176 Transient analysis
The sudden disconnection of a d.c. supply
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Figure 8.5V(a) (b)R i L
I ] .~ r-cw'v'~Suppose that the circuit of Fig. 8.5(a) has been in steady state operation for a
long time ('long' here means long compared to the time constant of the circuit),
when the switch S~ is suddenly opened and $2 is simultaneously closed. Let this
instant be t = 0 and at this instant the current will have its steady state value (I).
The circuit then becomes that of Fig. 8.5(b).
Applying KVL to the closed circuit we have
iR + Ldi/dt = 0
iR - -Ldi/dt
Separating the variables di and dt,
(R/L) dt = -(di/i)
f (R/L) dt = -f (1/0 di
Integrating gives
Rt//L = - In i + C (8.8)
where C is a constant. Now at the instant t = 0, i = I and so 0 = -In I + C, so
C= lnI
Substituting for C in Equation (8.8)
Rt/L - -ln i + In I = In I/i
Taking antilogs, we have exp (Rt/L) - I/i so that/= I/exp (Rt/L). Finally
i = I exp (-Rt//L) (8.9)
The potential difference across the resistor is given by VR = iR
= I exp (-Rt/L)R.
Since IR = V, then
VR = V exp (-Rt/L). (8.10)
The voltage across the inductor is vL - --VR -- --V exp (-Rt/L), so
vL = -V exp (-Rt/L) (8.11)