know A source of angular frequency 1 rad/s has a source impedance consisting of resistance in series with 1 H inductance. the load that will obtain the maximum power transfer is ? answers given below after 100 questions ?

1. For the circuit shown in figure, the time constant RC = 1 ms. the input voltage is VI(t) = 2sin 103t. the output voltage VO(t) is equal to

(a) sin (103t – 450)

(b) sin (103t + 450)

(c) sin (103t – 530)

(d) sin (103t + 530)

1. The minimum number of equations required to analyze the circuit shown in figure is

(a) 3

(b) 4

(c) 6

(d) 7

1. A source of angular frequency 1 rad/s has a source impedance consisting of resistance in series with 1 H inductance. the load that will obtain the maximum power transfer is

(a) 1 resistance

(b) 1 resistance in parallel with 1 H inductance

(c) 1 resistance in series with 1 F capacitor

(d) 1 resistance in parallel with 1 F capacitor

1. In the following circuit a network and its thevenin and norton equivalents are given.

The value of the parameter are VTH RTH IN RN

(a) 4 V, 2, 2 A, 2

(b) 4 V, 2, 2 A, 3

(c) 8 V, 1.2, 30/3 A, 1.2

(d) 8 V, 5, 8/5 A, 5

1. In the above circuit, the value of V1 is

(a) 6 V

(b) 7 V

(c) 8 V

(d) 10 V

1. The thevenin impedance across the terminals ab of the network shown below is

(a) 2

(b) 6

(c) 6.16

(d) 4/3

1. In the following circuit, the value of I1 is

(a) 3 A

(b) 0.75 mA

(c) 2 mA

(d) 1.75 mA

1. For the following circuit, the value of RTH is

(a) 3

(b) 12

(c) 6

(d) infinite

1. A series RLC circuit has a resonance frequency of 1 khz and a quality factor Q= 100. if each R,L and C is doubled from its original vaule, the new Q of the circuit is

(a) 25

(b) 50

(c) 100

(d) 200

1. The laplace transform of I(t) is given by I(s) = 2/s (1 + s) .

As t – 0 the value of I(t) tends to

(a) zero

(b) 1

(c) 2

(d) infinite

1. The differential equation for the current I(t) in the circuit of figure is

(a) 2d2l/dt2 + 2 dl/dt + l(t) = sin t

(b) d2l/dt2 + 2 dl/dt + 2l(t) = cos t

(c) 2 d2l/dt2 + 2 dl/dt + l(t) = cos t

(d) d2l/dt2 + 2 dl/dt + 2l(t) = sin t

1. The dependent current source shown in figure

(a) delivers 80 w

(b) absorbs 80 w

(c) delivers 40 w

(d) absorbs 40 w

1. In figure, the switch was closed for a long time before opening at t = 0. the voltage VX at t = 0+ is

(a) 25 v

(b) 50 v

(c) – 50 v

(d) zero

1. For the given the incidence matrix A is given by

(a) [-1  -1 0 /0 1 1/-1  0 -1]

(b) [1  0  -1/ 1  1  0/0 -1 1]

(c) [1  -1   0/ 0   1  1  -1  0   -1]

(d) none of these

1. Match the list 1 with list II and select the correct answer using the codes given below the lists

List 1                                  List II

1. 1, 2, 3, 8 1. twigs
2. 4, 5, 6, 7 2. links
3. 1, 2, 3, 4 3. Fundamental cut-set
4. 1, 4, 5, 6, 7 4. Fundamental loop

Codes

A         B        C         D

(a)   3         1         2          4

(b)   2         3        1            4

(c)   3       2           1            4

(d) 1        2           3             4

1. For a given network, the number of independent mesh equations (m) and the number of independent node equations (n) are related as follows

(a) m > n always

(b) m > n always

(c) m = n always

(d) m > or < n depending upon the form of the network

1. The natural frequency sn of the given circuit is

(a) 1

(b) -1

(c) – 1/2

(d) 2

1. Steady state is reached for the circuit. VC as marked in the circuit is

(a) zero

(b) 4

(c) 5

(d) 10/3

1. The natural frequency of the circuit is given by

(a) – 6

(b) -8

(c) 15/2

(d) -10

1. Initial voltage on capacitor V0 as marked |V0| = 5 V. VS = 8u(t), where u(t) is the unit step. the voltage marked V at = 0+ is given by

(a) 1 V

(b) -1 V

(c) 13/3 V

(d) -13/3 V

1. The voltage e0 in figure is

(a) 2 V

(b) 4/3 V

(c) 4 V

(d) 8 V

1. If each branch of a delta circuit has impedance 3z, then each branch of the equivalent Y-circuit has impedance

(a) Z/3

(b) 3Z

(c) 3/3Z

(d) Z/3

1. The admittance parameter Y12 in the two-port network in figure is

(a) -0.2 mho

(b) 0.1 mho

(c) -0.05 mho

(d) 0.05 mho

1. In the circuit of figure the voltage v(t) is

(a) eat – ebt

(b) eat – ebt

(c) aeat – bebt

(d) aeat + bebt

1. In the circuit of figure, the value of the voltage source E is

(a) -16 V

(b) 4 V

(c) – 6 V

(d) 16 V

1. The voltage across the 1 k resistor between A and B of the network shown in the given figure is

(a) 12 V

(b) 14/3 V

(c) 10 V

(d) 26/3 V

1. S is open, V0 = 6 V, I = 0. S is closed at t = 0, the values of I and dI/dt at t = 0+ are given by

(a) 0, 6

(b) 2, 0

(c) 1, 3

(d) 0, 3

1. S is in position (a) for a long time. S moved to position (b) at t = 0 , at t = 0+ , the values of IC and VL are

(a) – 5, 0

(b) 5, 0

(c) 0, 4

(d) – 2, 3

1. Given H(s) = X/Y = s + 1/s2 + s + 1 x(t) = cos t <00 . the phasor Y is given by

(a) 2 < 450

(b) 1/2 <450

(c) 1 <00

(d) 2/5 < tan-1 – <tan-2 2

1. In the figure VS = 10 cos t = 10 <00. the current drawn from VS is given by the phasor

(a) 2/5 < -450

(b) 2/2 <450

(c) 2 <00

(d) 2<900

1. Given, V2 = 2 cos 2t = 2 <00, the value of VS is given by

(a) 3 <00

(b) 29/ <tan-1 2.5

(c) 104 < tan-1 5

(d) 2/2 < 450

1. In the circuit VS = cos 2t. the value of C is chosen so that I from VS is phase with VS. then IC leads IL by an angle given by

(a) 450

(b) 900

(c) 1350

(d) 00

1. Pcomplex = 200 + j150 from a source VS = 200 cos ot the impedance across VS is given by

(a) 16 + j12

(b) 128 + j96

(c) 64 + j48

(d) 32 + j24

1. A source VS = 100 cos ot is connected to an impedance which drawn a power PAY = 200 W at power factor 0.8 lagging. the current drawn from the source is I = IM cos (0t + 0). the values of IM and 0 are given by

(a) 5 A and – cos-1 0.8

(b) 5 A and – cos-1 0.8

(c) 5/2 A and cos-1 0.8

(d) 5 A and 00

1. In the circuit shown below, if f = 50 Hz then, current I(t) is

(a) 14.14 cos (50 t – 450) A

(b) -11.4 cos (100t + 52.150) A

(c) 1.414 cos (100t + 450) A

(d) 1.414 cos (100t + 450) A

1. In the given circuit of figure, current I(t) is

(a) 10 cos (100t + 53.10)

(b) 14.14 cos (100t + 53.10)

(c) 10 cos (100 t – 53.10)

(d) 10 cos (100t – 53.10)

1. In the given circuit below, voltage VA is

(a ) 2 V

(b) (2 + 6j) V

(c) 1 V

(d) -2 V

1. In the given circuit of figure, current I(t) is

(a) 20 cos (300t + 68.20) A

(b) 20 cos (300t – 68.20) A

(c) 2.48 cos (300t + 68.20) A

(d) 2.48 cos (300t – 68.20) A

1. In the given circuit below, voltage VC (t) is

(a) 0.89 cos (103 t – 63.430) V

(b) 0.89 cos (103 t + 63.430) V

(c) 0.45 cos (103 t + 26.570) V

(d) 0.45 cos (103 t – 26.570) V

1. In the given circuit below, voltage VC (t) is

(a) 1/2 cos (2t – 450) V

(b) 1/2 cos (2t + 450) V

(c) 1/2 sin (2t – 450) V

(d) 1/2 sin (2t + 450) V

1. In the following circuit, the value of Z(s) is

(a) s2 + 1.5s + 1/s(s + 1)

(b) s2 + 3s + 1/s(s + 1)

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

(d) 2s2 + 3s + 1/2s(s + 1)

1. In the following circuit, the value of Z(s) is

(a) s2 + s + 1/s(s + 1)

(b) 2s2 + s + 1/s (s + 1)

(c) s(s + 1) 2s2 + s + 1

(d) s(s + 1)/s2 + s + 1

1. In the following circuit, the value of Z(s) is

(a) s2 + 1/s2 + 2s + 1

(b) 2 (s2 + 1)/(s + 1)2

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

(d) s2 + 1/3s + 2

1. In the following circuit, the value of Z(s) is

(a) 3s2 + 8s + 7/s (5s + 6)

(b) s(5s + 6)/3s2 + 8s + 7

(c) 3s2 + 7s + 6/s (5s + 6)

(d) s(5s + 6)/3s2 + 7s + 6

1. In the following circit, current I is

(a) 2.5 A

(b) (3 +2j) A

(c) (1 + j) A

(d) (2.5 – 1.75j) A

1. In the following circuit impedance Z is

(a) 92. 36 – j68.34

(b) 38.24 – j 49.88

(c) 42.16 + j18.46

(d) 13.16 + j17.33

1. In the given circuit below, voltage VC (t) is

(a) 2.25 cos (5t + 1500)

(b) 2.25 cos (5t – 1500) V

(c) 2.25 cos (5t + 140.710) V

(d) 2.25 cos (5t – 140.710) V

1. In the given circuit below, current I(t) is

(a) 2 sin (2t + 5.770) A

(b) cos (2t – 84.230) A

(c) 2sin (2t – 5.770) A

(d) cos (2t + 84.230) A

1. The s-domain equivalent of the following circuit is

(c) both (a) and (b)

(d) none of these

1. The s-domain equivalent of the following circuit is

(a) both (a) and (b)

(d) none of these

1. (a)

VO (s) = 1/joC/R + 1/JOC VI(s) = 1/1 + joRC VI(s)

given, VI(t) = 2 sin 103t, RC = 1 ms

in phasor form,

VI = 2<00

Putting values

VO(s) 1/1 + j x 103 x 10-3 vi (s)

= 1/1 + j VI (S)

In phasor form,

VO = 1/2 <-450 x 2 <00

1< – 450

hence, VA (t) = sin (t – 450) = sin (103t – 450)

1. (b)

number of branches in circuit = 8

number of nodes = 5

minimum number of equations = number of loops

= b – n + 1

b = number of branches

n = number of nodes

minimum number of equations = 8 – 5 + 1 = 4

1. (c)

ZN = Source impedance

R + JOL = 1 + J x 1 x 1 = 1 + j

for maximum power transfer to load,

ZL = ZS = 1 – J = R + 1/JOC

The load ZL is 1 resistor in series with 1 F capacitor.

1. (d)

VOC = 2 x 2 + 4 = 8 V = VTH

RTH = 2 + 3 + 5 = RN, IS = VTH/RTH = 8/5 A

1. (d)

4/1 + 1 + 12/1 + 2

V1 = 1      = 6 V

1/1 + 1 + 1/6 + 1/1 +

1. (b)

After replacing all the sources by their internal impedance.

RTH = (3||6) + (8||8) = 6

1. (b)

If we solve this circuit direct, we haven to deal with three variables. but by simple manipulation variable can be reduced to one. by changing the L.H.S and R.H.S in thevenin equivalent.

I1 = 20 – 6 – 8/2 + 4 + 2 = 0.75 mA

1. (c)

After killing the source, RTH = 6

1. (b)

Quality factor (Q) of series RLC circuit

= 1/R L/C

QNEW = 1/2R 2L/2C x 1/2 x 1/R 1/C = 1/2 Q = 1/2 x 100

QNEW = 50

1. (c)

Final value theorem states that if the laplace transform of f(t) is f(s) and if sF(s) is analytic on the imaginary axis and in the right-half of the s-plane, then

lim f(t) = lim sF (s)

lim I(t) = lim sI(s)

lim s x 2/s(1 + s) = 2

1. (c)

applying KVL in circuit

sin 2t = 2I(t) = 2 dI(t)/dt + 1 I(t) dt

differentiating and re-arranging,

2d2I(t)/dt2 = 2dI(t)/dt + I(t) = cos t

1. (a)

Resistance of current source is in parallel with it.

let voltage of node A is VA.

applying kCL at node A

V1 – VA/5 + V1/5 = VA/5

V1 = VA

VA = 20 V

As node A is at the voltage and current flowing from +ve of dependent source hence, power is delivered.

power delivered = VAI = VA. V1/5

VA2 /5 = 20 x 20/5 = 80 W

1. (c)
2. (d)

incidence matrix is given by

1. (a)
2. (d)

m > or < n depedanding upon the form of network.

1. (b)

the circuit is re-drawn to calculate the natural frequency given below.

SN = L/REQ

REQ = 4||4 = 2

L = 2 so, SN = -2/2 = -1

1. (b)

re-drawn circuit under steady state is given below.

VC = 4 x 5/4 + 1 = 4 V

1. (a)

Re-drawn circuit to calculate the natural frequency is given below.

here, natural frequency is given as

SN = – REQ.C

REQ = 6, C = 1

SN = – 6 x 1 = – 6

1. (c)

Case 1 when VS = 8U(t) = 8 v source is treate, the equivalent circuit becomes

Applying voltage divide method, we get

I = V/REQ = 8/1 + 1||1 = 8/1 + 1/2 = 16/3 A

I1 = 16/3 x 1/2 = 8/3 A

Case 2 when |VO| = 5 V source is treated, the equivalent circuit becomes

I2 = V/REQ = 5/1 + 1||1

= 5/1 + 1/2 = 10/3 A

I1 = I2 x 1/1 + 1

= 10/3 x 1/2

I1 = 5/3 A

The net current in 1 resistance

= I1 + I1 = 8/3 + 5/3 = 13/3 A

Voltage drop across 1 = 13 x 1/3 = 13/3 V

1. (c)

applying KCL at node A,

voltage of node A is eo with reference to ground.

12 – eo/4 = e0/4 + eo/4

3eo/4 = 12/4

eo = 4 V

1. (a)

ZA = ZAB . ZAC/ZAB + ZBC + ZAC

= (3Z) (3Z)/3Z + 3Z + 3Z

3Z2/3 3Z = Z/3

Similary, ZB = Z/3

ZC = Z/3

1. (c)

[I1 I2] = [Y11 Y12 Y21 Y22] = [V1 V2]

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

The parameter Y12 = I1/V2|V1 = 0

V2 = – 20 I1

Y12 – 1/20 = – 0.05 V

74 (d)

1. (d)

Applying KVL in path FDCB,

01 – E – 5 = 10

E = 16 V

1. (b)

In order to calcuate the voltage drop between A and B.

Applying superposition theorem,

Case 1 when 10 V source is taken, circuit becomes

VAB = 1/1 + 2 . 10 = 10/3 V (say V1)

Case 2 When current source is taken circuit becomes,

current in branch AB

I = 2 x 2/2 + 1 = 4/5 A

voltage drop across AB = 4/3 x 1 = 4/3 V

1. (d)

V = L dI/dt

6 = 2 dI/dt or dI/dt = 3

when the switch is closed at t = 0+, there will be only transient current.

IL (0+) = 0

1. (a)

where switch S is in position (a) for a long time in steady state in opened L is short. the circuit looks like

IL = 5 x 4/4 + 1 = 4 A

VC(0) = VC(0+) = 4 x 1 = 4 v

as the switch is moved to position (b) current through

4 resistor = 4/4 = 1 A

current thorugh capacitor = – (1 + 4) = – 5 A

VL = 4 x 1 = 4

as the two voltage are in opposite sign, so,

VL = 0 V

1. (b)

given  H(s) = X/Y = s + 1/s2 + s + 1 ……………(i)

X (t) = cos t = 1<00

from input x (t) = cos t, it is clear that 0 = 1, by putting s = j in eq. (1) with 0 = 1 we get

X/Y = j + /j202 + j + 1 = j + 1/-1 + j + 1

j + 1/j = 2 <450/<900 = j2 <-450

1<00/Y = 2 <- 450

Y = 1/2 <450

1. (b)

for the given circuit,

Z(s) = 5 ||1/1 5s = 5.5/s

Z(s) = 25/5s + 5 = 5/s + 1

I(s) = V2(S)/Z(S) = 10<00/5 + s + 1

I(s) = 2 <00 (j + 1)

= 2 <00. 2 <450 = 2 2<450

1. (d)

given that

V2 = 2 cos 2t = 2 <00

IL(s) = V2/IS = 4 V2/S

IC(s) = 1/3s/2 = 3s. v2/2

total current flowing through 1 resistance

= I2 (s) + IC (s)

= V2 [4/S + 3s/2

= V2 [8 + 3s/2/2s]

voltage drop across resistance

= 1.V2 = cos 2t = 2 <00

0 = 2

= v2 [8 – 3 x 4/2j x 2]

= V2 (-4/4J)

= – V2 <900

By using triangle law

VS = 2/2 <450

1. (c)

given that VS = cos 2t

IC = VS.gm

IC = cos 0t . s. 1

IC = cos 2t. j2

IC = 2 VS <900 ………….(1)

IL = VS/2 + Ls = VS/2 + s = VS/2 + J = VS/2 + J2

= VS/2 (1 + J) = VS/2 2 <450 = VS /2 2 = <- 450 = …………..(2)

From eq. (1) and (2) it is clear that the phase difference between the IC and IL = [90 – (-450)] = 1350

1. (c)
2. (a)

we know that

PAV = 1/2 VM IM cos 0

200 = 1/2 100 IM 0.8

IM = 5

I = IM cos (t + 0)

I = 5 cos (t + 0)

0 = – cos-1 (0.8)

1. (c)

by applying KCL,

3 + 4J + I = 2 + 5J + 2

I = 1 + J = 2<450

I (T) = 2 cos (2 x 50 + 450)

I (t) = 2 cos (100 + 450) A

1. (d)

by applying KCL at super node in following figure,

I = 4 + 1 + 2J + 3 + 4J

I = 8 + 6J

I = 10<53.10

I (t) = 10 cos (100t + 53.10)

1. (a)

applying KVL in loop ABGDEHFA is following figure,

– 3j + (2 + j) – VS + (1 + 2J) – (1 – J)- 1 + (1 – J) = 0

2 – VS = 0

VX = 2V

1. (d)

Z = 3 + J(25 MH) (300)

= 3 + J7.5 = 8.08 <68.20

I = 20 <0/8.08 <68.20 = 2.48 < – 68.20 A

I (t) = 2.48 cos (300t – 68.20) A

1. (d)

Y = 1/2 + J(1 mF) 103 = 0.5 + j = 1.12<63.430 V

VC  = (1 <0)/1.12 <63.430 = 0.89 <- 63. 430 V

1. (a)

Z = 5 + -J/(0.1) (2)

= 5 – J5 = 5 5 <- 450

VC = (1<0) (5 <-900)/5/2 <-450

= 1/2 <-450 V

VC (t) = 1/2 cos (2t – 450) V

1. (a)

Z(s) = 1/s + (1) (1 + 2s)/1 + 1 + 2s = s2 + 1.5s + 1/s(1 + s)

1. (c)

1/Z(s) = s /s + 1 + 1 + 1/s

Z(s) = s(s + 1)/2s2 + s + 1

1. (b)

1/Z(s) = 1/2 + s/s2 + 1 = s2 + 2s + 1/2 (s2 + 1)

Z(s) = 2(s2 + 1)/(s + 1)2

1. (d)

Z(s) = s||(1 + 2(1 + 2/s/2 + 1 + 2/s)

= s(5s + 6)/3s2 + 7s + 6

1. (a)

by applying KVL in the circuit

(1 + 2j) – 2I – (1 – j) + (10 – 3j) – 2I = 0

10 – 4i = 0

I = 2.5 A

1. (d)

0 = 2 x 10 x 103 = 2 x 104

Y = j(1uf) (2 x 104) + -j/(160uf) (2 x 104) = 1/36

= 0.0278 – j0.0366

Z = 1/Y = 13.16 + J17.33

1. (d)

Z = 9 + J(3) (5) + -J/(50UF)(5)

= 9 + J11

Z = 14.21 <50.710

VC = 8 <0 (4 < – 900)/14.21 <50.710

= 2.25 <140.710 V

VC (t) = 2.25 cos (5t – 140.710) v

1. (b)

the circuit is as shown below.

VA = 1/1 + 1 1/-J2 + 1/4 + J8

= 10 <0/1.05 +J0.4 V

I = VA/4 + J8 = 10<0/1 + J10 = 1 <- 84.23 A

I(t) = cos (2t – 84.230) A

1. (a)

the circuit is as shown below.

1. (c)

the circuit is as shown below.