# D.C. transients

(i) R-L transients :

In the R-L circuit shown in Fig. 76 i = V / R [1 –e –(R / L)t]

The plot of i (exponential rise equation) versus time is shown in Fig. 77.

The time constant (λ)for the above function is the time at which the exponent of e is unity. Thus in this case time constant(λ) is Ll R. At one time constant, the value of i will be i = (1- e-1) = 1-0.368 = 0.632

At this time current will be 63.2% of its final value.

The voltage across inductance,

vL = L di / dt = Ve –(R /L)t

and voltage across resistor,

vR = V[1- e – (RIL)t ] … (25)

The exponential rise of resistor voltage and exponential decay of inductor voltage are shown in Fig. 78.

Also,                VR + VL = V[1- e-(R/L)t + Ve-R/L)t = V

Power in the circuit elements is given by PR = V2 / R [ 1 – 2e –(R / Lt)t + e -2 (R / L)t

PL = V2 / R [ e –(R / L)t – e -2 (R / L)t

Total power,

P = PR + PL = V2 / R [1-e –(R / L)t

(ii) R-C transients :

In the R-C circuit shown in Fig. 79, i = V / R e – t / RC

Transients voltages across R and C are given by

u R = Ve –t / RC

u c = V(1 – e-t/RC)

Also, the power in circuit elements is given by

PR = V2 / R e -2t / RC

PL = V2 / V2 / R (e –t / RC –e 2t / RC)

(iii) R-L-C transients :

For R -L-C circuit shown in Fig. 81 the following integro differential equation can be written as follows : While solving for i, the following three cases are considered

Case I.                        (R / 2L) 2  > 1 / LC In this case the current is given by

i = C1e at C1(eβt + C2e -βt)

Case II.           (R / 2L)2 = 1 / LC

Here ,                             i = e at (C1 + C2 t)

Case III.         (R / 2L)2 < 1 / LC

Here  ,                         i  = e at (C1 cos βt + C2 sin βt) In all the above cases the current contains the factor eat and since ex = – R/2L the final value is zero, assuming that the complimentary function decays in a relatively short time. Fig, 82 shows the value of i for initial values zero and initial slope positive.  