Branch current and loop current relation and expressed in matrix form below where II represents branch current and IK loop current.

1. The rank of incidence matrix is

(a) 4

(b) 5

(c) 6

(d) 8

1. The directed graph will be

For a network branch voltage and node voltage relation are expressed in matrix form as follows

where VI is the branch voltage and VK is the node voltage with resect to datum node.

1. The independent mesh equations for this network are

(a) 4

(b) 5

(c) 6

(d) 7

1. The oriented graph for this network is

In the following circuit when R = 0, the current IR equals 10 A.

1. The value of R, for which it absorbs maximum power, is

(a) 4

(b) 3

(c) 2

(d) none of these

1. The maximum power will be

(a) 50 W

(b) 100 W

(c) 200 W

(d) 400 W

In the following circuit, the maximum power transfer condition is met for the load RL.

1. The value of RL will be

(a) 2

(b) 3

(c) 1

(d) none of these

1. The maximum power is

(a) 0.75 W

(b) 1.5 W

(c) 2.25 W

(d) 1.125 W

Common data for Questions 9 and 10

Consider the following circuit :

1. If VS1 = VS2 = 6 V then the value of VA is

(a) 3 V

(b) 4 V

(c) 6 V

(d) 5 V

1. If VS1 = 6 V and VS2 = – 6 V then the value of VA is

(a) 4 V

(b) 4 V

(c) 6 V

(d) – 6 V

Common data for Questions 11 and 12

A circuit is given in figure. find the thevenin equivalent as given in question.

1. As viewed from terminals x and x’ is

(a) 8 V, 6

(b) 5 V, 6

(c) 5 V, 32

(d) 8 V, 32

1. As viewed from terminals Y and Y’ is

(a) 8 V, 32

(b) 4 V, 32

(c) 5 V, 6

(d) 7 V, 6

Common data for Questions 13 and 14

The circuit shown below is at steady state before the switch closes at t = 0. the switch remains closed for 1.5 r and then opens.

1. At t = 1 s and V(t) will be

(a) -3.24 V

(b) 1.97 V

(c) 5.03 V

(d) 13.24 V

1. At t = 2s, V(t) will be

(a) 5.12 V

(b) 6.43 V

(c) 8.57 V

(d) 9.88 V

Common data for Questions 15 and 16

The circuit is as shown below.

1. In the circuit shown below, the current I1(t) is

(a) 2.36 cos (4t – 41.070) A

(b) 2.36 cos (4t + 41.070) A

(c) 1.37 cos (4t – 41.070) A

(d) 2.36 cos (4t + 41.070) A

1. In the circuit shown below, the current I2(t) is

(a) 2.04 sin (4t + 92.130) A

(b) -2.04 sin (4t + 2.130) A

(c) 2.04 cos (4t + 2.130) A

(d) -2.04 cos (4t + 92.130) A

In a balanced y-connected three phase generator, VAB = 400 VMS

1. If the phase sequence is abc then phase voltages VA VB and VC are, respectively

(a) 231<00 , 231<1200 , 231<2400

(b) 231< – 300, 231< – 1500, 231<900

(c) 231<300, 231<1500, 231<-900

(d) 231<600, 231<1800, 231<-600

1. If phase sequence is acb, then phase voltages are

(a) 231<00, 231<1200, 231<2400

(b) 231<-300, 231<-1500, 231<900

(c) 231<300, 231<1500 , 231<-900

(d) 231<600, 231<1800, 231<-600

Determine the complex power for the given values in questions.

1. P = 269 W, Q = 150 VAR (capacitive)

(a) 150 – j269 VA

(b) 150 + j269 VA

(c) 269 – j150 VA

(d) 269 + j150 VA

1. Q = 2000 VAR, PF = 0.9 (leading)

(a) 4129.8 +j2000 VA

(b) 2000 + j4129.8 VA

(c) 2000 – j4129.8 VA

(d) 4129.8 – j2000 VA

1. S = 60 VA, Q = 45 VAR (inductive)

(a) 39.69 + j45 VA

(b) 39.69 – j45 VA

(c) 45 + j39.69 VA

(d) 45 – j39.69 VA

1. VMS = 200 V, P = 1KW, |Z| = 40 (inductive)

(a) 1000 – j681.25 VA

(b) 1000 + j681.25 VA

(c) 681.25 + j1000 VA

(d) 681.25 – j1000 VA

1. VMS = 21<200 V, VRMS = 21<200 V, IRMS = 8.5 <-500 A

(a) 154.6 + j89.3 VA

(b) 154.6 – j89.3 VA

(c) 61 + j167.7 VA

(d) 61 – j167.7 VA

1. VRMS = 120 <300 V, Z = 40 + J80

(a) 72 + j144 VA

(b) 72 – j144 VA

(c) 144 + j72 VA

(d) 144 – j72 VA

Common data for Questions 25 and 26

Consider the circuit shown below. initial conditions are I1(0) = I2(0) = 11 A

1. The value of current I1(1s) is

(a) 0.78 A

(b) 1.46 A

(c) 2.56 A

(d) 3.62 A

1. The value of current I2(1s) is

(a) 0.78 A

(b) 1.46 A

(c) 2.56 A

(d) 3.62 A

In the circuit shown below, all initial conditions are zero.

1. If IS(t) = 1 A, then the inductor current IL(t)

(a) 1 A

(b) t A

(c) t + 1 A

(d) zero

1. If IS(t) = 0.5 t A, then IL(t) is

(a) 0.5t – 3.25 x 10-3 A

(b) 2t – 3250 A

(c) 0.5t – 0.25 x 10-3 A

(d) 2t + 3250 A

1. If IS(t) = 2e-250t A then IL (t) is

(a) 4000/3 te-250t A

(b) 4000/3 e-250t

(c) 200/7 e-250t A

(d) 200/7 te-250t A

The circuit is as shown below. solve the problem and choose correct option.

1. The transfer function H1(s) = VO(s)/VS(s) is

(a) s(s3 + 2s2  + 3s + 1)-1

(b) (s3 + 3s2 + 2s + 1)-1

(c) (s3 + 2s2 + 3s + 2)-1

(d) s(s3 + 3s2 + 2s2 + 2)-1

1. The transfer function H2(s) = IO(s)/VS(s) is

(a) -s/(s3 + 3s2 + 2s + 1)

(b) -(s3 + 3s2 + 2s + 1)-1

(c) -s/(s2 + 2s2 + 3s + 1)

(d) (s3 + 2s2 + 3s + 2)-1

Consider the circuit shown below. All initial conditions are zero.

1. The transfer function IO(s)/IIN(s) is

(a) (s + 1)/2s

(b) 2s(s + 1)-1

(c) (s + 1)s-1

(d) s(s + 1)-1

1. If IIN (t) = 48t then IO(t) will be

(a) 48(t) -e-t u(t) A

(b) 48(t) -4e-1 u(t) A

(c)4e-t u(t) -48(t) A

(d) e-1 u(t) -8(t) A

1. If IIN(t) = tu(t), then, IO(t) will be

(a) e-t u(t) A

(b) (1 – e-t) u(t) A

(c) u(t) A

(d) (2 – e-t) u(t) A

Consider the circuit shown below

1. The current ratio transfer function IO/IS is

(a) s(s + 4)/s2 + 3s + 4

(b) s(s + 4)/(s + 1) (s + 3)

(c) s2 + 3s + 4/s(s + 4)

(d) (s + 1) (s + 3)/s(s + 4)

1. The response is

(a) over damped

(b) under damped

(c) critically damped

(d) can not be determined

1. If input IS is 2u(t) A, the output current IO is

(a) (2e-t – 3te-3t) u(t) A

(b) (3te-1  e-3t) u(t) A

(c) (3e-1 – e-3t) u(t) A

(d) (e-3t – 3e-t) u(t) A

Common data for Questions 38 and 39

In the following circuit I1 = 4 sin 2t A and I2 = 0.

1. The voltage V1 is

(a) -16 cos 2t V

(b) 16 cos 2t V

(c) 4 cos 2t V

(d) -4 cos 2t V

1. The voltage V2 is

(a) 2 cos 2t V

(b) -2 cos 2t V

(c) 8 cos 2t V

(d) -8 cos 2t V

Common data for Questions 40 and 41.

Consider the following circuit shown below :

1. If I1 = 0 and I2 = 2sin 4t A, the voltage V1 is

(a) 24 cos 4t V

(b) -24 cos 4t V

(c) 1.5 cos 4t V

(d) -1.5 cos 4t V

1. If I1 = e-2t V and I2 = 0, the voltage V2 is

(a) -6e-2t V

(b) 6e-2t V

(c) 1.5e-2t V

(d) -1.5e-2t V

Commons data for Questions 42 and 43

Consider the circuit shown below:

1. If currents I2 = 3 cos 4t A and I2 = 0, then voltage V1 and V2 are

(a) V1 =- 24 sin 4t V, V2 = – 24 sin 4t V

(b) V1 = 24 sin 4t V, V2 = -36 sin 4t V

(c) V1 = 1.5 sin 4t V, V2 = sin 4t V

(d) V1 = – 1.5 sin 4t V, V2 = sin 4t V

1. If currents I1 = 0 and I2 = 4 sin 3t A, then voltage V1 and V2 are

(a) V1 = 24 cos 3t V, V2 = 36 cos 3t V

(b) V1 = 24 cos 3t V, V2 = – 36 cos 3t V

(c) V1 = – 24 cos 3t V, V2 = 36 cos  3t V

(d) V1 = – 24 cos 3t V, V2 = – 36 cos 3t V

Common data for Questions 44 and 45

In the circuit shown below,

I1 = 3 cos 3t A and I2 = 4 sin 3t A.

1. The voltage V1 is

(a) 6 (-2 cos t + 3 sin t) V

(b) 6 (2 cos t + 3 sin t) V

(c) – 6 (2 cos t + 3 sin t) V

(d) 6 (2 cos t – 3 sin t) V

1. The voltage V2 is

(a) 3(8 cos 3t – 3 sin t) V

(b) 6(2 cos t + 3 sin t) V

(c) 3(8 cos 3t + 3 sin 3t) V

(d) 6 (2 cos t – 3 sin t) V

Common data for Questions 46 and 47

In the circuit shown below, I1 = 5 sin 3t A and I2 = 3 cos 3t A.

1. The voltage V1 is

(a) 9 (5 cos 3t + 3 sin 3t) V

(b) 9 (5 cos 3t – 3 sin 3t) V

(c) 9 (4 cos 3t + 5 sin 3t) V

(d) 9 (5 cos 3t – 3 sin 3t) V

1. The voltage V2 is

(a) 9(-4 sin 3t + 5 cos 3t) V

(b) 9(4 sin 3t – 5cos 3t) V

(c) 9(-4 sin 3t – 5 cos 3t) V

(d) 9(4 sin 3t + 5 cos 3t) V

Common data for Questions 48 to 50

Consider the circuit shown below :

1. The voltage VAC of terminals AD is

(a) 60 V

(b) -60 V

(c) 180 V

(d) 240 V

1. The voltage VBC of terminals BD is

(a) 45 V

(b) 33 V

(c) 69 V

(d) 105 V

1. The voltage VCC of terminals CD is

(a) 50 V

(b) zero

(c) -36 V

(d) 36 V

1. (a)

number of branches b = 8

number of links I = 4

number of twigs t = b  – I = 4

Rank of matrix = n – 1 = t = 4

1. (b)

we know the branch current and loop current are related as

[IB] =[BT][IL]

So, fundamental loop-matrix is

f-loop 1 includes branch (2,4,6,7) and direction of branch 2 is opposite to other (b only).

1. (a)

There are 8 branches and 4 + 1 = 5 nodes

number of links = 8 – 5 + 1 = 4

so, independent mesh equation = number of links.

1. (d)

we know that [VB] = ART[VIN]

So, reduced incidence matrix is

at node 1, three branches leave so the only option is (d).

1. (c)

Thevenized the circuit across R, RTH = 2

1. (a)

the circuit is shown below.

ISC = 10 A, RTH = 2

PMAX = (10/2)2 x 2 = 50 W

1. (b)

for VOC the circuit is as follows

IX + 0.9 = 10IX

IX = 0.1 A

VOC = 3 x 10IX  30IX

VOC = 3 V

ISC = 10IX = 1 A

RTH = 3/1 = 3

1. (a)

VTH = VOC = 3 V

RL = 3

PMAX = 32 /4 x 3 = 0.75 W

1. (d)

the given circuit has mirror symmetry. it is modified and re-drawn as shown below.

Now, in this circuit all straigh connection have been cut as shown below.

VA = 6 x (2 + 3)/2 + 3 + 1 = 5 V

1. (b)

since, both sources have opposite hence short-circuit the all straight through connections as shown in figure below.

VA = 6 x (6||3)/2 + 1 = – 4V

1. (b)

we thevenized the left side of xx and source transformed right side of yy’

VXX’ = VTH

4/8 + 8/24 = 5 V

1/8 + 1/24

RTH = 8||(16 + 8) = 6

1. (d)

thevenin equivalent seen from terminal yy’ is

VYY’ = VTH = 4/24 + 8/8 = 7 V

1/24 + 1/8

RTH = (8 + 16)||8 = 6

1. (c)

V(0+) = 10 V = V(0+)

for 0 < t < 1.5 s,

= 8/2 x 0.05 = 0.2 s, VOC = 5 V

V(t) = 5 + (10 – 5) e-t/0.2

= 5 + 5e-5t

v(1s) = 5.03 v

1. (c)

V(1.5s) = 5.002 V

for t > 1.5s, 8 x 0.05 = 0.4

V(t) = 10 + (5 – 10)e(t – 1.5/0.4)

= 10 – 5e-2.5(t – 1.5)

for t > 1.5s, V(2s) = 8.57 V

1. (c)

the circuit is shown below

5<00 = I1(j4 + 1 + 1 – j/4) – I2(1 – J/4)

(8 + j15) I1 – (4 – J) I2 = 20 <0…………..(1)

-10 < -300 = I2 (1 + J4 + 1 – J/4) -I1 (1 – J/4)

(4 – j) I1 – (8 + J15)I2 = 40 < – 300 …………(2)

I1[(8 + J15)2 – (4 – J)2]

= (20 <0) (8 +J15) – (40 <-300) (4 – J)

I1 (-176 + J248) = 41.43 + J414.64

I1 = 1.03 – J0.9 = 1.37 <- 410.7

1. (a)

I2 = (8 + J15)(1.03 – J0.9) – 20 <00/4 – J

= – 0.076 + J2.04

I2 = 2.04 <92.130

1. (b)

the circuit is shown below.

VA = 400/3 <- 300 = 231 < -300 V

VA = 213 <-1500 V, VC = 231 < – 2700 V

1. (c)

for the acb sequence,

VAB = VA – VB = VP <00 – VP <1200

400 = VP (1 + 1/2 – J3/2) = VP 3 < – 300

VP = 400/3 <300

VA = VP <00 = 231 <300 V

VB = VP <1200 = 231 <1500 V

VC = VP <2400 = 231 < – 900 V

1. (c)

S = P – JQ = 269 – j150 VA

1. (d)

power factor = cos 0 = 0.9   0 = 25.840

Q = s sin 0

S = Q/sin0 = 2000/sin 25.840 = 4588.6 VA

P = S cos 0 = 4129.8

S = 4129.8 – j2000

1. (a)

Q = S sin 0  sin 0 Q /s = 45/60

0 = 48.590

P = S cos 0 = 39.69

S = 39.69 + J45 VA

1. (b)

S = |VRMS|2/|Z| = (2202)/40 = 1210

cos 0 = P/S = 1000/1210 = 0.8264

0 = 34.260

Q = S sin 0 = 681.25

S = 1000 + J681.25 VA

1. (c)

S = VRMSIRMS = (21<200) (85.<500)

= 61 + J167.7 VA

1. (a)

S = |V|2/Z* = (120)2/40 – J80 = 72 + J144 VA

1. (d)

I1 + 5dI1/dt – 3dI2/dt = 0

2I2 + 3dI2/dt – 3dI1/dt = 0

(1 + 5s)I1 – 3sI2 = 0, – 3sI1 + (2 + 3s)I2 = 0

(1 + 5s)I1 – (3s) (3s)I1/2 + 3s = 0

6s2 + 13s + 2 = 0

s = – 1/6 – 2

I1 = Ae-1/6t + Be-2t, I (0) = A + B = 11

In differential equation putting t = 0 and solving

dI1(0+)/dt = – 33/2, dI2(0+)/dt = 143/6

-A/6 – 2B = – 33/2   A = 3, B = 8

I1 = 3et/6 + 8e-2t

I1(1s) = 3e-1/6 + 8e-2 = 3.62 A

1. (a)

I2 = Cet/6 + De-2t

I2(0) = 11 = C + D dI2(0)/dt = -143/6 = – C/6 – 2D

C = – 1 and D = 12

I2 = – e-t/6 + 12e-2t A

I2(1s) + e-t/6 + 12e-2t = 0.78 A

1. (a)

IS = V/100/65 + 10-3 dV/dt + IL

V = 10 x 10-3 dIL/dt

IS = 65/100 (10 x 10-3) dIL/dt + 10-3 (10 x 10-3) d2IL/dt + IL = 0

d2IL/dt + 650 dIL/dt + 105IL = 105IS

Trying IL(t) = B,

0 + 0 + 105B = 105

B = 1, IL = 1A

1. (a)

trying IL(t) = At + B,

0 + 650 A + (at + B) 105 = 105(0.5t)

A = 0.5

650 x 0.5 + B.105 = 0

B = – 3.25 x 10-3

1. (b)

Trying IL(t) = (A1 + A2t)e-250t

105 x 2 x e-250t  = (-250)2e-250t (A1 + A2t)

-500A2e-250t + 650 [-250e-250t(A1 + A2t) + A2e-250t] + 105(A1 + A2)e-250t

equating the coefficient,

-500A2 + 650A2 = 2 x 105

A2 = 4000/3, A1 = 0

1. (c)

the circuit is shown below,

V(s) = VS(s) + VO(s)/s/1 + s + 1/s = sVS(s) + VO(s)/s2 + s + 1

(s2 + s + 1)V(s) = sVS(s) + VO(s)

VO(s) = V1(s)/s/s + 1/s + 1 = V1(s)/s2 + s + 1

V1(s) = (s2 + s + 1)VO(s)

(s2 + s + 1)2V0(s) = sVO(s) = sVS(s) + VO(s)

(s4 + s2 + 1 + 2s3 + 2s2 + 2s) VO (s) = sVS(s)

VO(s)/VS(s) = 1/s3 + 2s2 + 3s + 2

1. (d)

IO(s)/VS(s) = VO(s)/ VS(s)= H1(s)

= (s3 + 2s2 + 3s + 2)-1

1. (d)

IO(s)/IIN(s) = s/s+1/s /s + 1 + s +1 = s/s + 1

1. (b)

IIN(s) = 4

IO(s) = 4s/s + 1 = 4 – 4/s + 1

IO(t) = 48(t) -4e-t u(t)

1. (b)

IIN(s) = 1/s2

IO(s) = 1/s(s + 1) = 1/s – 1/s + 1

IO(t) = u(t) -e-t u(t) = (1 – e-t) u(t)

1. (b)

IO/IS = s + 4/s + 4 +3/s = s(s + 4)/(s + 1) (s +3)

1. (a)

the characteristic equation is (s + 1) (s + 3) = 0.

being real and unequal root, it is over damped response.

1. (c)

IS = 2u(t)    IS(s) = 2/s

IO(s) = 2(s + 4)/(s + 1) (s + 3) = 3/s + 1 – 1/s + 3

IO = (3e-t – e-3t) u(t) A

1. (b)

V1 = 2 dI1/dt + 1dI2/dt = 2dI1/dt = 16 cos 2t V

1. (c)

V2 = (1) dI2/dt + (1) dI1/dt = dI1/dt = 8 cos 2t V

1. (b)

V1 = 3 dI1/dt – 3 dI2/dt = – 3dI2/dt = – 24 cos 4t V

1. (b)

V2 = 4 dI2/dt – 3dI1/dt = – 3dI2/dt = 6e-2t V

1. (b)

V1 = 2 dI1/dt – 2dI2/dt = 2 dI1/dt = – 24 sin 4t V

V2 = – 3 dI2/dt + 2dI1/dt = 2dI1dt = – 24 sin 4t V

1. (d)

V1 = 2dI1/dt – 2dI2/dt = – 2dI2/dt

= – 24 cos 3t V

V2 = 3dI2/dt + 2 dI1/dt = – 3dI2/dt = – 36 cos 3t V

1. (d)

V1 = 2dI1/dt + 1dI2/dt

= – 18 sin t + 12 cos t = 6 (2 cos t – 3 sin t) V

1. (a)

V2 = 2dI2/dt + 1dI1/dt

= 24 cos 3t – 9 sin 3t = 3 (8 cos 3t – 3 sin 3t) V

1. (a)

V1 = 3 3dI1/dt – 3dI2/dt

= 45 cos 3t + 27 sin 3t = 9 (5 cos 3t + 3 sin 3t) V

1. (d)

V2 = – 4dI2/dt + 3dI1/dt

= 36 sin 3t + 45 cos 3t = 9 (4 sin 3t + 5 cos 3t) V

1. (c)

VAG = 20 d(6t)/dt + 4 d(15t)/dt = 180 V

1. (b)

VBG = 3 d(15t)/dt + 4d6(t)/dt – 6 d(6t)/dt

= 33 V

1. (c)

VCG = – 6d(6t)/dt = – 36 V