Introduction to Electric Circuits

8.3 Circuits containing resistance and capacitance 185

i= vc/R = (V/R)exp (-I/CR)

and since V/R = I then

i = I exp (-t/CR) (8.20)

The voltage across the resistor decays as the current through it decays so that

VR = iR in magnitude. Thus

VR = RI exp (-t/CR) - V exp (-t/CR)

Also, VR + VC = 0 SO that the voltage across the resistor is given by

VR = --Vc = -V exp (-t/cn) (8.21)

The waveforms described by Equations (8.19), (8.20) and (8.21) are shown in

Fig. 8.18(a), (b) and (a), respectively. Remember, i is discharging current.

Figure 8.18

-V

/C

(a)

t t

(b)

Example 8.7

A 10 ~F capacitor is charged to 150 V and then allowed to discharge through its

own leakage resistance. After 200 s, it is observed that the voltage, measured on

an electrostatic voltmeter, has fallen to 90 V. Calculate the leakage resistance

of the capacitor.

Solution

From Equation (8.19) we have that Vc = V exp (-t/CR). In this case R is the
leakage resistance of the capacitor. The resistance of the electrostatic voltmeter
which is in parallel with R can be taken to be infinite. Putting in the values we
have 90 = 150 exp [-200//(10 • 10 -6 • R)], so

-200/(10 • 10 -6 • R) = In (90/150)

and

R : -200/[10.~ X 10-6X In (90/150)] : 39.15 MD,