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Unit - 4

Electrical Circuit Analysis Using Laplace Transforms

Q1) For given RL circuit find the differential equation for current i?

Fig1: Series RL circuit

A1) After switch is closed applying KVL

This is first order homogeneous differential equation so

Integrating both sides

ln i= t+K

taking antilog of both sides

i=k

At t=0

i (0) ==I0

=ke0

The particular solution is given as

i= for t≥0

=for t<0

Q2) Find IL (0) for the given circuit below

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224139.jpg$

A2) From above circuit

L = 2H

LiL (0) = 25 mv

IL (0) = 12.5 m A

Q3) Find value across capacitor?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224353.jpg$

A3) From above circuit we can write

1/cs = 2/s

C=1/2f

Vc (0)/s = 0.25v/s

Vc (0) = .25v

Q4) Find voltage across capacitor?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224735.jpg$

A4) 1/cs = 1/2S

C=2F

CVc (0) = 0.02

Vc (0) = 0.01V

Q5)

(a) switch is at before moving to position = at t= 0, at t= 0+, i1(t) is

(b) –V/2R (b) –V/R (c) – V/4R

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160151.jpg$

A5)

I1(t) L.T I1(s)

I2(t) L.T I2(s)

then express in form

[ I1(s)/I2(s)] [:] = [:]

Drawing the ckl at t = 0-

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160708.jpg$

:. The capacitor is connected for long

Vc1(0-) = V

iL (0-) =0

Vc2(0-) =0

at t= 0+

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161211.jpg$

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161423.jpg$

i1 (0+) = -V/2R

for t>0 the circuit is

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161828.jpg$

Apply KVL

-R i1(t) -1/c dt -L=0

Also, we can deduce i1(t)- i2 (t) = iL(t)

-Ri1(t) -1/c - L = 0

-RI1(s)-[-]-L[sI1(s)-iL (0-)] =0

- RI1(s)- - -s L I2(s)=0

- s L I2(s)= - (1)

Again

-Ri2(t)) -1/c dt -L=0

-Ri1(t) -1/c - L = 0

Taking Laplace Transform

-RI2(s)- [ -]- L [s IL(s)- iL (0-)] =0

-RI2(s)- -Ls [ I1(s)- I2(s)] =0

- s L I1(s)=0 (2)

Hence in matrix representation

Q6) All initial conditions are zero i(t) L.T, I(s), then find I(s)?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162652.jpg$

A6) Drawing Laplace circuit

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162827.jpg$

(2+S) 11 2/S

Zeq = [(2+S)2/S]/[2+S+2/S] =[2(S+2)]/[S2+2s+2]

$C:\Users\TARUNA\Desktop\internship\29 dec 2019\IMG_20191228_132621.jpg$

V1(s) = [5/S{2(S+2)/S2+2S+2}]/[2+{2(S+2)/ S2+2S+2}]

V1(s) = [5(S+2)/S]/ [S2+ 3S +4]

I(s) = V1(s)/(S+2) = (5/S)/(S2+35+4)

Q7) Write equation for given waveform

A7)

The final equation is=u(t)+u(t-1) +u(t-2) +u(t-3)-4u(t-4)

As the waveform is stopped at t=4 with amplitude 4, so we have to balance that using -4u(t-4).

Q8) Write the equation for the given waveform?

A8) Calculating slope of above waveform

Slope for (0,0) and (1,2)

a1 slope==2

a2 slope==-4

a3 slope =4 and a4= -2

At t=1 slope changes from 2 to -4 so we need to add -6r(t-1) to 2r(t) to get the slope of -4.

At t=1.5 slope changes from -4 to 4 so adding 8r(t-1.5) to get slope of 4

At t=2 slope changes from +4 to -2 so we add -6r(t-2). But we need to stop the waveform at t=3. Hence, we have to make slope 0. Which can be done by adding +2r(t-3). Final equation is given as 2r(t)-6r(t-1.5) +8r(t-1.5)-6r(t-2) +2r(t-3)

Q9) All initial conditions are zero i(t) I(s), then find I(s)?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162652.jpg$

A9) Drawing Laplace circuit

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162827.jpg$

(2+S) 11 2/S

Zeq = [(2+S)2/S]/[2+S+2/S] =[2(S+2)]/[S2+2s+2]

$C:\Users\TARUNA\Desktop\internship\29 dec 2019\IMG_20191228_132621.jpg$

V1(s) = [5/S{2(S+2)/S2+2S+2}]/[2+{2(S+2)/ S2+2S+2}]

V1(s) = [5(S+2)/S]/ [S2+ 3S +4]

I(s) = V1(s)/(S+2) = (5/S)/(S2+35+4)

Q10) A series RL low-pass filter cut-off frequency is 2KHz. R=5KΩ Find the inductor value. Find |H(jω) | at 50 KHz?

A10) The cut-off frequency is c = R/L

L= 5000/2000 =2.5Hz

|H(jω) | =

Substitute values in above equation for 50KHz we get

|H(jω) | = 0.0635

Q11) A coil takes a current of 6A when connected to 24V dc supply. To obtain the same current with 50HZ ac, the voltage required was 30V. Calculate inductance and p.f of coil?

A11) The coil will offer only resistance to dc voltage and impedance to ac voltage

R =24/6 = 4ohm

Z= 30/6 = 5ohm

XL =

= 3ohm

cosφ = = 4/5 = 0.8 lagging

Q12) The potential difference measured across a coil is 4.5V, when it carries a dc current of 8A. The same coil when carries ac current of 8A at 25Hz, the potential difference is 24V. Find current and power when supplied by 50V,50Hz supply?

A12) R=V/I= 4.5/8 = 0.56ohm

At 25Hz, Z= V/I=24/8 =3ohms

XL =

= 2.93ohm

XL = 2fL = 2x 25x L = 2.93

L=0.0187ohm

At 50Hz

XL = 2x3 =6ohm

Z = = 5.97ohm

I= 50/5.97 = 8.37A

Power = I2R = 39.28W

Q13) A coil having inductance of 50mH an resistance 10ohmis connected in series with a 25F capacitor across a 200V ac supply. Calculate resonant frequency and current flowing at resonance?

A13) f0= = 142.3Hz

I0 = V/R = 200/10 = 20A

Q14) A 15mH inductor is in series with a parallel combination of 80ohm resistor and 20F capacitor. If the angular frequency of the applied voltage is 1000rad/s find admittance?

A14) XL = 2fL = 1000x15x10-3 = 15ohm

XL = 1/C = 50ohm

Impedance of parallel combination Z = 80||-j50 = 22.5-j36

Total impedance = j15+22.5-j36 = 22.5-j21

Admittance Y= 1/Z = 0.023-j0.022 siemens

Q15) A circuit connected to a 100V, 50 Hz supply takes 0.8A at a p.f of 0.3 lagging. Calculate the resistance and inductance of the circuit when connected in series and parallel?

A15) For series Z =100/0.8 = 125ohm

cosφ =