Physics · Electric Fields · A2 Level

Charging Capacitors

Charging a capacitor through a resistor is the mirror image of discharging: instead of dying away, charge, potential difference and current all creep toward a final value, fastest at the start and ever more slowly thereafter, without quite reaching it. The same time constant that governed discharge governs charging too.

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Charging through a resistor

Connect an uncharged capacitor in series with a resistor and a fixed-voltage supply, and charge begins to build on the plates. As charge $Q$ builds up, the potential difference $V=Q/C$ across the capacitor rises toward the supply e.m.f. $\varepsilon$, which shrinks the potential difference driving current round the resistor and so reduces the charging current — so, just as in discharge, the rate of charging continuously slows. This produces a rising exponential: charge and potential difference climb quickly at first, then level off, approaching their final values ever more slowly but never quite reaching them.

DerivationQ = Q₀(1 − e^(−t/RC)), from the charging equation

At any instant during charging, applying Kirchhoff's voltage law around the circuit: $\varepsilon = IR + V$, where $\varepsilon$ is the supply e.m.f., $I$ the charging current, and $V=Q/C$ the potential difference across the capacitor.

The supply e.m.f. splits between driving current through the resistor and building up potential difference on the capacitor.

The charging current is the rate at which charge builds up on the plates, $I=\dfrac{dQ}{dt}$. Substituting into the loop equation: $\varepsilon = R\dfrac{dQ}{dt} + \dfrac{Q}{C}$.

This is a first-order differential equation whose solution (beyond A-level to derive fully, but standard to quote and use), given $Q=0$ at $t=0$, is $Q=Q_0\left(1-e^{-t/RC}\right)$, where $Q_0=C\varepsilon$ is the final (maximum) charge as $t\to\infty$.

The same exponential time-dependence as discharge, but now building up toward a limit rather than decaying away from one.

Dividing both sides by $C$ gives the equivalent statement in potential difference: $V=V_0\left(1-e^{-t/RC}\right)$, where $V_0=\varepsilon$. The charging current, by contrast, still decays exponentially from its initial maximum: $I=I_0e^{-t/RC}$.

Current falls as the capacitor becomes more charged and offers more opposing potential difference to the supply — it's driven by the shrinking gap $\varepsilon-V$, which itself decays exponentially.

▪ derived

The time constant means the same thing here

The time constant $\tau=RC$ sets the timescale of charging exactly as it does discharging. Substituting $t=\tau$ into $V=V_0(1-e^{-t/RC})$ gives $V=V_0(1-e^{-1})\approx0.63V_0$ — after one time constant, the capacitor has charged to about 63% of its final value. A capacitor is usually considered practically fully charged after about five time constants, by which point it's within 1% of $V_0$.

Charge and potential difference rise as identically-shaped curves, just rescaled, while current still decays — each graph below is marked at one time constant $\tau=RC$.

Animated graph of charge Q against time t during charging. The curve starts at zero and rises steeply at first, then levels off and approaches a maximum value asymptotically, with a dashed marker showing the value at one time constant. — $Q$ against $t$ during charging: $Q=Q_0(1-e^{-t/RC})$, reaching about 63% of its final value after one time constant $\tau=RC$.

Animated graph of potential difference V against time t during charging. The curve starts at zero and rises steeply at first, then levels off and approaches a maximum value asymptotically, with a dashed marker showing the value at one time constant. — $V$ against $t$ during charging: $V=V_0(1-e^{-t/RC})$, the same rising-exponential shape as $Q$, since $V=Q/C$.

Animated graph of current I against time t during charging. The curve starts at its maximum value and decays exponentially toward zero, with a dashed marker showing the value at one time constant. — $I$ against $t$ during charging: $I=I_0e^{-t/RC}$ — unlike $Q$ and $V$, the charging current decays, reaching about 37% of its initial value after one time constant.

Charge and voltage rise, but current still falls

It's easy to assume everything behaves the same way during charging, but the charging current $I=I_0e^{-t/RC}$ is a decaying exponential, identical in shape to discharge current — only $Q$ and $V$ show the rising form. The current is largest at the instant charging begins (when the capacitor has zero opposing p.d.) and falls steadily to zero as the capacitor approaches full charge.

Key takeaways

Charging: $Q=Q_0(1-e^{-t/RC})$ and $V=V_0(1-e^{-t/RC})$ rise as "rising exponentials", approaching their final values asymptotically but never quite reaching them.
The charging current still decays exponentially, $I=I_0e^{-t/RC}$ — identical in shape to the discharge current, even though $Q$ and $V$ behave oppositely to discharge.
The time constant $\tau=RC$ has the same meaning as in discharge: after one time constant, the charging quantity has risen to about 63% of its final value; a capacitor is practically fully charged after about five time constants.
$Q_0=C\varepsilon$ and $V_0=\varepsilon$, where $\varepsilon$ is the e.m.f. of the supply charging the capacitor.

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Discharging Capacitors

Magnetic Fields