Introduction to Electric Circuits

(Michael S) #1
8.3 Circuits containing resistance and capacitance 183

Figure 8.15

0 %

1,' R = V exp (-t/cn)
This describes an exponential decay of voltage and is shown in Fig. 8.14.

Example 8.6


(8.16)

In the circuit of Fig. 8.16, the switch S is closed at t = 0. Determine the value of
the current (is) drawn from the supply after 30 ms.

10 1-
0

C = 200pF
II .... i
r'-i
R1 = 50s U R2 350~

,.
Figure 8.16

Solution
From Equation (8.15) the current through the branch containing C and R 2 is
given by
i2 = 12 exp (-t/CR2)
The time constant of this branch is
7" = CR 2 = 200 • 10 -6 X 350 = 0.07 s
The steady state value of the current through the branch is
12 = V/R 2 = 100/350 = 0.286 A.

After 30 ms,


i2 = 0.286 exp (-0.03/0.07) - 0.186 A


The current I~ - V/R1 = 100,/50 - 2 A and it reaches this value immediately S
is closed because the branch is purely resistive. After 30 ms, therefore,


is = I~ + i 2 = 2 + 0.186 = 2.186 A

Free download pdf