Unit 3 | unit 3 transients in rlc circuit

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Unit - 3

Transients in RLC circuit

Q1) For the circuit given below Find V0 (t) for t =o if the 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 the 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 the capacitor acts as an 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 the 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 the capacitor and current through capacitor at t= 0+?

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A6) At t= 0- the 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 a 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 the 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 the RC circuit?

A9) FOR RC Circuit

Fig 2: Series RC circuit

For t>0 applying KVL

Ri(t)++V=0

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

i=k

i(0)=-

Hence, a particular solution for the network is given as

i=- tfor t≥0

=for t<0

The time constant is given as

T=RC

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

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

First-order homogenous differential equation.

Integrating both sides

y(t)=k

First-order non-homogenous differential equation

Integrating the above equation, we get

y(t)=∫ Q

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

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