Introduction to Electric Circuits

8.3 Circuits containing resistance and capacitance 181

8.3 CIRCUITS CONTAINING RESISTANCE AND
CAPACITANCE

The sudden application of a step voltage

Let the step function shown in Fig. 8.2 be applied to the circuit of Fig. 8.13.
Applying KVL we see that

V = VR + VC (8.12)

v(

Figure 8.13

i R C

:! II

VR VC

)

Now VR = iR and from Equation (2.1), Chapter 2, i= dq/dt where i is the
current through the capacitor and q is its charge at any instant. Also, from
Equation (2.18), q = Cvc where C is the capacitance of the capacitor and Vc is
the potential difference between its plates at any instant, so that i- Cdvc/dt.
Therefore VR = iR = RCdvc/dt and Equation (8.12) becomes

V = RCdvc/dt + vc (8.13)

Rearranging, we have

V- Vc = RCdvc/dt

Separating the variables,

dvc/(V- Vc)- dt/(RC)

Integrating we get

In (V- Vc) = t//(RC) + C

where C is the constant of integration. Taking antilogs we have

V- Vc- exp [(-t/RC) + C]

= exp (-t/RC) exp (C)

When t = 0, Vc = 0 and so exp (C) = V. Therefore

V-vc-Vexp(-t/RC) and vc-V-Vexp(-t/RC)

Finally

Vc = V[1 - exp (-t/RC)] (8.14)