Unit 3 | unit 3 solution of first and second order networks

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Unit - 3

Solution of First and Second order Networks

Q1) V0 (t) = ? t =o if capacitor is uncharged?

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A1) We know i(t) = i()+[ i(0)- i()]

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Fig 1 Finding Req

Req = R1R2 /R1+R2=(5x5)/(5+5)=5/2ohm

=L/Req=4/5 sec

i(0)=0

i()=u(t)+u(t)=1.4u(t)

i(t)=1.4[1-]u(t)

Q2) Find Vo(t) for the circuit below?

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A2)

Time constant =L/Req = L/R

V () = 0, V (0) =VR (supply voltage)

i () = VR/R, i(0) =0

By KVL,

VR=i(t) R + L di/dt(t)

On solving

V0(t)= VR e-tR/L

Q3) In network below switch is moved from 1 to 2 at t=0 a steady state is achieved already. Fond capacitor voltage Vc(t) and current i(t) through 200Ω?

Fig 2. Circuit Diagram

A3) The Voltage across capacitor is given by

Vc(t)= K1+K2

The steady state is reached when t= the above equation becomes

Vc()= Lt t--> K = K1

At t=0

Vc(0)= K1+K2

K2= Vc(0)- Vc()

The final response for the circuit is given as

Vc(t)= Vc()-[Vc()-Vc()]

When switch is at position 1

Vc(0)= x 50 = 100V

So, voltage of capacitor at t=0 and t=0+ will remain same.

TO find Req we replace voltage source by short circuit

Req = 60||200||50 = 24 Ω

The time constant will be T = ReqC = 24x0.05 = 1.2sec

Switch at position 2

The steady state voltage across capacitor is given by

= +

Vc() = 20V

The complete response will be

Vc(t)= Vc()-[Vc()-Vc()]

Vc(t)= 20 –[20-100]

Vc(t)= 20+ 80

The value of current at switch position 1

I= 50/(200+60) = 0.192A

When switch is at 2

i(0+) = 100/200= 0.5A

i() = Vc() /200 = 0.1A

The complete response of current is

i(t)= i()-[i()-i()]

i(t)= 0.1-[0.1-0.5]

i(t)=0.1+0.4

Q4) Find the value of V0(t) for circuit below?

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A4)

Time constant, c= L/Req = L/R

V() = 0

V(0) = VR

i() = VR/R

i(0) =0

By KVL,

VR=i(t) R + L di/dt(t)

On solving

V0(t)= VR e-tR/L

Q5) Find iL(t) for the circuit below. For t>0?

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A5)

The circuit for Req will be

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Fig 3. Circuit for Req

Req = 6+3

= 9ohm

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Fig 4. Circuit when inductor acts as short circuit

T= L/Req = 2/9 sec

i(0) =0

i() = 3*3/6+3 = 1A

i(t) = 1 [1-e4.5t]

Q6) Capacitor is initially uncharged find Vc (t); t >0

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Fig 5. Circuit Diagram

A6)

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Fig 6. Circuit for Req

Req = (3116)+2

=2+2 = 4ohm

T= C Req= 2 sec

Clearly Vc(0) =0

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Fig 7. Circuit when capacitor acts as open circuit

By voltage divides, VA = 5*6/3+B = 30 v/9

Clearly by ohms low

VB = 2*2 = 4v

Apply KVL, VA-Vc-VB =0

Vc() = 30/9 – 4 = -2/3 V

Vc (t) = -2/3 [1- e0.25t]

CURVE:

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Vc(t) =2/3 [1-e-0.5t]

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Fig 8 Response for above circuit

Vc(t) = -2/3 [1-e-0.5t]

Q7) Find i(t) for the given circuit below?

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A7)

We know i(t) = i () +[i(0)- i()] e-t/T

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Fig 9. Circuit for Req

Req = 5*5/5+5 = 5/2 ohm

T=L/Req = 4/5 sec

Also

i(0) =0

i() = u(t) +2/5 u(t)

i(t) = 1.4 [1-e-5/4t)]u(t)

Q8) If switch ‘s’ closed for long time before opening at t=0-. Find Vx (0+)?

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Fig 10. Circuit diagram

A8) At t= 0- switch was closed,

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Fig 11. Switch is open

iL(0-) = 2.5A

At t= 0+

Inductor is initially charged so iL(0-) = iL(0+)

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Fig 12. When switch is closed for long

-Vx = 20x2.5

Vx = -50A

Q9) Find VR(0+) and dil/dt(t) = 0+ If switch is opened at t= 0?

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A9)

At time t=0-

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Fig 13. Capacitor SC switch closed for long

At t=0-

Vc(0-) = 10V

IL(0-) = 10/10 = 1 A

Also, iL(0-) = iL(0+) = 1A

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VR(0+) = 5V

Ldi(t)/dt = VL (t)

DiL(t)/dt = VL(t)/L

At t= 0+

d iL (0+) /dt = VL(0+)/L

-VL (0+)-VR =0

VL (0+) = -5V

DiL(t)/dt at t=0+

VL(0+)/L =-5/2

Dil/dt(o+) = 2.5 A/s

Q10) Find Vc(o+) and iR (o+) it switch‘s’ is closed at t=0?

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A10) At t=0-

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Fig 14. When switch opened at t=0

As capacitor is fully charged so acts as open circuit

Vc (0-) = 7*10/7+2 = 70/9v

At t=0+

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Fig 15. At t=0+ circuit

Vc(0+) = Vc (0) = 40/9v

By applying kcl at A,

VA-10/2+ VA/5+VA-7019/2 =0

VA [1/5+1/2+1/2] = 5+35/9

Va [6/5] = 80/9

Va= 400/54v

IR (0+) = Va/5 = 400/54*5 = 80/54 A

Q11) Find diL(t)/dt and dVc(t)/dt at t=0+. If switch is opened at t=0?

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A11)

At t=0-

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Fig 16. Switch closed for long

Vc(0-) =10V = Vc(0+)

iL(0-) = 2A

DiL(t)/dt = VL(t)/L

At t=0+

DiL(0+)/dt = VL(0+)/L

Similarly, C = ic(t)

= ic(t)/C

At t=0+

= ic(0+)/C

ic(0+) = -2A

Q12) Find whether the following network function represent the driving point function.

a) f(s)= (s+1)/(s2+1) b) f(s) = 3s2+2s+1/ss3+9s2+3s+2 c) f(s)=(s2+1)2/s2(s+3)

A12)

a) f(s)=s+1/s2+1

It represents the driving point function:

b) F(S)= 3S2+2S+1 / 5S3+9S2+3S+2

It represents the driving point function.

c) f(s) = (s2+1)2/s2(5+3)

It has representative zeros, hence not valid

(s2+1)2=0

s2=+-1

s2=+-j,+-j

Q13) For network below 1), pole - zero pattern is represented in fig 2) , find numerical value of R,L and c for z (0)=1?

A13)

Calculating z(s) by taking L.T of fig 1)

z(s) = (R+SL)*1/CS

(R+SL)+1/CS = R+SL / SRC+S2LC+1

Z(S)= S+R/L / C [S2+R/L S+1/LC]

Given z(0)=1

z(0)=R/L / C/LC =1

Z(0)=R=1

.: R=1

Polls are given as

S2+RS/L+1/LC=0

S=-R/2L +J

Zeros are given as

S+R / L=0

From pole zero plot value of pole location is at -1 so

-R / L = -1

-1/ L=-1

L=1H

Also imaginary part of pole from plot is

1/LC-(R/2L)2 =11/4

1/C-(1/2)2=11/4

1/C-1/4=11/4

1/C=11/4+1/4

1/C=12/4 =3

C=1/3 F

Hence, value of R=1L=14 and c=1/3 f

Q14) FOR the following circuits, FIND G12(S) =v2/v1?

A14)

Taking Laplace transform of above we have

Applying KCL at node v0

V0-V1(S) / S*1/2S/(S+1/2S) +V0+V2(S)/ (S*1/2S)/(S+1/2S)+V0/1/S=0

For Porr 2)We have,

v2-v0/(s*1/2s)/(s+1/2s)+v2/1/s =0 ---------(ii)

(v2-v0)(2s2+1) / s +sv2=0

v2 [2s2+1/s+s] = v0( 2s2+1)/ s

v0=v2(3s2+1) / (2s2+1)-----------(iii)

From (i) we have,

(v0-v1) /s (2s2+1) *sv0+(v0-v2)/5 (s2+1) =0

(2s2+1/5)v0-(2s2+1)/sv1+sv0+(2s2+1)/sv0-(s2+1)/s v2=0

(2s2+1)v1=(5s2+2)v0-(2s2+1)v2

Substitute value v0 from (iii) above

(2s2+1)v1 =[( 5s2+2)(3s2+1) /(2s2+1)-(2s2+1)]v2

(2s2+1)v1=(15s4+5s2+6s2+2-4s4+4s2) / (2s2+1 *v2

G12(s) = v2/v1 = (2s2+1) 2/11s4+15s2+1

Q15) Find the transfer function admittance ratio

Y12(s) = I2(S) / V1(S)

A15)

Taking Laplace transform of above figure

Applying KCL we get,

(V2-V1) / (3*1/2S) / (3+1/2S) * V2/ 1/2S+ V2/1/6 =0

(Y2-V1)(6S+1) / (6S+4) * (2S+6)V2V=0

BUT I2=-6V2

V2=-I2/6

We have,

(6s+1) / (6s+4)*v2-(6s+1) / (6s+4)*v1 +(2s+6)v2=0

-[6s+1 / 6s+s+2s+6]I2 / 6 =V1(6S+1) / 6S+4

-[6S+1(2S+6)(6S+4)]I2= 6(6S+1)V1

-[(6S+1)*12S2+36S+8S+24] I2=6(6S+1)

I2 / V1 =-6(6S+1) / 12S2+50S+25

Y12(S) = I2(S) / V1(S) = -6(6S+1) / 12S2+50S+25

Q16) Find overall Y-parameter?

A16)

V1 – I1R – V2 = 0

V1 – V2 = I1R

I1 = V1 - V2

V2 = I2R + V1

I2 = - V1 + V2

Y11 =

Y12 = Y21 =

Y22 =

Q17) Find overall Y-parameter

A17)

Y-parameter does not exist as V1 = V2

Q17) Find overall Y-parameter?

A18)

I1 = V1Ya + (V1 – V2)Yc

I1 = (Ya + Yc)V1 - YcV2

I2 = V2Yb + (V2 – V1)Yc

I1 = (Yb + Yc)V2 - YcV1

Y11 = Yb + Yc

Y12 = Y21 = - Yc

Y22 = Yb + Yc

Q18) Find overall Y-parameter?

A19)

Q19) Find Z11, Z12, Z22, Z21?

A20)

V1 = I1Za + I1Zc + I2Zc

= (Za + Zc)I1 + ZcI2

V2 = I2Zb + I2Zc + I1Zc

= (Zb + Zc)I1 + ZcI1

Z11 = (Za + Zc)

Z12 = Zc = Z21

Z22 = (Zb + Zc)

Q20) Find all z-parameters?

A21)

V1 = Za(I1 - I)

(I - I1)Za+ IZc+ Zb(I + I2) = 0

I(Za + Zb + Zc) – I1Za + I2Zb = 0

I =

V1 = ZaI1 - Za

= I1 + I2

V2 = Zb(I2 + I)

= ZbI2 + Zb

= I2 + I2

Z11 =

Z12 = Z21 =

Z22 =

Q21) Find all the Transmission parameters?

A22)

-3V1 – I1 + + = 0

+ = I1

V1 – V2 = I1

V1 = V2 + I1

V1= V2- I1----------------(1)

I2 = 3V1 + V2 + V2 – V1

I2 = 2V1 + 2V2

2V1 = I2 - 2V2

2V1 = - 2V2 + I2

V1 = -V2 + I2 ----------------(2)

A = -1

B =

From (1) & (2)

-V2 + I2 = I1 - V2

V2 - V2 + I2 = I1

I1 = V2 + V2 - I2

I1 = V2 - I2

C =

D =

Q22) Find all the Transmission parameters?

A23)

V1 = RI1 + V2 ----------------(1)

I2R = V2 – V1

I2 = V2 - V1

I2R = V2 – V1

V1 = I2R - V2 --------------------(2)

A = 1

B = R

From (2) in (1)

V2 - I2R = V2 + RI1

I2 = -I1

C = 0

D = 1

Q23) Find out overall transmission parameter?

A24)

Q24) 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?

A25)

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

Q25) 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?

A26)

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

Q26) 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?

A27)

f0= = 142.3Hz

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

Q27) 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?

A28)

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

Q28) 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?

A29)

For series Z =100/0.8 = 125ohm

Cosφ =