Introduction to Electric Circuits

184 Transient analysis

The discharge of a capacitor

Once the capacitor in Fig. 8.13 has been charged to V volts, it will remain at that

voltage so long as the input voltage remains at V volts. Even if the supply is

disconnected, as shown in Fig. 8.17(a), the capacitor (provided that it has no

leakage resistance between its plates) will remain charged to V volts. If,

however, the terminals A and B are now short circuited as shown in Fig. 8.17(b)

the capacitor will begin to release its stored energy. The current will now be

leaving its positive plate.

R J,,C i C

Ao | I " ~it- ..... ~" +ll- II

VC VC

Bo

Figure 8.17

R

A l 1

MR

B

(a) (b)

Applying KVL to the circuit of Fig. 8.17(b) we see that

Vc - iR = 0 (8.17)

Now i= -(dq/dt), the minus sign indicating that the charge on the plates is

decaying. Since Q - CV then q = Cvc and

• (dq/dt) = - C(dvc/dt)
  Substituting in Equation (8.17) we get
  v c - [-C(dvc/dt)]R = 0
  Vc = -CR(dvc/dt)
  Separating the variables we have dvc/vc = -dt/CR. Integrating gives
  In Vc + C = -t/CR (8.18)
  where C is the constant of integration. At t- 0, Vc - V so that In V + C = 0
  and C - -ln V. Substituting in Equation (8.18) we obtain
  In Vc - In V = -t/CR ~ In (vc/V) = -t/CR

Taking antilogs

vc/V- exp (-(t/CR))

Finally

Vc = V exp (-t/CR) (8.19)

The current at any instant is given by