Introduction to Electric Circuits

8.2 Circuits containing resistance and inductance 179

consider the period 0-T (0-40 Ixs). The time constant of the circuit is

z = L/R = 8 • 10-3fl2 • 10 3 = 4 x lO-6s = 4 Ixs

Now 5~- = 20 ixs and since the pulse width is twice as long as this the steady state

conditions will have been reached well before the pulse is removed.

The steady state value of the current is V/R = 5//2 x 103=

2.5 X 10 -3 A = 2.5 mA and this is considered to have been reached in 20 Ixs.

The instantaneous value of the current is given by Equation (8.2) to be

i= I[1- exp (-Rt/L)]. After the time constant (4 ~s) the current will have

reached 0.632 1 = 0.632 x 2.5 x 10 -3 A --- 1.58 mA. The output voltage is given

by Equation (8.5) to be Vo = VR = V[1 - exp (-Rt/L)] and after 4 I~s this will

also have reached 63.2 per cent of the steady state value (5 V). The output

voltage will then be 0.632 x 5 = 3.16 V.

Next we consider the period subsequent to the removal of the pulse (i.e. after

40 Ixs). During this period the current will be decaying in accordance with

Equation (8.9) while the output voltage is also decaying according to Equation

(8.10). After 5~" = 20 txs (i.e. a total of 60 I~s after t = 0), both will have

died away to zero. After r = 4 Ixs (44 Ixs after t = 0) the current will have

reached I exp (-1) so that i = 0.368 1 = 0.368 x 2.5 = 0.92 mA. Similarly

Vo = VR = 0.368 V = 0.368 X 5 = 1.84 V.

The graphs of current and output voltage are now drawn as shown in

Fig. 8.10.

i(mA);~ Vo (V)+

2.5- 5 --

1.58 ---3.16

0.92 ---1.84 I

O- 0 20 I

Figure 8.10

I I

40 (40+1:) 60 "t(~s)

The RL differentiator circuit

If the output voltage is taken to be that across the inductor as shown in Fig. 8.11
the circuit is called a differentiator because the output voltage waveform
approximates to the differential of the input voltage waveform.

Figure 8.11

v(

i i R i

"~! I O

L Vo =

-O