NAME: EUGUNE CHOW MUN WAI

MATRICS CARD NO: A15MJ0033

Question 1

1. By using nodal analysis, analyse the circuit of Fig. 1(a) and Fig. 1(b) for v1, v2 and v3.

Figure 1(a)

Solution :

At supernode v1-v2,

I1=I2+I3+I4+I5

2=V1-V3 + V1 + V2-V3 + V2

0.5 0.125 0.25

2=2V1-2V3+V1+8V2-8V3+4V2

2+3V1+12V2-10V3 : equation 1

By using KVL at loop :

-V1+2V0+V0=0

V1=3V0 , V0=V2

V1=3V2-: equation 2

V3=13V: equation 3

Sub equation 2 and 3 into equation 1 :

2=3(3V2)+12V2-10(13)

2=9V2+12V2-130

132=21V2

V2=6.286V

When V2=6.286V , sub into equation 2 , so V1 = 18.858V

Answer : V1 = 18.858V , V2=6.286V , V3=13V

Figure 1(b)

Solution :

Figure A

Figure B

At supernode V1,V2,V3 (shown at Figure A):

We cannot use KCL because it able cancel each other .But from the circuit , we able know that V2 =

12V

So to find V1 and V3 , we use KVL (shown at Figure B):

Loop 1 : -V1-10+12=0

V1=2V

Loop 2 : -12+20+V3=0

V3=-8V

Answer : V1=2V , V2=12V , V3=-8V

Question 2

11kΩ and 10kΩ resistors are parallel. Therefore,

R11||10 =

11 x 1011+10 = 5.238kΩ as shown in the figure below.

By Using supermesh:

Applying mesh to Loop 1 and Loop 2:

-25 + 2i +4i2 + 5.238i = 0

-25 + 7.238i + 4i2 = 0 ………… (1)

Applying KCL at the node:

i = 20 + i2

Substitute i = 20 + i2 into Equation (1):

-25 + 7.238(20 + i2) + 4i2 = 0

119.76 + 11.238i2 = 0

i2= -10.66mA

v = 4i2

= 4(-10.66) = -42.64V

Substitute i2 = -10.66 into Equation (1):

-25 + 7.238i + 4(-10.66) = 0

i = 9.345mA

Question 3

3) (a)

Combining the 3A ammeter and the 2 ohm resistor by using ohm’s law,

V=IR

V=3× 2=6V

The voltage and the resistor becomes series after source transformation. The diagram is

shown below:

Combining the two resistors in series,

R=2+4=6 Ω

The diagram is shown below:

By using the loop, combining all the voltage, and the resistors are combining as parallel:

V T=10−1=9 V

Req=[ 13+ 1

6]−1

Req=[ 12]−1

Req=2Ω

By using the source transformation, the current can be calculated:

I=VR

V=3× 2=6V

I=93=3 A

The two voltage and current diagram is shown after source transformation,

(b)

By using source transformation, the voltage of 4 ohm resistor and 1 ampere current is found:

V=IR

V=1 × 4=4 V

The diagram is shown as below,

Combining the voltage between 1V and 4V, and the 2 ohm and 4 ohm resistors:

V=1+4=5V

R=[ 12+ 1

4]−1

R=[ 34]−1

R=43

Ω

The diagram is shown as below after combining,

Combining to find the total resistance and the voltage as well as the current:

V=5+4=9V

R=1.333+3=4.333 Ω

I=VR

I= 94.333

=2.077 A

After the source transformation, the final diagram is shown as below,

4.333Ωa

b

9 Vrms 60 Hz 0°

4.333Ω

a

b

2.077 A 1kHz 0°

Question 4

Solution:

Using current division:

Current flowing through 5Ω resistor is, I = 6

5+6(2)

= 1211

A

Vx = IR

= (12/11)(5)

= 5.455V

The 5Ω and 6Ω resistors will be ignored because of short circuit.

Therefore, Vx = 0

The 5Ω and 6Ω resistors will be ignored because of short circuit.

Therefore, Vx = 0

Vx = 5.455 + 0 + 0

= 5.455V.

Question 5

When the switch is on (t<0), the inductor acts as a straight wire while the capacitor acts as an open circuit. Thus, the circuit can be redrawn to circuit 1a.

Circuit 1a

From the circuit, it could be seen that:

R2 ||(R3 +R4)||(R5 +R6)

Req =( 115

+ 140+40

+ 110+10

)kΩ = 7.742kΏ

Rtotal = 10kΏ +7.742kΏ = 17.742kΏ

V(0-)= V(0) =

7 .74217 . 742

(30)=13. 09 V

i(0-)= i(0) =

13 . 09

17 . 742×103=0 . 738 mA

When t > 0, the switch is opened and the circuit becomes as follows in circuit 2a and can be redrawn into the diagram in circuit 2b:

Circuit 2a Circuit 2b

Circuit 2b can then be further simplified to become circuit 2c.

where Req = 25.413kΏ

α=25 . 413×103

2(2 )=6353 . 25

ωo=1

√2(120

)=√10

α > ωo Hence, the circuit is overdamped.

S1,2=−α±√α 2−ωo

2=−(6353 . 25)±√(6353 .25 )2−(√10)2

S1=−7 . 879×10−4

S2=−12706 .5

i( t )=A1 es1 t

+ A 2 es2 t

15kΏ

∴i( t )=A1e−7.879 t+ A 2 e−12706.5 t

When t = 0

i(0) = A1+A2 =0.738×10-3

A2 =0.738×10-3－A1

di(0 )dt

=−7 . 879×10−4 A1−12706 .5 A 2

1L

( RI o+V o )=−7 .879×10−4 A1−12706 .5 A 2

12

((25. 413×103 )(0 . 738×10−3)+13 . 09 )=−7 . 879×10−4 A1−12706 .5 A 2

15 . 922=−7 . 879×10−4 A1−12706 . 5(0 .738×10−3−A 1 )15 . 922=12706 . 499 A1−9 .377

A1=1 .991×10−3

A2=0 . 738×10−3−1 . 991×10−3=−1 . 253×10−3

∴i( t )=1 . 991×10−3e−7. 879 t−1 . 253×10−3e−12706 .5 t

V ( t )=I ( t ) R=(1.991×10−3e−7.879×10−4

−1 .253×10−3 e−12706 .5 t )(25 .413×103)

∴V ( t )=50 .6 e−7 .879×10−4

−31 . 842 e−12706 .5 t

Question 6

Analyze the calculate io(t) and vo(t) of the circuit shown in Figure 4 for t>0 .

For t < 0,

At t = 0- ,

vo(0) = 8

8+12(30 )

= 24V

For t > 0,

α = 1

(2 RC ) 1

√ LC

= 1

2(8)(14) =

1√ 1 (1/4 )

= ¼ = 2

Since α less than we have an under-damped response .

d = √ (ω )2−α2

= √4−( 116

)

= 1.9843

vo(t) = (A1cosdt + A2sindt)e-αt

A1= 24

io(t) = C(dvo/dt) = 0 when t = 0

dvo/dt = -α(A1cosd dt + A2sind dt)e-αt + (-d A1sind dt + d A2cosd dt)e-αt

at t = 0, dvo(0)/dt = 0 = -αA1 + d A2

Thus, A2 = (α/d)A1 = (1/4)(24)/1.9843 = 3.024

vo(t) = (24cos1.9843t + 3.024sin1.9843t)e-t/4 volts.

i0(t) = Cdv/dt

= 0.25[–24(1.9843)sin1.9843t + 3.024(1.9843)cos1.9843t –0.25(24cos1.9843t) 0.25(3.024sin1.9843t)]e-t/4

= [0.000131cos1.9843t – 12.095sin1.9843t]e-t/4A.

Question 7

7. Referring to the circuit shown in Figure 7, analyse the value of v(t) for all time.

Solution:

When t < 0, u(t) = 0, LC circuit is disconnected from the voltage source.

i(0) = 0A, v(0) = 0V

When t > 0, u(t) = 1, capacitor acrs like open circuit and inductor acts like short circuit.

v(t) = 50V

i(t) = 505

i(t) = 10A