**(2 Marks Questions)**

- In the given circuit below, values of V
_{C}(0^{+}) and dI_{L}(0^{+})/dt are

(a) zero, zero

(b) 15 V, -10 A/s

(c) zero, – 10 A/s

(d) 15 V, zero

- The transfer function H(s) = V
_{O}(s)/V_{I}(s) of an RLC circuit is given by H(s) = 10^{6}/s^{2}+ 20s + 10^{6}the quality factor (Q-factor) of this circuit is

(a) 25

(b) 50

(c) 100

(d) 5000

- For the circuit shown in figure, the initial conditions are zero. its transfer function H(s) = V
_{C}(s)/V_{I}(s) is

(a) 1/s^{2} + 10^{6} s + 10^{6}

(b) 10^{6}/s^{2} + 10^{3} s + 10^{6}

(c) 10^{3}/s^{2} + 10^{3} s + 10^{6}

(d)10^{6}/s^{2} + 10^{6} s + 10^{6}

- Conider the following s
_{1}and s_{2}: s_{1}: At the resonant frequency the impedance of a series RLC circuit is zero. S_{2}: in a parallel GLC circuit, increasing the conductance G results in increase in its Q-factor.

Which one of the following is correct?

(a) s_{1} is false and s_{2} true

(b) both s_{1 }and s_{2} are true

(c) s_{1} is true and s_{2} is false

(d) both s_{1} and s_{2} are false

- 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.

- At t = 0
^{+}, the current I_{1}is

(a) -V/2R

(b) -V/R

(c) -V/4R

(d) zero

- I
_{1}(s) and I_{2}(s) are laplace transforms of I_{1}(t) and I_{2}(t) respectively. the equations for the loop currents I_{1}(s) and I_{2}(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][I_{1}(s) I_{2}(s)] = [V/s 0]

(b) [R + Ls + 1/Cs -Ls/-Ls R + 1/Cs] [I_{1}(s) I_{2}(s)] = [-V/s 0]

(c) [R + Ls + 1/Cs -Ls /-Ls R + lS + 1/Cs] [I_{1}(s) I_{2}(s)] = [V/s 0]

(d) [R + Ls + 1/Cs -Ls /-Ls R + Ls + 1/Cs] [I_{1}(s) I_{2}(s) ] = [V/-S 0]

- An input voltage V(t) = 10 2cos (t + 10
^{0}) + 10 5 cos (2t + 10^{0}) 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 + 55^{0}) + 10 cos (2t + 10^{0} + tan^{-1} 2)

(b) 10 cos (t + 55^{0}) + 10 3/2 cos (2t + 55^{0})

(c) 10 cos (t – 35^{0}) + 10 cos (2t + 10^{0} – tan^{-1} 2)

(d) 10 cos (t – 35^{0}) + 10 3/2 cos (2t – 35^{0})

- 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)/s^{2} + 2s + 3

(b) 2(s + 3)/s^{2} + 2s + 2

(c) 3(s – 3)/s^{2} – 2s – 2

(d) 2(s -3)/s^{2} – 2s – 3

- The impedance parameters Z
_{11}and Z_{12}of the two-port network in figure are

(a) Z_{11} and 2.75 and Z_{12} = 0.25

(b) Z_{11} = 3 and Z_{12} = 0.5

(c) Z_{11} = 3 and Z_{12} = 0.25

(d) Z_{11} = 2.25 and Z_{12} = 0.5

- In the network of figure, the maximum power is delivered to R
_{L}if its value is

(a) 16

(b) 40/3

(c) 60

(d) 20

- If the 3-phase balanced source in figure delivers 1500 W at a leading power factor of 0.844, then the value of Z
_{L}(in ohm) is approximately

(a) 90 < 32.44^{0}

(b) 80 < 32.44^{0}

(c) 80 < -32.44^{0}

(d) 90 < -32.44^{0}

- The voltage e
_{0}in figure is

(a) 48 V

(b) 24 V

(c) 36 V

(d) 28 V

- 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

- When the angular frequency 0 in figure is varied from 0 to 00, the locus of the current phasor I
_{2}is given by - The Z-parameters Z
_{11}and Z_{21}for the two-port network in figure are

(a) Z_{11 }= – 6/11, Z_{21} = 0 16/11

(b) Z_{11} = 6/11, Z_{21} = 4/11

(c) Z_{11} = 6/11, Z_{21} = – 16/11

(d) Z_{11} = 4/11, Z_{21} = 4/11

- 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

- For the circuit in figure the voltage V
_{O}is

(a) 2 V

(b) 1 V

(c) -1 V

(d) none of these

- A linear time invariant system has an impulse response e
^{2t}, t > 0. if the initial conditions are zero and the input is e^{3t}, the output for t > 0 is

(a) e^{3t} – e^{2t}

(b) e^{5t}

(c) e^{3t} + e^{2t}

(d) none of these

- 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

- The incidence matrix of a graph is as given below

the number of possible trees are

(a) 11

(b) 14

(c) 16

(d) 8

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

The oriented graph of the network is

- The fundamental cut-set matrix of a graph is

The oriented graph of the network is

- 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

- 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

- 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

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

The fundamental loop matrix is

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

Where, I_{J} represent branch current and I_{K} loop current. the number of independent node equations are

(a) 4

(b) 5

(c) 6

(d) 7

**Unit Exercise – 2**

- (b)

for t < 0, equivalent circuit is

u(t) = 0,t < 0

u(-t) = 1, t < 0

I_{L} (0) = 40/5 + 3 = 5 A

V_{C}(0) = 5 x 3 = 15 V

at t = 0, I_{L} and V_{C} cannot change instantaneously, so the circuit is

u(-t) = 0, t > 0

u(t) = 1 t > 0

V_{C}(0^{+}) = V_{C}(0^{–}) = 15

I_{L} (0^{+}) = I_{L} (0^{–}) = 5 A

Voltage across 5 resistor at t = 0^{+} is

V_{R}(0^{+}) = – 5 x I_{L} (0^{+})

= – 5 x 5 = – 25 v

voltage across inductor at t = 0^{+}

V_{L}(0^{+}) = V_{R} (0^{+}) – V_{C}(0^{+})

= – 25 – 15 = – 40 V

V_{L}(0^{+}) = – 40 = L dI_{L}(0^{+})/dt

dI_{L}(0^{+})/dt = -40/4 = -10 a/s

- (b)

H(s) = 10^{6}/s^{2} + 20s + 10^{6}

the characteristic equation is

s^{2} + 20s + 10^{6} = 0

comparing with general expression,

s^{2} + 2 ns + o^{2}_{n} = 0

0_{n} = 10^{3}

2 = 20

20/2 x 10^{3} = 0.01

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

- (d)

H(s) = V_{O}(s)/V_{I}(s) = 1/sC/R + sL + 1/sC = 1/S^{2}LC + sRC + 1

= 1/S^{2}(10 x 10^{-3} x 100 x 10^{-6} + (s(10 + 10^{3 }x 100 x 10^{-6}) + 1

= 1/s^{2} x 10^{-6} + s + 1

= 10^{6}/s^{2} + 10^{6}s + 10^{6}

- (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.

- (a)
- (a)

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

V_{C}(0) = V = V_{C}(0^{+})

I_{L}(0^{+}) = 0 = I_{L}(0^{+})

At t = 0,

at t = 0^{+},

as current through inductor is zero

I_{1} = I_{2}

I_{1} = I_{2} = – V/2R

- (c)

when switch is in position (2)

appliying KVL in loop 1,

RI_{1}(t) + 1/C [I_{1}(t) dt L d/dt [I_{1} (t) – I_{2}(t)] = 0

take laplace transform

RI_{1}(s) + V/s + 1/sC I_{1}(s) + sL (I_{1}(s) – I_{2}(s) = 0

I_{1}(s) [R + 1/sC + sL] – I_{2}(s) sL = – V/s …………(i)

applying KVL in loop 2,

sL[I_{1}(s) – I_{2}(s)] + RI_{2}(s) + 1/sC I_{2}(s) = 0

-sL I_{1}(s) + [R + sL + 1/sC] I_{2}(s) = 0…………….(ii)

hence, we can write, [R + sL + 1/sC -sL/-sL R + sL + 1/sC] [I_{1}(s)/I_{2}(s)] = [-V/S 0]

- (c)

V(t) = 10 2 cos (t + 10^{0}) + 10 5 cos x + 10^{0 }) V

I(t) = 10 2 cos (t + 10^{0})/R + jo_{1}L + 10 5 cos (2t + 10^{0})/R + j_{o2}L

= 10 2 cos (t + 10^{0})/1 + j + 10 5 cos (2t + 10^{0})/1 + 2j

I(t) = 10 2/2 cos (t + 10^{0} – tan^{-1} + 10 5/5 cos (2t + 10^{0} + tan^{-1}2)

= 10 cos (t – 35^{0}) + 10 cos (2t + 10 – tan^{-1}2)

- (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)/s^{2} + 2s + 2

- (a)

V_{1} = Z_{11}I_{1} + Z_{12}I_{2}

V_{2} = Z_{21}I_{1} + Z_{22}I_{2}

Z_{11} = V_{1}/I_{2}|_{I1 = 0}

R_{EQ} leaving 3 resistor

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

Z_{12} = V_{1}/I_{2}|_{I1 = 0}

From current division rule,

I = 1/3 + 1 I_{2} = 1/4 I_{2}

I = V_{1}/1

V_{1} = 1/4 I_{2}

V_{1}/I_{2} |_{I1 = 0} = 0.25

- (d)

the maximum power delivered to eR_{R}, if

R_{L} = R_{TH}

R_{TH} = V_{OC}/I_{SC}

To calculate V_{OC}

Applying KCL at node A,

0.5I_{1} = V_{OC}/20 + I_{1}

V_{OC} = – 10I_{1}

I_{1} = V_{OC} – 50/40

V_{OC} = – 1.0 (V_{OC} – 50/40)

V_{OC} + V_{OC}/4 = 50/4

V_{OC} = 10 V

The 20 resistor is removed.

0.5I_{1} = I_{SC} + I_{1}

I_{SC} = 0.5 I_{1}

I_{1 }= – 50/40 (Applying KVL in side loop)

I_{SC} = – 0.5 x 50/40 = 0.625 A

R_{L} = R_{TH} = V_{OC}/I_{SC} = 10/0.625 = 16

- (d) 3V
_{P}I_{P}cos 0 = 1500

3(V_{L}/3) (V_{L}/3 Z_{L}) cos 0 = 1500

Z_{L} = V^{2}_{L} cos 0/1500 = 400^{2} 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,

- (d)

Applying KCL at node A.

80 – e_{o}/12 = e_{o} – 16/6 + e_{o}/12

e_{o} = 28 v

- (a)

Z_{S} = R + J_{O}L [Given V = E_{M} cos 10t, = 10]

= 10 x j x 10 x 1

Z_{S} = 10 + J10

|Z_{S}| = 10 2

To maximise power

R_{L} = |Z_{S}| = 10 2 = 14.14

- (a)
- (c)

from Z-parameters

E_{1} = Z_{11}I_{1} + Z_{12}I_{2}

E_{2} = Z_{21}I_{1} + Z_{22}I_{2}

Z_{11} = E_{1}/I_{1}|_{I2 = 0}

Applying KVL in loop 1,

E_{1} – I_{1} x 2 – 4I_{1} + 10E_{1} = 0

Z_{11} = E_{1}/I_{1} = 6/11

Z_{21} = E_{2}/I_{1}|_{I2 = 0}

E_{2} = 4I_{1} – 10E_{1} …………(1)

I_{1} = E_{1} – E_{2}/2

2I_{1} + E_{2} = E_{1} …………….(2)

Substituting the value of E_{1},

E_{2} = 4I_{1} – 10 (I_{1} + E_{2})

E_{2} = 4I_{1} – 20I_{1} – 10E_{2}

Z_{21} = E_{2}/I_{1} = – 16/11

- (d)

from reciprocity theorem

10 V – 2 A

20 V – 4 A

- (d)

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

applying KCL at node A,

4 – V_{A}/2 = V_{A}/2 + V_{A} + 2/2

2 – 1 = 3V_{A}/2

V_{A} = 2/3

V_{O} = – V_{A} = – 2/3

- (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) = (e^{3t} – e^{2t})

- (a)

V_{I} = 3 + 4 sin 100t

transfer function

H(s) = 1/sc R + 1/sC = 1/1 + sRC = 1/1 + J_{O}RC

|H(J_{O})| = 1/ 1 + O^{2}R^{2}C^{2}

= 1/1 + 100^{2} x 10^{6} x 10 x 10 x 10^{-12} = 1/2

<H (J_{O}) = – <tan^{-1} RC

= – tan^{-1}(100 x 10^{3} x 10^{-6}) = – 45^{0}

V_{O}(t) = 3 + 4/2 sin (100t – 45^{0})

- (a)

the incideance matrix is complete because sum of every column is 0. number of possible trees is det {A_{R}A_{R}^{T}] where A_{r} is reduced incidence matrix.

- (d)

C_{1}(1,5) : in opposite direction (c, d)

C_{2} (2, 5, 7) in all same direction (d)

- (c)

the figure is shown below,

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

- (a)
- (b)
- (d)
- (a)

the figure is shown below

- (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.