USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

A.C. transients

(i) R-L sinusoidal transient :Refer Fig. 83

Here the voltage function could be at any point in the period at the instant of closing the switch and therefore the phase angle ϕ can take any values from 0 to 2 π rad /sec.

In this case, the current (i) is given by

i = e –(R / L)t [ -V max / √ R2 + w2 L2 sin  (ψ – tan  -1 wL  /R)]

+ V max / √ R2 + w2 L sin (wt + ψ – tan -1 wL /R)

where tan -1 (wL / R) = ϕ

It may be noted that:

The first part (transient component of current, it, of the above equation contains the

factor e-(R / L)t which has a value of zero in a relatively short time.

 

The second part of the above equation is the steady current (is) which lags the applied

voltage by tan-1 wL / R

Here ,              V max / R2 + w2 L2 = I max , tan -1 (wL/R) = ϕ

 

(ii) R-C sinusoidal transient :

For R-C circuit shown in Fig. 84 the basic equation is :

Ri + 1 / C ʃ vdt = V max sin (wt + ψ)

Here the current i is given by,

It may be noted that :

-The first part of the above equation is the transient with decay factor e-t/ RC

-The second part is the steady current which leads the applied voltage by tan-1 1 /wCR.

 

Example 51. A series circuit has R = 10 n and L = 0.1 H. A direct voltage of200 Vis suddenly applied to it. Calculate the following :

(i) The voltage drop across the inductance at the instance of switching on and tLt 0. 01 second,

(ii) The flux linkages at these instants.

Solution. Given: R = 10 n; L = 0.1 H; V = 200 volts

(i) The voltage drop across the inductance :

(a) Switching instant. At the instant of switching on, i = 0, so that i R = 0 hence all the

applied voltage must drop across the inductance only. Therefore, voltage drop across inductance

= 200 V. (Ans.)

(b) When t = 0.01 second

i = V / R [ 1-e –(R / L)y ]

= 200 / 10 [ 1- e –(10 / 0.1) x 0.01 ] = 20 (1-e -1) = 12.64 A.

and                  i R = 12.64 x 10 = 126.4 V.

The voltage drop across the inductance = √ ( 200)2 – (126 .4)2 = 155 v.

Now ,              L = Nϕ / i       or         Nϕ = Li

Flux linkages (Nϕ) = Li = 0.1 x 12.64 = 1.264 Wb-turns. (Ans.)
 

Example 52. A choke has a resistance of 50 Q and inductance of 1. 0 H. It is supplied with an A. C. voltage given by 141 sin 314t. Find the expression for transient component of the current flowing through the choke after the voltage is suddenly switched on.

Solution. Given: R = 50 Q; L = 1.0 H; e = 141 sin 314t

Expression for transient component of the current :

The equation of the transient component of the current is given by:

it = I max sin ϕ e –(R / L)t

Here    I max – V max / Z = V max / √ R2 + w2 L2 = 141 / √ (50)2 + (3.14)2 (1.0)2 = 1.443 A

and                                 ϕ = tan -1 (wL / R) = tan -1 (314 x 1.0 / 500 = 80.95

i  = 0.443 sin 80.95 e –(50 / 1.0) t = 0.437 e – 50t

 

Example 53. A series circuit has R = 10 Q and L = 0.1 H. A 50 Hz sinusoidal voltage of maximum value of 400 Vis applied across this circuit. Find an expression for the value of current at any instant after the voltage is applied, assuming that the voltage is zero at the instant of application.  Calculate its value 0.02 second after switching on.

Solution. Given: R = 10 Q; L = 0.1 H ;f= 50 Hz; V max = 400 V; t = 0.02 s.

The current consists of transient component and steady-state component. The equation of

the resultant current is given by :

i = e – (R / L)t [ -V max / √ R2 + w2L2 sin (ψ – tan -1 wL / R)

+ V max / √ R2 + w2L2 sin (wt + ψ – tan  -1 wL / R)

 

where              V max / √ R2+ w2 L2 = I max , tan -1 ( wL / R) = ϕ

Here, ψ = 0 as per given condition, then

Now,               V max / R2 + w2L2 = I max = 400 / √ 102 + (314)2 (0.1)2 = 12.14 A

Tan -1 wL / R = ϕ = tan -1 (314 x 0.1 / 10) = 73.3 = 1.262 rad

 

Substituting the value in eqn. (i), we get

i = e-(10/0.1) x 0.02 [- 12.14 sin (- 72.3°)] + 12.14 sin (314 x 0.02 – 1.262)

= e-2 [12.14 sin (72 .3°)] + 12.14 sin (5.018)

(rad)

= 0.1353(12.14 x 0.9527) + 12.14 sin (287.5°)

(deg)

= 1.56 – 11.58 = – 10.02 A. (Ans.)

 

i = 6 –(10 / 0.1) x [ -12.14 sin (-72.3 )] + 12.14 sin (314 x 0.02 -1.262)

= e -2 [ 12.14 x 0.9527 x 0.9527 ) + 12.14 sin (287.5)

= 1.56 – 11.58 = – 10.02 A.