Question

Question B2 Figure 8 shows a DC electric circuit with a 2-way switch. The capacitor is initially not charged. At t= Os, switch is at position and the capacitor is then charged up 1000 Position1 RT Position2 (20V R 500 0.02F (NV Figure 8 (a) Whent > sec, the capacitor is charged through R, to steady state, 0 Find the complete response of y(t) fort > 0s. (5 marks) (m) Sketch the response of (t) for t> Os. (3 marks) (iii) Calculate energy stored in the capacitor at steady state. (2 marks) (iv) Find the time X needed for the voltage v(t) to increase to 4V. (3 marks) (b) Att > T sec, the circuit is at steady state and then the switch is turned to position 2. The capacitor is now being discharged through Rz (as N<20V). () Find the complete response of y(t) fort > Ts. (4 marks) (ii) Sketch the response for this discharge cycle (from t > Ts) (3 marks)

Answer 1

Ans:(a)

(i)

Draw the circuit for t>0s

R1 100 2 Vo(t) + 20v +1 0.02 F

Apply nodal analysis at Vc(t)

V c(t) - 20 R1 +C dV c(t) dt -0

dV c(t) Vct) + dt R1C 20 RC

Substitute:

R1 100Ω

C 0.02

20 dV c(t) dt + Vct) 100 X 0.02 100 X 0.02

20 dV c(t) V c(t) + dt 2 2

dV c(t) dt +0.5V c(t) = 10

Solve the above differential equation:

This is a linear differential equation

Integrating Factor

LE=e/0.5dt

    

0.5t =e

L.F. V c(t) = 10 x I.F.dt+C

0.5t x Vc(t) 10 x 0.5t + c

0.5t 0.5t XV c(t) = 10 x + 0.5

0.5 e x V c(t) = 20 x 0.5t +

V c(t) = 20 + Ce-0.5t

Apply initial condtion , at t=0s Vc=0V

0 = 20 + Ce -0.5x0

>>C= -20

V c(t) = 20 - 20e -0.5

  
= 20(1 - e-0.5) V
,  
t>

(ii)

V c(t) = 20(1 – e-0.5)

This is exponential graph:

(i) at t=0 ,

V c(0) = 20(1 - e-0.5x0V

20(1 - 1)

10 三

(ii)at t=2s

  

V c(2) = 20(1 - e-0.5x2)V

= 12.641

(iii) at

= 7

  

V c00) = 20(1 – e-0.5x)

  

20(1-0)V

  

10,三

20F 12.64 Vot) 2 t(s)

(iii)

At steady state Vc=20V

Recall the formula of energy stored by a capacitor

2 E CV 2

Substitute

C= 0.2F, V = 20V

2 E CV 2

x 0.02 x 102

  

1J

(iv)

V c(t) = 20(1 – e-0.5)

Given:

V () = 4V

4= 20(1 – e-0.52

0.2 = 1 -0.52 e

-0.5.1 e 0.8

-0.5.x = ln (0.8)

c= 0.446s

Therefore, Vc(t) will be 4V after 0.446s.

(b)

For t>Ts , circuit is at steady state.

Lets us shift our frame after t>Ts,

So, instant after t>Ts, will be considered t=0 in new frame

Redraw the circuit after t>Ts

R2 502 Vot) + N+ 0.02.:

Apply nodal analysis at Vc

dV c Vct) - N R2 +C = 0 dt

N dV c(t) V c(t) + dt R2C RC

Substitute:

R2 50Ω

C 0.02

dV c(t) dt + V ct) 50 X 0.02 N 50 X 0,02

N dV c(t) V c(t) dt 1

dV c(t) dt +Vc(t) = N

Solve the above differential equation:

This is a linear differential equation

Integrating Factor

LE = el ldt

    

t =e

$I.F.\times Vc(t)=\int N\times I.F.dt+C$

$e^{t}\times Vc(t)=\int N\times e^{t}+C$

$e^{t}\times Vc(t)= Ne^{t}+C$

Apply initial condtion , at t=0s Vc=20V

$\Rightarrow C=20-N$

(ii)

This is exponentialy decaying curve

(i) at t=0s

$=N+(20-N)$

  

(ii) at

= 7

8 V c(0) = N + (20 – Ne

  $=N+0$

  $=NV$

20 N Vc(t)