Charging an RC circuit

Put an emf ε, a resistor R and an uncharged capacitor C in one series loop, and close the switch at t = 0. Kirchhoff's loop rule is a statement about a moment, not about a steady state, so it applies here unchanged: travel round the loop in the direction the current flows and the potential changes must sum to zero. Across the battery you gain ε, across the resistor you drop iR, and across the capacitor you drop q/C, where q is the charge already sitting on the plate you meet first. So ε − iR − q/C = 0. Two terms in that equation are functions of time, and that is both the difficulty and the interest. At the first instant q = 0, so the capacitor drops nothing and the loop rule reads ε = iR. Long afterwards the current has stopped, so the resistor drops nothing and it reads ε = q/C. The circuit must travel from one of those statements to the other, and the loop rule by itself cannot say how quickly.

The current here is nothing other than the rate at which charge lands on the capacitor plate, i = dq/dt. Substituting that removes i and leaves a single unknown function: R dq/dt = ε − q/C, or R dq/dt = (Cε − q)/C. Read the right-hand side as a net driving voltage — the battery pushes with ε, the charge already stored pushes back with q/C, and whatever is left is what R has to work with. Separate the variables: dq/(Cε − q) = dt/(RC). Integrate from the closed switch, q = 0 at t = 0, out to a general time, and the left side gives a logarithm: ln[Cε/(Cε − q)] = t/(RC). Exponentiate and rearrange for the transient, q(t) = Cε(1 − e^(−t/RC)), then differentiate it back for the current, i(t) = (ε/R) e^(−t/RC). Notice what the initial condition bought: q(0) = 0 is what fixed the amplitude at Cε. A capacitor that already carried charge would give the same exponential with a different amplitude in front of it.

At t = 0 the plates are bare, so v_C = 0 and the capacitor behaves exactly like a piece of wire: the whole of ε appears across R, and the current is as large as it will ever be, i(0) = ε/R. With ε = 12.0 V and R = 47 kΩ that is 0.255 mA, and changing C does not move it at all. As t → ∞ the exponential dies, dq/dt → 0, and a branch carrying no current behaves like a break: the resistor drops nothing, all of ε appears across the capacitor, and q → Cε. With C = 2.2 μF that final charge is 26.4 μC, and changing R does not move it at all. These two limits are worth more than the formula, because you can write them down for any RC network in a single line — an uncharged capacitor is a short circuit, a fully charged one is an open circuit — and they hand you the start and the end of the transient before you solve anything.

Check first that RC is a time: Ω × F = (V A⁻¹)(C V⁻¹) = C A⁻¹ = s. It is a duration, and it is the only duration the circuit contains, since no other combination of ε, R and C has the dimension of time. Put t = τ into the solution: the exponential becomes e⁻¹ = 0.368, so q = 0.632 Cε. One time constant reaches 63.2% of the final charge whatever the component values; two reach 86.5%, three 95.0%, five 99.3%, which is why 5RC is the usual engineering shorthand for finished. The time constant also has a geometric meaning. The initial slope of q(t) is i(0) = ε/R, and if charge kept arriving at that rate it would reach Cε after a time Cε ÷ (ε/R) = RC. Draw the tangent at the origin and it cuts the final level exactly at t = τ. That is the construction to use on an oscilloscope trace, because the curve itself flattens far too gradually to read a time off, while its initial tangent does not.

Rearrange the loop rule and read it as it stands: v_R + v_C = ε at every instant. That single constraint means the two curves are not independent, but reflections of one another in the line v = ε/2. The capacitor voltage v_C = ε(1 − e^(−t/RC)) grows from 0 towards ε; the resistor voltage v_R = iR = ε e^(−t/RC) decays from ε towards 0; their sum is ε, with no t left in it. Take ε = 12.0 V, R = 47 kΩ and C = 2.2 μF, so τ = 0.103 s. At t = τ, v_C = 12.0 V × 0.6321 = 7.59 V and v_R = 12.0 V × 0.3679 = 4.41 V, and those add to 12.00 V. At t = 3τ they are 11.40 V and 0.60 V. Since v_R and i differ only by the constant factor R, putting a scope across the resistor is the cleanest way to record the current without breaking the circuit, and it is the sharper trace too: it falls from its largest value instead of creeping up to its last few percent.

Four checks cost a minute between them and catch almost every slip. Substitute back: with q = Cε(1 − e^(−t/RC)) the derivative is dq/dt = (ε/R)e^(−t/RC), so R dq/dt + q/C = ε e^(−t/RC) + ε(1 − e^(−t/RC)) = ε, which is the original equation with no stray t surviving. Check the two ends: q(0) = Cε(1 − 1) = 0, as the switch condition demands, and q(∞) = Cε, as the long-time argument demands. Check the total charge that passed through the resistor: ∫₀^∞ (ε/R) e^(−t/RC) dt = (ε/R)(RC) = Cε, the same final charge, as it has to be, since every coulomb that reached the plates went through the resistor first. Check the exponent last: t/RC is a time divided by a time, so the exponential takes a pure number, as it must. A solution that survives all four is very unlikely to be wrong, and any one of them catches a stray factor of R or C.