Charging Capacitors

The reverse of discharging is charging. Connect an uncharged capacitor to a battery through a resistor and the capacitor fills up — slowly at first, then faster, then tailing off as it approaches the supply voltage. This lesson derives the charging equations, shows how they differ subtly from the discharge equations, and shows how to read charging curves in exam questions.

This lesson continues OCR H556 Module 6.1.

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The Charging Circuit

graph LR
    B((V₀ battery)) --- R1[Resistor R] --- C1[Capacitor C] --- B

When the switch closes at t = 0, the capacitor is empty: its voltage is 0 and the full e.m.f. V₀ appears across the resistor. The initial current is therefore the maximum possible,

I₀ = V₀ / R

As charge builds up on the capacitor, the voltage across it rises and the voltage across the resistor falls. The current drops because it is proportional to the resistor voltage. Eventually the capacitor voltage equals V₀, no more current flows, and the capacitor is fully charged.

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Deriving the Charging Equations

By Kirchhoff's voltage law around the loop,

V₀ = v_R + v_C = iR + q/C

With i = dq/dt this becomes

V₀ = R(dq/dt) + q/C

Rearranging,

dq/dt = (CV₀ − q) / (RC)

Let Q₀ = CV₀ be the final charge on the capacitor. Substituting u = Q₀ − q gives du/dt = −u/(RC), the familiar exponential decay equation, whose solution is u = Q₀ e^(−t/RC). Therefore

q(t) = Q₀ (1 − e^(−t/RC))

and correspondingly

v_C(t) = V₀ (1 − e^(−t/RC))

The current, however, decays from I₀ just as in a discharge circuit:

i(t) = I₀ e^(−t/RC) with I₀ = V₀ / R

This difference is important:

• Charge and voltage across the capacitor: 1 − e^(−t/RC) — rising from 0 to the final value.
• Current: e^(−t/RC) — falling from I₀ to 0.

Time	Capacitor voltage	Current
0	0	I₀
τ	0.632 V₀	0.368 I₀
2τ	0.865 V₀	0.135 I₀
3τ	0.950 V₀	0.050 I₀
5τ	0.993 V₀	0.007 I₀

The capacitor is essentially fully charged after about 5 time constants.

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Charging Curves

V_c
V₀ |      ,,,***************
   |  ,**
   | *
   |*
   |
 0 +-----------------------> t
   0   τ   2τ   3τ   4τ   5τ

I
I₀ |*
   | *
   |  *
   |   **
   |     ****
 0 |--------********----------> t
   0   τ   2τ   3τ   4τ   5τ

Notice the two curves are mirror images about the horizontal line V₀/2: as the capacitor voltage rises from 0 to V₀, the current falls from I₀ to 0, and at all times

V₀ = v_C + v_R = v_C + iR

confirming energy conservation.

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Worked Example — Charging Through a Resistor

A 220 μF capacitor is charged from 0 V through a 10 kΩ resistor by a 9.0 V battery. Calculate (a) the initial current, (b) the time constant, (c) the voltage after 3.0 s, (d) the current at the same moment, (e) the time for the voltage to reach 8.0 V.

(a) Initial current

I₀ = V₀/R = 9.0 / 10 000 = 9.0 × 10⁻⁴ A = 0.90 mA

(b) Time constant

τ = RC = 10 × 10³ × 220 × 10⁻⁶ = 2.2 s

(c) Voltage after 3 s