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.

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:

where V

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

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

When t = RC, V = V

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

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 = 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 10

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 V

Therefore:

ln(1/0.8) = 20 000 x 0.0001

This gives t= 2xln(1/0.8) = 0.45 s.

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.

**WarningSome 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.**

1. A 4000 μF capacitor is charged through a 2.5 kΩ resistor using a 15 V supply.

Calculate:

(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 = V

(b) 10/15 = 1 – e

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.

Calculate:

(a) the charging current after 2.5s

(b) the charge on the plates after 2.5 s

(a) Initial charging current (I

Current after 2.5 s = I

(b) Q = Q

Final charge (Q

Charge after 2.5 s = 0.012[1 - 0.287] = 0.012x0.713 = 0.0086 C