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Unit - 2

Solution of First and Second order networks

Q1) For the circuit given below Find V0 (t) for t =o if capacitor is uncharged?

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A1) Finding Req

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Fig. Circuit for Req

Req = R1R2 /R1+R2

V0(0) =0

V0 (infinity) = VR R2/R1+R2

V0 (t) = V0 (infinity) + [Vo (0) – V0 (infinity)] e-t/c

=VR R2/R1+R2 *[ 1- e+(R1+R2)/R1R2]

Q2) Find the value of V0(t) for circuit below?

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A2)

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

V(infinity) = 0

V (0) = VR

I(infinity) = VR/R, I (0) =0

By KVL,

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

On solving

V0(t)= VR e-tR/L

Q3) Find i(t) for the circuit below for t>0?

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A3) The circuit for Req will be

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Fig. Circuit for Req

Req = 6+3

= 9ohm

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Fig. Circuit when inductor acts as short circuit

T= L/Req = 2/9 sec

I (0) =0

I(infinity) = 3*3/6+3 = 1A

i (+) = 1 [1-e4.51]

Q4) Capacitor is initially uncharged find Vc (t); t >0

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A4)

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Fig. Circuit for Req

Req = (3116) +2

=2+2 = 4ohm

C= c Req

= 2 sec

Clearly Vc (0) =0

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Fig. 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(infinity) = 30/9 – 4 = -2/3 V

Vc (t) = -2/3 [1- e0.5t]

CURVE:

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

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Fig. Response for above circuit

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

Q5) Find i(t) for the given circuit below?

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A5) We know i(t) = I (infinity) + [i (0)- i(infinity)] e-t/c

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Fig. Circuit for Req

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

:. T=L/Req = 4/5 sec

Also, i (0) =0

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

:. I(t) = 1.4 [1-e-5/4t)] u(t)

Q6) If switch ‘s’ closed at t= 0 find out the voltage across capacitor and current through capacitor at t= 0+?

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A6) At t= 0- switch was open so,

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:. There is no powntial so Vc (0-) = 0

At t= 0+, Vc (0-) = Vc (0+) = 0

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Isc = 10/5 = 2 A

Q7) The switch was closed for long time before

Opening at t=0- find VX (0+)?

(a) 25V (b) 50 V (c)-50V (d) 0V

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A7) At time t=0-

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clearly, il (0-) = 2.5 A

at t= 0+, as inductor is initially charges

so il (0-) = il (0+)

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-VX= 20/2.5 20*2.5

VX= -50A

Q8) Find VR (0+) and dil/dt(t) = 0+ If switch is opened at t= 0?

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A8)

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

-VL (0+)-VR =0

VL (0+) = -5V

dt(t) at t = ot = vl (ot)/L

=-5/2

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

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

At t=0

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As capacitor is fully charged so acts as

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

At t=0+

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

By applying kcl at A,

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

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

Va [6/5] = 80/9

Va= 400/54v

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

Q9) Derive the time constant for RC circuit?

A9) FOR RC Circuit

Fig. Series RC circuit

For t>0 applying KVL

Ri(t)++V=0

Hence general solution of above equation is calculated same as for RL circuit

i=k

i (0) =-

Hence, particular solution for network is given as

i=- for t≥0

=for t<0

The time constant is given as

T=RC

Q10) Write the equation for complete solution of a differential equation?

A10) The complete or total response of network is the sum of the transient response and steady state response which is represented by general solution of differential equation.

First order homogenous differential equation.

Integrating both sides

y(t)=k

First order non homogenous differential equation

Integrating above equation we get

y(t)=∫ Q

The first term of above solution is known as particular Integral and second is known as complementary function.

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