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

Common Data/Linked Answer Questions

Statement for Linked Answer Questions 1 and 2

Branch current and loop current relation and expressed in matrix form below where I_I represents branch current and I_K loop current.

1. The rank of incidence matrix is

(a) 4

(b) 5

(c) 6

(d) 8

2. The directed graph will be

Statement for Linked Answer Questions 3 and 4

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

where V_I is the branch voltage and V_K is the node voltage with resect to datum node.

3. The independent mesh equations for this network are

(a) 4

(b) 5

(c) 6

(d) 7

4. The oriented graph for this network is

Statement for Linked Answer Questions 5 and 6

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

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

(a) 4

(b) 3

(c) 2

(d) none of these

6. The maximum power will be

(a) 50 W

(b) 100 W

(c) 200 W

(d) 400 W

Statement for Linked Answer Questions 7 and 8

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

7. The value of R_L will be

(a) 2

(b) 3

(c) 1

(d) none of these

8. 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 :

9. If V_S1 = V_S2 = 6 V then the value of V_A is

(a) 3 V

(b) 4 V

(c) 6 V

(d) 5 V

10. If V_S1 = 6 V and V_S2 = – 6 V then the value of V_A 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.

11. As viewed from terminals x and x’ is

(a) 8 V, 6

(b) 5 V, 6

(c) 5 V, 32

(d) 8 V, 32

12. 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.

13. 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

14. 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.

15. 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

16. 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

Statement for linked Answer Questions 17 and 18

In a balanced y-connected three phase generator, V_AB = 400 V_MS

17. If the phase sequence is abc then phase voltages V_A V_B and V_C 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

18. 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

Statement for linked Answer Questions 19 to 24

Determine the complex power for the given values in questions.

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

(a) 150 – j269 VA

(b) 150 + j269 VA

(c) 269 – j150 VA

(d) 269 + j150 VA

20. 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

21. 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

22. V_MS = 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

23. V_MS = 21<20⁰ V, V_RMS = 21<20⁰ V, I_RMS = 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

24. V_RMS = 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

25. The value of current I₁(1s) is

(a) 0.78 A

(b) 1.46 A

(c) 2.56 A

(d) 3.62 A

26. The value of current I₂(1s) is

(a) 0.78 A

(b) 1.46 A

(c) 2.56 A

(d) 3.62 A

Statement for linked Answer Questions 27 to 29

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

27. If I_S(t) = 1 A, then the inductor current I_L(t)

(a) 1 A

(b) t A

(c) t + 1 A

(d) zero

28. If I_S(t) = 0.5 t A, then I_L(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

29. If I_S(t) = 2e^-250t A then I_L (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

Statement for linked Answer Questions 30 and 31

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

30. The transfer function H₁(s) = V_O(s)/V_S(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

31. The transfer function H₂(s) = I_O(s)/V_S(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

Statement for linked Answer Questions 32 to 34

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

32. The transfer function I_O(s)/I_IN(s) is

(a) (s + 1)/2s

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

(c) (s + 1)s^-1

(d) s(s + 1)^-1

33. If I_IN (t) = 48t then I_O(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

34. If I_IN(t) = tu(t), then, I_O(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

Statement for Linked Answer Questions 35 to 37

Consider the circuit shown below

35. The current ratio transfer function I_O/I_S 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)

36. The response is

(a) over damped

(b) under damped

(c) critically damped

(d) can not be determined

37. If input I_S is 2u(t) A, the output current I_O 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.

38. The voltage V₁ is

(a) -16 cos 2t V

(b) 16 cos 2t V

(c) 4 cos 2t V

(d) -4 cos 2t V

39. 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 :

40. 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

41. 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:

42. 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

43. 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.

44. 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

45. 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.

46. 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

47. 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 :

48. The voltage V_AC of terminals AD is

(a) 60 V

(b) -60 V

(c) 180 V

(d) 240 V

49. The voltage V_BC of terminals BD is

(a) 45 V

(b) 33 V

(c) 69 V

(d) 105 V

50. The voltage V_CC of terminals CD is

(a) 50 V

(b) zero

(c) -36 V

(d) 36 V

Common Data/Linked Answer Questions

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

2. (b)

we know the branch current and loop current are related as

[I_B] =[B^T][I_L]

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).

3. (a)

There are 8 branches and 4 + 1 = 5 nodes

number of links = 8 – 5 + 1 = 4

so, independent mesh equation = number of links.

4. (d)

we know that [V_B] = A_R^T[V_IN]

So, reduced incidence matrix is

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

5. (c)

Thevenized the circuit across R, R_TH = 2

6. (a)

the circuit is shown below.

I_SC = 10 A, R_TH = 2

P_MAX = (10/2)² x 2 = 50 W

7. (b)

for V_OC the circuit is as follows

I_X + 0.9 = 10I_X

I_X = 0.1 A

V_OC = 3 x 10I_X  30I_X

V_OC = 3 V

I_SC = 10I_X = 1 A

R_TH = 3/1 = 3

8. (a)

V_TH = V_OC = 3 V

R_L = 3

P_MAX = 3² /4 x 3 = 0.75 W

9. (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.

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

10. (b)

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

V_A = 6 x (6||3)/2 + 1 = – 4V

11. (b)

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

V_XX’ = V_TH

4/8 + 8/24 = 5 V

1/8 + 1/24

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

12. (d)

thevenin equivalent seen from terminal yy’ is

V_YY’ = V_TH = 4/24 + 8/8 = 7 V

1/24 + 1/8

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

13. (c)

V(0⁺) = 10 V = V(0⁺)

for 0 < t < 1.5 s,

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

V(t) = 5 + (10 – 5) e^-t/0.2

= 5 + 5e^-5t

v(1s) = 5.03 v

14. (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

15. (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

16. (a)

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

= – 0.076 + J2.04

I₂ = 2.04 <92.13⁰

17. (b)

the circuit is shown below.

V_A = 400/3 <- 30⁰ = 231 < -30⁰ V

V_A = 213 <-150⁰ V, V_C = 231 < – 270⁰ V

18. (c)

for the acb sequence,

V_AB = V_A – V_B = V_P <0⁰ – V_P <120⁰

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

V_P = 400/3 <30⁰

V_A = V_P <0⁰ = 231 <30⁰ V

V_B = V_P <120⁰ = 231 <150⁰ V

V_C = V_P <240⁰ = 231 < – 90⁰ V

19. (c)

S = P – JQ = 269 – j150 VA

20. (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

21. (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

22. (b)

S = |V_RMS|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

23. (c)

S = V_RMSI_RMS = (21<20⁰) (85.<50⁰)

= 61 + J167.7 VA

24. (a)

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

25. (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 = 3e^t/6 + 8e^-2t

I₁(1s) = 3e^-1/6 + 8e^-2 = 3.62 A

26. (a)

I₂ = Ce^t/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

27. (a)

I_S = V/100_/65 + 10^-3 dV/dt + I_L

V = 10 x 10^-3 dI_L/dt

I_S = 65/100 (10 x 10^-3) dI_L/dt + 10^-3 (10 x 10^-3) d²I_L/dt + I_L = 0

d²I_L/dt + 650 dI_L/dt + 10⁵I_L = 10⁵I_S

Trying I_L(t) = B,

0 + 0 + 10⁵B = 10⁵

B = 1, I_L = 1A

28. (a)

trying I_L(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

29. (b)

Trying I_L(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

30. (c)

the circuit is shown below,

V(s) = V_S(s) + V_O(s)/s/1 + s + 1_/s = sV_S(s) + V_O(s)/s² + s + 1

(s² + s + 1)V(s) = sV_S(s) + V_O(s)

V_O(s) = V₁(s)/s/s + 1_/s + 1 = V₁(s)/s² + s + 1

V₁(s) = (s² + s + 1)V_O(s)

(s² + s + 1)²V₀(s) = sV_O(s) = sV_S(s) + V_O(s)

(s⁴ + s² + 1 + 2s³ + 2s² + 2s) V_O (s) = sV_S(s)

V_O(s)/V_S(s) = 1/s³ + 2s² + 3s + 2

31. (d)

I_O(s)/V_S(s) = V_O(s)/ V_S(s)= H₁(s)

= (s³ + 2s² + 3s + 2)^-1

32. (d)

I_O(s)/I_IN(s) = s/s+1/s /s + 1 + s +1 = s/s + 1

33. (b)

I_IN(s) = 4

I_O(s) = 4s/s + 1 = 4 – 4/s + 1

I_O(t) = 48(t) -4e^-t u(t)

34. (b)

I_IN(s) = 1/s²

I_O(s) = 1/s(s + 1) = 1/s – 1/s + 1

I_O(t) = u(t) -e^-t u(t) = (1 – e^-t) u(t)

35. (b)

I_O/I_S = s + 4/s + 4 +3_/s = s(s + 4)/(s + 1) (s +3)

36. (a)

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

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

37. (c)

I_S = 2u(t)    I_S(s) = 2/s

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

I_O = (3e^-t – e^-3t) u(t) A

38. (b)

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

39. (c)

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

40. (b)

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

41. (b)

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

42. (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

43. (d)

V₁ = 2dI₁/dt – 2dI₂/dt = – 2dI₂/dt

= – 24 cos 3t V

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

44. (d)

V₁ = 2dI₁/dt + 1dI₂/dt

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

45. (a)

V₂ = 2dI₂/dt + 1dI₁/dt

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

46. (a)

V₁ = 3 3dI₁/dt – 3dI₂/dt

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

47. (d)

V₂ = – 4dI₂/dt + 3dI₁/dt

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

48. (c)

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

49. (b)

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

= 33 V

50. (c)

V_CG = – 6d(6t)/dt = – 36 V