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Mathematical treatment of charging and discharging a capacitor


The area under the current-time discharge graph gives the charge held by the capacitor. The gradient of the charge-time graph gives the current flowing from the capacitor at that moment.

Discharge of a capacitor through a resistor

In Figure 1 let the charge on a capacitor of capacitance C at any instant be q, and let V be the potential difference across it at that instant.

The current (I) in the discharge at that instant is therefore:
I = - dq/dt

But V = IR and q = CV so dq/dt = d(CV)/dt = C dV/dt
Therefore we have V = -CR dV/dt Rearranging and integrating gives:

Capacitor discharge (voltage decay): V = Voe-(t/RC)

where Vo is the initial voltage applied to the capacitor. A graph of this exponential discharge is shown below in Figure 2.

Since Q = CV the equation for the charge (Q) on the capacitor after a time t is therefore:

Capacitor discharge (charge decay): Q = Qoe-(t/RC)

V = Voe-(t/RC)    and also I = Ioe-(t/RC)    Q = Qoe-(t/RC)

You should realise that the term RC governs the rate at which the charge on the capacitor decays.

When t = RC, V = Vo/e = 0.37 Vo and the product RC is known as the time constant for the circuit. The bigger the value of RC the slower the rate at which the capacitor discharges.

The value of C can be found from this discharge curve if R is known.

Example problems
1. A capacitor of 1000 μF is with a potential difference of 12 V across it is discharged through a 500 Ω resistor.
Calculate the voltage across the capacitor after 1.5 s

V = Voe-(t/RC) so V = 12e-1.5/[500 x 0.001] = 0.6 V

2. A capacitor is discharged through a 10 MΩ resistor and it is found that the time constant is 200 s.
Calculate the value of the capacitor.
RC= 200
Therefore C = 200/10 x 106 = 20 μF.

3. Calculate the time for the potential across a 100 ΨF capacitor to fall to 80 per cent of its original value if it is discharged through a 20 kΩ resistor.
V = 0.8 Vo. Therefore 0.8 = et/20000x0.0001
ln(1/0.8) = 20 000 x 0.0001
This gives t= 2xln(1/0.8) = 0.45 s.

Charging a capacitor

When a capacitor (C) is being charged through a resistance (R) to a final potential Vo the equation giving the voltage (V) across the capacitor at any time t is given by:

Capacitor charging (potential difference): V = Vo[1-e-(t/RC)]

and the variation of potential with time is shown in Figure 2.

As the capacitor charges the charging current decreases since the potential across the resistance decreases as the potential across the capacitor increases.

Figure 4 shows how both the potential difference across the capacitor and the charge on the plates vary with time during charging.

The charging current would be given by the gradient of the curve in Figure 2 at any time and the graph of charging current against time is shown in Figure 3.

The area below the current-time curve in both charg ing and discharging represents the total charge held by the capacitor.

Some badly made power supplies have a capacitor connected across their outputs and so remain live even after the power supply has been switched off. Always be careful when handling apparatus containing capacitors.

Example problems
1. A 4000 μF capacitor is charged through a 2.5 kΩ resistor using a 15 V supply.
(a) the potential difference across the plates after 5s
(b) the time taken for the potential difference across the plates to reach 10 V

(a) V = Vo[1 e-t/RC] = 15[1 e-5/(2500x0.004)] = 5.9 V
(b) 10/15 = 1 e-t/RC and so e-t/RC = 0.33 therefore t/RC = 1.099
This gives: t = 1.0986x2500x0.004 = 11 s

2. A 2000 μF capacitor is charged through a 1 kΩ resistor using a 6 V supply.
(a) the charging current after 2.5s
(b) the charge on the plates after 2.5 s

(a) Initial charging current (Io) = V/R = 6/1000 = 6 mA
Current after 2.5 s = Ioe-t/RC = 6x10-3xe-(2.5/1000x0.002) = 6x10-3 x 0.287 = 1.7 mA
(b) Q = Qo [1 - e-t/RC]
Final charge (Qo) = CV = 0.002x6 = 0.012 C
Charge after 2.5 s = 0.012[1 - 0.287] = 0.012x0.713 = 0.0086 C

© Keith Gibbs 2013