RC Circuits: Charging and Discharging a Capacitor

Capacitors Do Not Charge Instantly

A camera flash takes a few seconds to "charge up" before it can fire. Some car interior lights fade out slowly after the door closes. Windscreen wipers can pause for a set delay between sweeps. Each of these uses a capacitor charging or discharging through a resistor, in what is called an RC circuit. The resistor slows the flow of charge, and the result is a smooth exponential curve with a characteristic time, the time constant. This tutorial covers RC circuits for A-level-style WAEC questions and first-year university physics: charging and discharging equations, the time constant RC, the half-time, the current during charging, energy, how to measure the time constant from a graph, and practical uses, with worked examples. It builds on the tutorial on capacitors.

Discharging a Capacitor

Suppose a capacitor of capacitance C is charged to a voltage V0, so it holds charge Q0 = CV0. It is then connected across a resistor R. Charge flows from one plate, through the resistor, to the other plate.

At first the voltage is large, so the current is large and the charge drops quickly. As the charge falls, the voltage falls, the current falls, and the discharge slows down. The rate of loss of charge is always proportional to the charge remaining, which is the signature of exponential decay:

Q = Q0 e^(−t/RC)

Because V = Q/C and I = V/R, the voltage and current follow exactly the same shape:

V = V0 e^(−t/RC) and I = I0 e^(−t/RC), where I0 = V0/R.

The current during discharge flows in the opposite direction to the charging current.

Deriving the equation

Around the circuit, the capacitor voltage drives the current through the resistor: Q/C = IR. Since the charge is decreasing, I = −dQ/dt. So dQ/dt = −Q/(RC). This is a first-order differential equation whose solution is Q = Q0 e^(−t/RC), as shown in the tutorial on first-order differential equations.

The Time Constant

The product RC has units of time (ohms × farads = seconds) and is called the time constant, τ (tau):

τ = RC

After one time constant of discharging, t = RC, the charge has fallen to e⁻¹ ≈ 0.37 (37%) of its initial value.
After 2RC it is about 13.5%, and after 5RC it is under 1%. In practice, a capacitor is treated as fully discharged after about 5 time constants.
A larger R or a larger C gives a slower discharge.

Worked example 1

Find the time constant for a 470 μF capacitor discharging through a 100 kΩ resistor.

τ = RC = 100 × 10³ × 470 × 10⁻⁶ = 47 s.

The half-time

The time for the charge (or voltage, or current) to halve is constant, like the half-life in radioactivity:

t½ = RC ln 2 ≈ 0.693 RC

Worked Example: Discharge

A 1000 μF capacitor is charged to 9.0 V and then discharged through a 4.7 kΩ resistor. Find (a) the time constant, (b) the initial current, (c) the time for the voltage to fall to 3.0 V, and (d) the half-time.

(a) τ = 4.7 × 10³ × 1000 × 10⁻⁶ = 4.7 s.

(b) I0 = V0/R = 9.0 ÷ 4700 ≈ 1.9 mA.

(d) t½ = 4.7 × ln 2 ≈ 3.3 s.

Charging a Capacitor

Now connect an uncharged capacitor C in series with a resistor R across a battery of e.m.f. V0. At the instant of switching on, the capacitor has no charge and no voltage across it, so the full battery voltage is across the resistor and the current is maximum, I0 = V0/R.

As charge builds up on the plates, the capacitor's voltage rises and opposes the battery. The voltage across the resistor falls, so the current falls. Charging slows down as it approaches the full charge Q0 = CV0:

Q = Q0 (1 − e^(−t/RC)) and V_C = V0 (1 − e^(−t/RC))

The current, however, still decays:

I = I0 e^(−t/RC)

After one time constant, the capacitor has reached 1 − e⁻¹ ≈ 63% of its final charge. After 5RC it is over 99% charged.

Worked example 2

A 100 μF capacitor is charged through a 20 kΩ resistor from a 12 V supply. Find (a) the time constant, (b) the initial current, (c) the capacitor voltage after 2 s and after 5 s, and (d) the current after 2 s.

(a) τ = 20 × 10³ × 100 × 10⁻⁶ = 2.0 s.

(b) I0 = 12 ÷ 20,000 = 0.6 mA.

(d) I = 0.6 × e⁻¹ ≈ 0.22 mA.

At every moment the resistor voltage plus the capacitor voltage equals 12 V. After 2 s, the resistor has 0.22 mA × 20 kΩ ≈ 4.4 V, and 4.4 + 7.6 = 12 V.

Graphs of Charging and Discharging

Discharging: the Q–t, V–t and I–t graphs all start high and decay exponentially towards zero.
Charging: the Q–t and V_C–t graphs rise steeply at first, then level off towards Q0 and V0. The I–t graph decays exponentially from I0, just as in discharging.
The area under an I–t graph equals the charge that has flowed, because Q = It for each small time interval.
The gradient of a Q–t graph at any time equals the current at that time.

Measuring the Time Constant

In the laboratory, a capacitor is discharged through a known resistor while the voltage is recorded at regular intervals, using a voltmeter and stopwatch or a data logger.

Read from the curve: find the time at which V has fallen to 0.37 V0. That time is RC.
Straight-line graph: taking natural logs of V = V0 e^(−t/RC) gives ln V = ln V0 − t/RC. A graph of ln V against t is a straight line with gradient −1/RC and intercept ln V0. This method uses all the readings and is more accurate.
Half-time: measure the time to halve, then divide by 0.693.

Precautions: use a high-resistance (digital) voltmeter so it does not discharge the capacitor itself, and choose R and C so the time constant is at least 20 to 30 seconds when timing by hand.

Energy in an RC Circuit

A capacitor charged to V0 stores energy E = ½CV0². During discharge, all of this energy is dissipated as heat in the resistor.

During charging, the battery supplies energy Q0V0 = CV0², but only half of this is stored in the capacitor. The other half is dissipated in the resistor, whatever the value of R. For example, a 1000 μF capacitor charged to 9 V stores ½ × 1000 × 10⁻⁶ × 81 ≈ 0.041 J, and the same amount was lost as heat while charging it.

Uses of RC Circuits

Timing circuits: the delay before a security light turns off, wiper delays, and the 555 timer chip all use the time taken for a capacitor to charge to a threshold voltage.
Camera flash: a capacitor charges slowly from a small battery, then discharges rapidly through the flash tube (small R, short time constant) to give a brief, intense burst of light.
Smoothing: after rectification, a capacitor discharges slowly between peaks, smoothing the output of a power supply. A large RC gives smoother d.c.
Filters: RC combinations block or pass signals of certain frequencies in audio and radio circuits.
Defibrillators store energy in a capacitor and release it as a controlled pulse through the patient's chest.

RC Circuits: Practice Questions

Find the time constant of a 220 μF capacitor and a 10 kΩ resistor.
That capacitor is charged to 10 V and discharged through the resistor. What is its voltage after 4.4 s?
To what percentage of the final voltage does a capacitor charge in one time constant?
A capacitor's voltage halves every 10 s during discharge. Find the time constant.
A graph of ln V against t for a discharging capacitor has a gradient of −0.05 s⁻¹. Find RC, and find C if R = 40 kΩ.

Answers: 1. RC = 2.2 s. 2. 4.4 s is two time constants, so V = 10e⁻² ≈ 1.35 V. 3. About 63%. 4. RC = 10 ÷ ln 2 ≈ 14.4 s. 5. RC = 1/0.05 = 20 s, so C = 20 ÷ 40,000 = 5.0 × 10⁻⁴ F = 500 μF.

Conclusion

In an RC circuit the capacitor charges and discharges exponentially, never instantly. The time constant τ = RC sets the pace: after one time constant a discharging capacitor keeps 37% of its charge, and a charging capacitor reaches 63% of its final charge. The current always decays as I0 e^(−t/RC). A graph of ln V against t gives a straight line of gradient −1/RC, which is the best way to measure the time constant, and RC circuits are the basis of timers, flashes, smoothing circuits and filters.