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

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 ?

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

(a) sin (10³t – 45⁰)

(b) sin (10³t + 45⁰)

(d) sin (10³t + 53⁰)

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

(a) 3

(b) 4

(d) 7

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

(d) 1 resistance in parallel with 1 F capacitor

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

The value of the parameter are V_TH R_TH I_N R_N

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

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

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

In the above circuit, the value of V₁ is

(a) 6 V

(b) 7 V

(d) 10 V

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

(a) 2

(b) 6

(d) 4/3

In the following circuit, the value of I₁ is

(a) 3 A

(b) 0.75 mA

(d) 1.75 mA

For the following circuit, the value of R_TH is

(a) 3

(b) 12

(d) infinite

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

(d) 200

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

(d) infinite

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

(d) d²l/dt² + 2 dl/dt + 2l(t) = sin t

The dependent current source shown in figure

(a) delivers 80 w

(b) absorbs 80 w

(d) absorbs 40 w

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

(a) 25 v

(b) 50 v

(d) zero

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]

(d) none of these

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

List 1 List II

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

Codes

A B C D

(a) 3 1 2 4

(b) 2 3 1 4

(d) 1 2 3 4

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

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

The natural frequency s_n of the given circuit is

(a) 1

(b) -1

(d) 2

Steady state is reached for the circuit. V_C as marked in the circuit is

(a) zero

(b) 4

(d) 10/3

The natural frequency of the circuit is given by

(a) – 6

(b) -8

(d) -10

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

(a) 1 V

(b) -1 V

(d) -13/3 V

The voltage e₀ in figure is

(a) 2 V

(b) 4/3 V

(d) 8 V

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

(d) Z/3

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

(a) -0.2 mho

(b) 0.1 mho

(d) 0.05 mho

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

(a) e^at – e^bt

(b) e^at – e^bt

(d) ae^at + be^bt

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

(a) -16 V

(b) 4 V

(d) 16 V

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

(d) 26/3 V

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

(d) 0, 3

S is in position (a) for a long time. S moved to position (b) at t = 0^– , at t = 0⁺ , the values of I_C and V_L are

(a) – 5, 0

(b) 5, 0

(d) – 2, 3

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

(d) 2/5 < tan^-1 – <tan^-2 2

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

(a) 2/5 < -45⁰

(b) 2/2 <45⁰

(d) 2<90⁰

Given, V₂ = 2 cos 2t = 2 <0⁰, the value of V_S is given by

(a) 3 <0⁰

(b) 29/ <tan^-1 2.5

(d) 2/2 < 45⁰

In the circuit V_S = cos 2t. the value of C is chosen so that I from V_S is phase with V_S. then I_C leads I_L by an angle given by

(a) 45⁰

(b) 90⁰

(d) 0⁰

P_complex = 200 + j150 from a source V_S = 200 cos ot the impedance across V_S is given by

(a) 16 + j12

(b) 128 + j96

(d) 32 + j24

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

(a) 5 A and – cos^-1 0.8

(b) 5 A and – cos^-1 0.8

(d) 5 A and 0⁰

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

(d) 1.414 cos (100t + 45⁰) A

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

(a) 10 cos (100t + 53.1⁰)

(b) 14.14 cos (100t + 53.1⁰)

(d) 10 cos (100t – 53.1⁰)

In the given circuit below, voltage V_A is

(a ) 2 V

(b) (2 + 6j) V

(d) -2 V

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

(a) 20 cos (300t + 68.2⁰) A

(b) 20 cos (300t – 68.2⁰) A

(d) 2.48 cos (300t – 68.2⁰) A

In the given circuit below, voltage V_C (t) is

(a) 0.89 cos (10³ t – 63.43⁰) V

(b) 0.89 cos (10³ t + 63.43⁰) V

(d) 0.45 cos (10³ t – 26.57⁰) V

In the given circuit below, voltage V_C (t) is

(a) 1/2 cos (2t – 45⁰) V

(b) 1/2 cos (2t + 45⁰) V

(d) 1/2 sin (2t + 45⁰) V

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)

(d) 2s² + 3s + 1/2s(s + 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)

(d) s(s + 1)/s² + s + 1

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

(a) s² + 1/s² + 2s + 1

(b) 2 (s² + 1)/(s + 1)²

(d) s² + 1/3s + 2

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

(a) 3s² + 8s + 7/s (5s + 6)

(b) s(5s + 6)/3s² + 8s + 7

(d) s(5s + 6)/3s² + 7s + 6

In the following circit, current I is

(a) 2.5 A

(b) (3 +2j) A

(d) (2.5 – 1.75j) A

In the following circuit impedance Z is

(a) 92. 36 – j68.34

(b) 38.24 – j 49.88

(d) 13.16 + j17.33

In the given circuit below, voltage V_C (t) is

(a) 2.25 cos (5t + 150⁰)

(b) 2.25 cos (5t – 150⁰) V

(d) 2.25 cos (5t – 140.71⁰) V

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

(a) 2 sin (2t + 5.77⁰) A

(b) cos (2t – 84.23⁰) A

(d) cos (2t + 84.23⁰) A

The s-domain equivalent of the following circuit is

(d) none of these

The s-domain equivalent of the following circuit is

(a) both (a) and (b)

(d) none of these

Answers with Solutions

V_O (s) = 1/j_oC/R + 1/J_OC V_I(s) = 1/1 + j_oRC V_I(s)

given, V_I(t) = 2 sin 10³t, RC = 1 ms

hence, 0 = 10³ rad/s

in phasor form,

V_I= 2<0⁰

Putting values

V_O(s) 1/1 + j x 10³ x 10^-3 v_i (s)

= 1/1 + j V_I (S)

In phasor form,

V_O = 1/2 <-45⁰ x 2 <0⁰

1< – 45⁰

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

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

Z_N = Source impedance

R + J_OL = 1 + J x 1 x 1 = 1 + j

for maximum power transfer to load,

Z_L = Z_S = 1 – J = R + 1/J_OC

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

V_OC = 2 x 2 + 4 = 8 V = V_TH

R_TH = 2 + 3 + 5 = R_N, I_S = V_TH/R_TH = 8/5 A

4/1 + 1 + 12/1 + 2

V₁ = 1 = 6 V

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

After replacing all the sources by their internal impedance.

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

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

After killing the source, R_TH = 6

Quality factor (Q) of series RLC circuit

= 1/R L/C

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

Q_NEW = 50

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

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

Resistance of current source is in parallel with it.

let voltage of node A is V_A.

applying kCL at node A

V₁ – V_A/5 + V₁/5 = V_A/5

V₁ = V_A

V_A = 20 V

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

power delivered = V_AI = V_A. V₁/5

V_A² /5 = 20 x 20/5 = 80 W

incidence matrix is given by

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

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

S_N = L/R_EQ

R_EQ = 4||4 = 2

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

re-drawn circuit under steady state is given below.

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

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

here, natural frequency is given as

S_N = – R_EQ.C

R_EQ = 6, C = 1

S_N = – 6 x 1 = – 6

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

Applying voltage divide method, we get

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

I₁ = 16/3 x 1/2 = 8/3 A

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

I₂ = V/R_EQ = 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

applying KCL at node A,

voltage of node A is e_o with reference to ground.

12 – e_o/4 = e₀/4 + e_o/4

3e_o/4 = 12/4

e_o = 4 V

Z_A = Z_AB . Z_AC/Z_AB + Z_BC + Z_AC

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

3Z²/3 3Z = Z/3

Similary, Z_B = Z/3

Z_C = Z/3

Admittance matrix equation

[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)

Applying KVL in path FDCB,

01 – E – 5 = 10

E = 16 V

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

Applying superposition theorem,

Case 1 when 10 V source is taken, circuit becomes

V_AB = 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

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.

I_L (0⁺) = 0

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

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

V_C(0^–) = V_C(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

V_L = 4 x 1 = 4

as the two voltage are in opposite sign, so,

V_L = 0 V

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

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

given that

V₂ = 2 cos 2t = 2 <0⁰

I_L(s) = V₂/I_S = 4 V₂/S

I_C(s) = 1/3s_/2 = 3s. v₂/2

total current flowing through 1 resistance

= I₂ (s) + I_C (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

V_S = 2/2 <45⁰

given that V_S = cos 2t

I_C = V_S.g_m

I_C = cos 0t . s. 1

I_C = cos 2t. j2

I_C = 2 V_S <90⁰ ………….(1)

I_L = V_S/2 + Ls = V_S/2 + s = V_S/2 + J = V_S/2 + J2

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

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

we know that

P_AV = 1/2 V_M I_M cos 0

200 = 1/2 100 I_M 0.8

I_M = 5

I = I_M cos (t + 0)

I = 5 cos (t + 0)

0 = – cos^-1 (0.8)

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

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)

applying KVL in loop ABGDEHFA is following figure,

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

2 – V_S = 0

V_X = 2V

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

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

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

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

= 5 – J5 = 5 5 <- 45⁰

V_C = (1<0) (5 <-90⁰)/5/2 <-45⁰

= 1/2 <-45⁰ V

V_C (t) = 1/2 cos (2t – 45⁰) V

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

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

Z(s) = s(s + 1)/2s² + s + 1

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

Z(s) = 2(s² + 1)/(s + 1)²

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

= s(5s + 6)/3s² + 7s + 6

by applying KVL in the circuit

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

10 – 4i = 0

I = 2.5 A

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

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

= 9 + J11

Z = 14.21 <50.71⁰

V_C = 8 <0 (4 < – 90⁰)/14.21 <50.71⁰

= 2.25 <140.71⁰ V

V_C (t) = 2.25 cos (5t – 140.71⁰) v

the circuit is as shown below.

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

= 10 <0/1.05 +J0.4 V

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

I(t) = cos (2t – 84.23⁰) A

the circuit is as shown below.