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

(2 Marks Questions)

76. 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

77. The transfer function H(s) = V_O(s)/V_I(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

78. 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/s² + 10⁶ s + 10⁶

(b) 10⁶/s² + 10³ s + 10⁶

(c) 10³/s² + 10³ s + 10⁶

(d)10⁶/s² + 10⁶ s + 10⁶

79. 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

80. 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.

81. At t = 0⁺, the current I₁ is

(a) -V/2R

(b) -V/R

(c) -V/4R

(d) zero

82. 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]

83. 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)

84. 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

85. 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

86. 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

87. 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 < 32.44⁰

(b) 80 < 32.44⁰

(c) 80 < -32.44⁰

(d) 90 < -32.44⁰

88. The voltage e₀ in figure is

(a) 48 V

(b) 24 V

(c) 36 V

(d) 28 V

89. 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

90. When the angular frequency 0 in figure is varied from 0 to 00, the locus of the current phasor I₂ is given by
91. 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

92. 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

93. For the circuit in figure the voltage V_O is

(a) 2 V

(b) 1 V

(c) -1 V

(d) none of these

94. 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

95. 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

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

the number of possible trees are

(a) 11

(b) 14

(c) 16

(d) 8

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

The oriented graph of the network is

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

The oriented graph of the network is

99. 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

100. 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

101. 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

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

The fundamental loop matrix is

103. 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

76. (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

77. (b)

H(s) = 10⁶/s² + 20s + 10⁶

the characteristic equation is

s² + 20s + 10⁶ = 0

comparing with general expression,

s² + 2 ns + o²_n = 0

0_n = 10³

2 = 20

20/2 x 10³ = 0.01

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

78. (d)

H(s) = V_O(s)/V_I(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

79. (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.

80. (a)
81. (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

I₁ = I₂

I₁ = I₂ = – V/2R

82. (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]

83. (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 + j_o2L

= 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)

84. (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

85. (a)

V₁ = Z₁₁I₁ + Z₁₂I₂

V₂ = Z₂₁I₁ + Z₂₂I₂

Z₁₁ = V₁/I₂|_{I1 = 0}

R_EQ 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

86. (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,

0.5I₁ = V_OC/20 + I₁

V_OC = – 10I₁

I₁ = 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.

0.5I₁ = I_SC + I₁

I_SC = 0.5 I₁

I₁ = – 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

87. (d) 3V_PI_P cos 0 = 1500

3(V_L/3) (V_L/3 Z_L) cos 0 = 1500

Z_L = V²_L 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,

88. (d)

Applying KCL at node A.

80 – e_o/12 = e_o – 16/6 + e_o/12

e_o = 28 v

89. (a)

Z_S = R + J_OL [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

90. (a)
91. (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

92. (d)

from reciprocity theorem

10 V – 2 A

20 V – 4 A

93. (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

94. (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)

95. (a)

V_I = 3 + 4 sin 100t

transfer function

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

|H(J_O)| = 1/ 1 + O²R²C²

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

<H (J_O) = – <tan^-1 RC

= – tan^-1(100 x 10³ x 10^-6) = – 45⁰

V_O(t) = 3 + 4/2 sin (100t – 45⁰)

96. (a)

the incideance matrix is complete because sum of every column is 0. number of possible trees is det {A_RA_R^T] where A_r is reduced incidence matrix.

97. (d)

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

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

98. (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)

99. (a)
100. (b)
101. (d)
102. (a)

the figure is shown below

103. (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.