9478 / 2027
Circuits overview

Topic 8 of 9

RC charging and discharging

Trace the switch connection first. Then keep one sign convention as the capacitor charges or discharges, so the current graph agrees with the change in its plate charge.

Use a resistor R between switch terminal S and capacitor plate B. The other plate connects to common return node C. The source positive terminal is A, at Es above C. Here the letter C labels a node in the drawing; the italic quantity C in an equation is capacitance.

Charge: connect the resistor to the source

Charge: connect the resistor to the sourceThe SPDT common S feeds a two-hundred-kilohm resistor leading to B. The ten-microfarad capacitor joins B to return node C. The ten-volt source joins C to A. The blade joins S to A; the lower C contact is open. Conventional charging current flows from S towards B. No conducting line crosses the capacitor gap. In both experiments, positive I is defined from S towards B and positive capacitor voltage is V at B minus V at C.+-ASBCC200 kΩ10 µF10 VActual current: S towards BFixed reference: I positive S to B

The current reference stays S to B in both experiments. B/C names the capacitor terminals; the value 10 microfarad is its capacitance. Each experiment starts its own clock.

Discharge: isolate the positive source terminal

Discharge: isolate the positive source terminalThe SPDT common S feeds a two-hundred-kilohm resistor leading to B. The ten-microfarad capacitor joins B to return node C. The ten-volt source joins C to A. The blade joins S to C; A is left open. The capacitor discharges through the resistor and the return wire. Actual conventional current goes from B towards S, opposite to the fixed positive current reference. In both experiments, positive I is defined from S towards B and positive capacitor voltage is V at B minus V at C.+-ASBCC200 kΩ10 µF10 VA is isolatedActual current: B towards SFixed reference: I positive S to B

The current reference stays S to B in both experiments. B/C names the capacitor terminals; the value 10 microfarad is its capacitance. Each experiment starts its own clock.

Charging connects S to A; discharging connects S to C and isolates A. The fixed positive current reference is S towards B in both states. Actual discharge current through the resistor runs B towards S.

Define Q as the charge on plate B and VC = VB - Vnode C = Q/C. Define positive I through the resistor from S towards B. Then I = dQ/dt: positive current increases Q and negative current reduces it. In the examples Q stays positive while falling during discharge.

A switch that simply disconnects every conducting path does not discharge an ideal capacitor. The discharge state shown provides a closed resistor-capacitor loop without connecting the source across a short circuit.

Charging from zero

Take an initially uncharged capacitor, a constant ideal source e.m.f. Es, fixed R and C, and negligible leakage and meter loading. At each instant, Es = IR + Q/C. Initially the capacitor p.d. is zero, so current is Es/R. As Q and its p.d. grow, less voltage remains across R and current decreases.

τ = RC
Q = CEs(1 - e-t/τ)
VC = Es(1 - e-t/τ)
I = (Es/R)e-t/τ

Charge and capacitor p.d. rise towards their final values; charging current decreases towards zero. The source voltage stays constant in this model. A finite resistor prevents capacitor voltage from jumping instantaneously, though the current can change abruptly when the switch changes a connection.

Discharging from a specified initial voltage

Disconnect the source and connect R across a capacitor initially at V0, with Q0 = CV0. The stored charge now drives current through R. Keep the same positive current reference:

Q = Q0e-t/τ
VC = V0e-t/τ
I = -(V0/R)e-t/τ

The current is negative and approaches zero from below. Its magnitude decreases. Calling it negative does not mean that the capacitor has reversed plate polarity. If a question chooses the opposite current reference, reverse the current sign while keeping the physical behaviour unchanged.

In the general rising form x = x0(1 - e-t/τ), x0 is the final limit when the initial value is zero. In the decreasing form x = x0e-t/τ, x0 is the initial value. Check which quantity the symbol describes.

Compare matched charge and discharge models

Let Es = V0 = 10 V, R = 200 kΩ and C = 10 µF. These are two separate runs, each with its own t = 0: charging starts uncharged, while discharging starts fully at 10 V.

τ = (200000)(10 × 10-6) = 2.0 s
CEs = CV0 = 100 µC
Es/R = V0/R = 50 µA

Charge on the B-side plate

Charge on the B-side plateCharge on the B-side plate. The solid blue curve is the charging experiment and dashed brown curve the separate discharging experiment; both start their own clocks at zero. The horizontal scale is zero to ten seconds, and the time constant is two seconds. Charging starts at zero and approaches 100; discharging starts at 100 and approaches zero, in the labelled vertical units. Dots mark each curve at two seconds.02468100255075100Charge Q / µCElapsed time t / sτ = 2.00 s for both experiments

Solid: charging. Dashed: discharging. These are separate experiments with separate time zeros; neither exponential reaches its final limit at a finite time.

Capacitor p.d. between B and C

Capacitor p.d. between B and CCapacitor p.d. between B and C. The solid blue curve is the charging experiment and dashed brown curve the separate discharging experiment; both start their own clocks at zero. The horizontal scale is zero to ten seconds, and the time constant is two seconds. Charging starts at zero and approaches 10; discharging starts at 10 and approaches zero, in the labelled vertical units. Dots mark each curve at two seconds.02468100246810Capacitor p.d. / VElapsed time t / sτ = 2.00 s for both experiments

Solid: charging. Dashed: discharging. These are separate experiments with separate time zeros; neither exponential reaches its final limit at a finite time.

Signed current with one reference direction

Signed current with one reference directionSigned current with one reference direction. The solid blue curve is the charging experiment and dashed brown curve the separate discharging experiment; both start their own clocks at zero. The horizontal scale is zero to ten seconds, and the time constant is two seconds. Current is positive from S to B: charging starts at plus fifty microampere and discharging at minus fifty. Both approach zero without changing sign. Dots mark each curve at two seconds.0246810-50-2502550Current I / µAElapsed time t / sτ = 2.00 s for both experiments

Solid: charging, I > 0. Dashed: discharging, I < 0. Both use the same S-to-B reference; these are separate experiments.

The separate quantity plots compare the two supplied runs using the same time scale. Charge and voltage remain positive, while discharge current is negative under the fixed S-to-B reference. The initial conditions belong to separate runs, not one uninterrupted curve.
Charging model: values at zero, one and two time constants
t / sVC / VQ / µCI / µA
000+50
2.06.32163.21+18.394
4.08.64786.47+6.767
Discharging model: the same times with a charged initial state
t / sVC / VQ / µCI / µA
010100-50
2.03.67936.79-18.394
4.01.35313.53-6.767

Extra digits show the model comparison; final experimental precision would depend on the actual component values and readings. Use micro = 10-6 consistently for both µF and µA.

What the time constant does and does not mean

At t = τ, a decaying quantity is e-1 = 0.368 of its initial value, while a rising quantity is 1 - e-1 = 0.632 of its final value. RC has units of time:

Ω F = (V/A)(C/V) = C/A = s

The time constant is not a half-life and is not the time for mathematically complete charging. At 5τ, charging reaches about 99.33% of its final value. To find a different fraction, solve the exponential:

Discharge to half: ½ = e-t/τ
t = τ ln 2 = 1.386 s
Charge to 90%: 0.10 = e-t/τ
t = -τ ln(0.10) = 4.605 s

Doubling R alone makes τ = 4.0 s and halves the initial charging current to 25 µA, while final charge stays 100 µC. Doubling C alone also makes τ = 4.0 s, but now final charge is 200 µC and initial current remains 50 µA. An equal change in time constant does not imply identical changes in every quantity.

Energy and the actual resistor path

For charging from zero in this ideal constant-voltage model, the source supplies energy EsQfinal = CEs2. The capacitor stores ½CEs2; the rest is transferred to internal energy in the resistance. Here the source supplies 1.00 mJ, the capacitor stores 0.500 mJ and the resistor receives 0.500 mJ. In a complete discharge through R, the stored energy is transferred through that resistance.

Choose R from the path present in the relevant state. Source internal resistance can add to a charging path while being absent from the separate discharge loop. A finite-resistance voltmeter adds a parallel path; leakage can also matter. An arbitrary product of a whole network's named resistance and capacitance is not automatically its time constant.

Optional check A 10.0 microfarad capacitor initially at 10.0 V discharges through 200 kilohm. Current I is still defined positive towards its initially positive B plate. What are the values at t = tau = 2.00 s?
A 10.0 microfarad capacitor initially at 10.0 V discharges through 200 kilohm. Current I is still defined positive towards its initially positive B plate. What are the values at t = tau = 2.00 s?