In the given circuit below, values of VC(0+) and dIL(0+)/dt are solution and answer

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(2 Marks Questions)

  1. In the given circuit below, values of VC(0+) and dIL(0+)/dt are

(a) zero, zero

(b) 15 V, -10 A/s

(c) zero, – 10 A/s

(d) 15 V, zero

  1. The transfer function H(s) = VO(s)/VI(s) of an RLC circuit is given by H(s) = 106/s2 + 20s + 106 the quality factor (Q-factor) of this circuit is

(a) 25

(b) 50

(c) 100

(d) 5000

  1. For the circuit shown in figure, the initial conditions are zero. its transfer function H(s) = VC(s)/VI(s) is

(a) 1/s2 + 106 s + 106

(b) 106/s2 + 103 s + 106

(c) 103/s2 + 103 s + 106

(d)106/s2 + 106 s + 106

  1. Conider the following s1 and s2 : s1 : At the resonant frequency the impedance of a series RLC circuit is zero. S2 : in a parallel GLC circuit, increasing the conductance G results in increase in its Q-factor.

Which one of the following is correct?

(a) s1 is false and s2 true

(b) both s1 and s2 are true

(c) s1 is true and s2 is false

(d) both s1 and s2 are false

  1. The current flowing through the resistance R in the circuit in figure has the form P cos 4t, where P is

(a) (0.18 + j0.72)

(b) (0.46 + j1.90)

(c) -(0.18 + j1.90)

(d) -(0.192 + j0.144)

Common data for Questions 81 and 82

For both the questions, assume that the switch s is in position 1 for a long time thrown to position  2 at t = 0.

  1. At t = 0+, the current I1 is

(a) -V/2R

(b) -V/R

(c) -V/4R

(d) zero

  1. I1(s) and I2(s) are laplace transforms of I1(t) and I2 (t) respectively. the equations for the loop currents I1(s) and I2 (s) for the circuit shown in figure after the switch is brought from position 1 to position 2 at t = 0, are

(a) [ R + Ls + 1/Cs  -Ls /-Ls  R + 1/Cs][I1(s) I2(s)] = [V/s 0]

(b) [R + Ls + 1/Cs -Ls/-Ls  R + 1/Cs] [I1(s) I2(s)] = [-V/s  0]

(c) [R + Ls + 1/Cs    -Ls /-Ls  R + lS + 1/Cs] [I1(s) I2(s)] = [V/s  0]

(d) [R + Ls + 1/Cs  -Ls /-Ls  R + Ls + 1/Cs] [I1(s) I2(s) ] = [V/-S  0]

  1. An input voltage V(t) = 10 2cos (t + 100) + 10 5 cos (2t + 100) V is applied to a series combination of resistance R = 1 and an inductance L = 1 H. the resulting steady state current I(t) in ampere is

(a) 10 cos (t + 550) + 10 cos (2t + 100 + tan-1 2)

(b) 10 cos (t + 550) + 10 3/2 cos (2t + 550)

(c) 10 cos (t – 350) + 10 cos (2t + 100 – tan-1 2)

(d) 10 cos (t – 350) + 10 3/2 cos (2t – 350)

  1. The driving ponit impedance Z(s) of a network has the pole zero locations as shown in figure. if Z(0) = 3, then Z(s) is

(a) 3(s + 3)/s2 + 2s + 3

(b) 2(s + 3)/s2 + 2s + 2

(c) 3(s – 3)/s2 – 2s – 2

(d) 2(s -3)/s2 – 2s – 3

  1. The impedance parameters Z11 and Z12 of the two-port network in figure are

(a) Z11 and 2.75 and Z12 = 0.25

(b) Z11 = 3 and Z12 = 0.5

(c) Z11 = 3 and Z12 = 0.25

(d) Z11 = 2.25 and Z12 = 0.5

  1. In the network of figure, the maximum power is delivered to RL if its value is

(a) 16

(b) 40/3

(c) 60

(d) 20

  1. If the 3-phase balanced source in figure delivers 1500 W at a leading power factor of 0.844, then the value of ZL (in ohm) is approximately

(a) 90 < 32.440

(b) 80 < 32.440

(c) 80 < -32.440

(d) 90 < -32.440

  1. The voltage e0 in figure is

(a) 48 V

(b) 24 V

(c) 36 V

(d) 28 V

  1. In figure, the value of the load resistor R which maximizes the power delivered to it is

(a) 14.14

(b) zero

(c) 200

(d) 28.28

  1. When the angular frequency 0 in figure is varied from 0 to 00, the locus of the current phasor I2 is given by
  2. The Z-parameters Z11 and Z21 for the two-port network in figure are

(a) Z11 = – 6/11, Z21 = 0 16/11

(b) Z11 = 6/11, Z21 = 4/11

(c) Z11 = 6/11, Z21 = – 16/11

(d) Z11 = 4/11, Z21 = 4/11

  1. Use the data of fig. (a). the current I in the circuit of figure. (b) is

(a) -2 A

(b) 2 A

(c) -4 A

(d) 4 A

  1. For the circuit in figure the voltage VO is

(a) 2 V

(b) 1 V

(c) -1 V

(d) none of these

  1. A linear time invariant system has an impulse response e2t, t > 0. if the initial conditions are zero and the input is e3t, the output for t > 0 is

(a) e3t – e2t

(b) e5t

(c) e3t + e2t

(d) none of these

  1. In figure the steady state output voltage corresponding to the input voltage 3 + 4 sin 100t V is

(a) 3 + 4/2 sin (100t – 4) v

(b) 3 + 4/2 sin (100t – 4) V

(c) 3/2 + 4/2 sin (100t – 4) v

(d) 3 + 4 sin (100t – 4) v

  1. The incidence matrix of a graph is as given below

the number of possible trees are

(a) 11

(b) 14

(c) 16

(d) 8

  1. The f-cut-set matrix of a graph is given as

The oriented graph of the network is

  1. The fundamental cut-set matrix of a graph is

The oriented graph of the network is

  1. A graph is shown below in which twigs are solid line and links are dotted line. for this chosen tree, fundamental set matrix is given below.

The oriented graph will be

  1. A graph is shown below in which twigs are solid line and links are dotted line. for this tree fundamental loop matrix is given as below

The oriented graph will be

  1. Consider the graph shown below in which twigs are solid line and links are dotted line

A fundamental-loop matrix for this tree is given as below

The oriented graph will be

  1. In the graph shown below solid lines are twigs and dotted lines are linkes

The fundamental loop matrix is

  1. Branch-current and loop-current relation are expressed in matrix form as

Where, IJ represent branch current and IK loop current. the number of independent node equations are

(a) 4

(b) 5

(c) 6

(d) 7

Unit Exercise – 2

  1. (b)

for t < 0, equivalent circuit is

u(t) = 0,t < 0

u(-t) = 1, t < 0

IL (0) = 40/5 + 3 = 5 A

VC(0) = 5 x 3 = 15 V

at t = 0, IL and VC cannot change instantaneously, so the circuit is

u(-t) = 0, t > 0

u(t) = 1 t > 0

VC(0+) = VC(0) = 15

IL (0+) = IL (0) = 5 A

Voltage across 5 resistor at t = 0+ is

VR(0+) = – 5 x IL (0+)

= – 5 x 5 = – 25 v

voltage across inductor at t = 0+

VL(0+) = VR (0+) – VC(0+)

= – 25 – 15 = – 40 V

VL(0+) = – 40 = L dIL(0+)/dt

dIL(0+)/dt = -40/4 = -10 a/s

  1. (b)

H(s) = 106/s2 + 20s + 106

the characteristic equation is

s2 + 20s + 106 = 0

comparing with general expression,

s2 + 2 ns + o2n = 0

0n = 103

2 = 20

20/2 x 103 = 0.01

quality factor Q = 1/2 = 1/0.02 = 50

  1. (d)

H(s) = VO(s)/VI(s) = 1/sC/R + sL + 1/sC = 1/S2LC + sRC + 1

= 1/S2(10 x 10-3 x 100 x 10-6 + (s(10 + 103 x 100 x 10-6) + 1

= 1/s2 x 10-6 + s + 1

= 106/s2 + 106s + 106

  1. (d)

impedance of series RLC circuit is resistance.

Quality factor (Q) = R C/L

Conductance G = 1/R

R = 1/G

Q = 1/G C/L

Q 1/G

Increasing conductance decreases Q.

  1. (a)
  2. (a)

As the switch in position 1 for long time, steady 2-bite reached inductor becomes short-circuit and capacitor becomes open-circuit.

VC(0) = V = VC(0+)

IL(0+) = 0 = IL(0+)

At t = 0,

at t = 0+,

as current through inductor is zero

I1 = I2

I1 = I2 = – V/2R

  1. (c)

when switch is in position (2)

appliying KVL in loop 1,

RI1(t) + 1/C [I1(t) dt L d/dt [I1 (t) – I2(t)] = 0

take laplace transform

RI1(s) + V/s + 1/sC I1(s) + sL (I1(s) – I2(s) = 0

I1(s) [R + 1/sC + sL] – I2(s) sL = – V/s …………(i)

applying KVL in loop 2,

sL[I1(s) – I2(s)] + RI2(s) + 1/sC I2(s) = 0

-sL I1(s) + [R + sL + 1/sC] I2(s) = 0…………….(ii)

hence, we can write, [R + sL + 1/sC   -sL/-sL  R + sL + 1/sC] [I1(s)/I2(s)] = [-V/S 0]

  1. (c)

V(t) = 10 2 cos (t + 100) + 10 5 cos x + 100 ) V

I(t) = 10 2 cos (t + 100)/R + jo1L + 10 5 cos (2t + 100)/R + jo2L

= 10 2 cos (t + 100)/1 + j + 10 5 cos (2t + 100)/1 + 2j

I(t) = 10 2/2 cos (t + 100 – tan-1 + 10 5/5  cos (2t + 100 + tan-12)

= 10 cos (t – 350) + 10 cos (2t + 10 – tan-12)

  1. (b)

poles at (-1 + j) and (-1 – j) and zeros at -3

Z(s) = K (s + 3)/s + 1 – j) (s + 1 + j) = K(s + 3)/ (s + 1)2 + 1

Z(0) = 3

K x 3/2 = 3  k = 2

Z(s) = 2(s + 3)/s2 + 2s + 2

  1. (a)

V1 = Z11I1 + Z12I2

V2 = Z21I1 + Z22I2

Z11 = V1/I2|I1 = 0

REQ leaving 3 resistor

3 x 1/3 + 1 + 2 = 11/4 = 2.75

Z12 = V1/I2|I1 = 0

From current division rule,

I = 1/3 + 1 I2 = 1/4 I2

I = V1/1

V1 = 1/4 I2

V1/I2 |I1 = 0 = 0.25

  1. (d)

the maximum power delivered to eRR, if

RL = RTH

RTH = VOC/ISC

To calculate VOC

Applying KCL at node A,

0.5I1 = VOC/20 + I1

VOC = – 10I1

I1 = VOC – 50/40

VOC = – 1.0 (VOC – 50/40)

VOC + VOC/4 = 50/4

VOC = 10 V

The 20 resistor is removed.

0.5I1 = ISC + I1

ISC = 0.5 I1

I1 = – 50/40 (Applying KVL in side loop)

ISC = – 0.5 x 50/40 = 0.625 A

RL = RTH = VOC/ISC = 10/0.625 = 16

  1. (d) 3VPIP cos 0 = 1500

3(VL/3) (VL/3 ZL) cos 0 = 1500

ZL = V2L cos 0/1500 = 4002 x 0.844/1500 = 90

0 = cos-1 (0.844) = 32.44

as power factor is leading load is capacitive so angle will be negative,

  1. (d)

Applying KCL at node A.

80 – eo/12 = eo – 16/6 + eo/12

eo = 28 v

  1. (a)

ZS = R + JOL [Given V = EM cos 10t, = 10]

= 10 x j x 10 x 1

ZS = 10 + J10

|ZS| = 10 2

To maximise power

RL = |ZS| = 10  2 = 14.14

  1. (a)
  2. (c)

from Z-parameters

E1 = Z11I1 + Z12I2

E2 = Z21I1 + Z22I2

Z11 = E1/I1|I2 = 0

Applying KVL in loop 1,

E1 – I1 x 2 – 4I1 + 10E1 = 0

Z11 = E1/I1 = 6/11

Z21 = E2/I1|I2 = 0

E2 = 4I1 – 10E1 …………(1)

I1 = E1 – E2/2

2I1 + E2 = E1 …………….(2)

Substituting the value of E1,

E2 = 4I1 – 10 (I1 + E2)

E2 = 4I1 – 20I1 – 10E2

Z21 = E2/I1 = – 16/11

  1. (d)

from reciprocity theorem

10 V – 2 A

20 V – 4 A

  1. (d)

as the diode is forward biased, it gets short-circuited

applying KCL at node A,

4 – VA/2 = VA/2 + VA + 2/2

2 – 1 = 3VA/2

VA = 2/3

VO = – VA = – 2/3

  1. (a)

output Y(s) = h(s) R(s)

given, H(s) = 1/s – 2

R(s) = 1/s – 3

Y(s) = 1/(s – 2) (s – 3) = – 1/s – 2 + 1/s – 3

taking inverse laplace transform,

y(t) = (e3t – e2t)

  1. (a)

VI = 3 + 4 sin 100t

transfer function

H(s) = 1/sc R + 1/sC = 1/1 + sRC = 1/1 + JORC

|H(JO)| = 1/ 1 + O2R2C2

= 1/1 + 1002 x 106 x 10 x 10 x 10-12 = 1/2

<H (JO) = – <tan-1 RC

= – tan-1(100 x 103 x 10-6) = – 450

VO(t) = 3 + 4/2 sin (100t – 450)

  1. (a)

the incideance matrix is complete because sum of every column is 0. number of possible trees is det {ARART] where Ar is reduced incidence matrix.

  1. (d)

C1(1,5) : in opposite direction (c, d)

C2 (2, 5, 7) in all same direction (d)

  1. (c)

the figure is shown below,

C1 (1,2) This separates the node a and directions of both branches are opposite from node (b, c)

  1. (a)
  2. (b)
  3. (d)
  4. (a)

the figure is shown below

  1. (a)

number of branches is 8 and number of links is 4, thus number of twigs is 8-4 = 4. number of independent node equations is equal to number of teigs.