltage, the charging current dies away to zero at steady state.
  • Confusing the dielectric with an open wire during the transient: No conduction charge crosses an ideal dielectric, but changing electric fields accompany the transient current in the external circuit.
  • Expecting a real capacitor to have exactly zero steady current: Real capacitors have finite insulation resistance, so a small leakage current can flow. Its size depends on the capacitor type, voltage, temperature, age, and condition.

References

  1. University Physics Volume 2, Section 10.6: RC Circuits — OpenStax
electric circuitselectric currentcapacitance

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