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 10^{3}t. the output voltage V_{O}(t) is equal to

(a) sin (10^{3}t – 45^{0})

(b) sin (10^{3}t + 45^{0})

(c) sin (10^{3}t – 53^{0})

(d) sin (10^{3}t + 53^{0})

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

(a) 3

(b) 4

(c) 6

(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

(c) 1 resistance in series with 1 F capacitor

(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

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

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

- In the above circuit, the value of V
_{1}is

(a) 6 V

(b) 7 V

(c) 8 V

(d) 10 V

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

(a) 2

(b) 6

(c) 6.16

(d) 4/3

- In the following circuit, the value of I
_{1}is

(a) 3 A

(b) 0.75 mA

(c) 2 mA

(d) 1.75 mA

- For the following circuit, the value of R
_{TH}is

(a) 3

(b) 12

(c) 6

(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

(c) 100

(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

(c) 2

(d) infinite

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

(a) 2d^{2}l/dt^{2} + 2 dl/dt + l(t) = sin t

(b) d^{2}l/dt^{2} + 2 dl/dt + 2l(t) = cos t

(c) 2 d^{2}l/dt^{2} + 2 dl/dt + l(t) = cos t

(d) d^{2}l/dt^{2} + 2 dl/dt + 2l(t) = sin t

- The dependent current source shown in figure

(a) delivers 80 w

(b) absorbs 80 w

(c) delivers 40 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

(c) – 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]

(c) [1 -1 0/ 0 1 1 -1 0 -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

(c) 3 2 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

(c) 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

(c) – 1/2

(d) 2

- Steady state is reached for the circuit. V
_{C}as marked in the circuit is

(a) zero

(b) 4

(c) 5

(d) 10/3

- The natural frequency of the circuit is given by

(a) – 6

(b) -8

(c) 15/2

(d) -10

- Initial voltage on capacitor V
_{0}as marked |V_{0}| = 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

(c) 13/3 V

(d) -13/3 V

- The voltage e
_{0}in figure is

(a) 2 V

(b) 4/3 V

(c) 4 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

(c) 3/3Z

(d) Z/3

- The admittance parameter Y
_{12}in the two-port network in figure is

(a) -0.2 mho

(b) 0.1 mho

(c) -0.05 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}

(c) ae^{at} – be^{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

(c) – 6 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

(c) 10 V

(d) 26/3 V

- S is open, V
_{0}= 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

- 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

(c) 0, 4

(d) – 2, 3

- Given H(s) = X/Y = s + 1/s
^{2}+ s + 1 x(t) = cos t <0^{0}. the phasor Y is given by

(a) 2 < 45^{0}

(b) 1/2 <45^{0}

(c) 1 <0^{0}

(d) 2/5 < tan^{-1} – <tan^{-2} 2

- In the figure V
_{S}= 10 cos t = 10 <0^{0}. the current drawn from V_{S}is given by the phasor

(a) 2/5 < -45^{0}

(b) 2/2 <45^{0}

(c) 2 <0^{0}

(d) 2<90^{0}

- Given, V
_{2}= 2 cos 2t = 2 <0^{0}, the value of V_{S}is given by

(a) 3 <0^{0}

(b) 29/ <tan^{-1} 2.5

(c) 104 < tan^{-1} 5

(d) 2/2 < 45^{0}

- 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) 45^{0}

(b) 90^{0}

(c) 135^{0}

(d) 0^{0}

- 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

(c) 64 + j48

(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

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

(d) 5 A and 0^{0}

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

(a) 14.14 cos (50 t – 45^{0}) A

(b) -11.4 cos (100t + 52.15^{0}) A

(c) 1.414 cos (100t + 45^{0}) A

(d) 1.414 cos (100t + 45^{0}) A

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

(a) 10 cos (100t + 53.1^{0})

(b) 14.14 cos (100t + 53.1^{0})

(c) 10 cos (100 t – 53.1^{0})

(d) 10 cos (100t – 53.1^{0})

- In the given circuit below, voltage V
_{A}is

(a ) 2 V

(b) (2 + 6j) V

(c) 1 V

(d) -2 V

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

(a) 20 cos (300t + 68.2^{0}) A

(b) 20 cos (300t – 68.2^{0}) A

(c) 2.48 cos (300t + 68.2^{0}) A

(d) 2.48 cos (300t – 68.2^{0}) A

- In the given circuit below, voltage V
_{C}(t) is

(a) 0.89 cos (10^{3} t – 63.43^{0}) V

(b) 0.89 cos (10^{3} t + 63.43^{0}) V

(c) 0.45 cos (10^{3} t + 26.57^{0}) V

(d) 0.45 cos (10^{3} t – 26.57^{0}) V

- In the given circuit below, voltage V
_{C}(t) is

(a) 1/2 cos (2t – 45^{0}) V

(b) 1/2 cos (2t + 45^{0}) V

(c) 1/2 sin (2t – 45^{0}) V

(d) 1/2 sin (2t + 45^{0}) V

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

(a) s^{2} + 1.5s + 1/s(s + 1)

(b) s^{2} + 3s + 1/s(s + 1)

(c) 2s^{2} + 3s + 2/s (s + 1)

(d) 2s^{2} + 3s + 1/2s(s + 1)

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

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

(b) 2s^{2} + s + 1/s (s + 1)

(c) s(s + 1) 2s^{2} + s + 1

(d) s(s + 1)/s^{2} + s + 1

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

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

(b) 2 (s^{2} + 1)/(s + 1)^{2}

(c) 2s^{2} + 1/s^{2} + 2s + 2

(d) s^{2} + 1/3s + 2

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

(a) 3s^{2} + 8s + 7/s (5s + 6)

(b) s(5s + 6)/3s^{2} + 8s + 7

(c) 3s^{2} + 7s + 6/s (5s + 6)

(d) s(5s + 6)/3s^{2} + 7s + 6

- 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

- 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

- In the given circuit below, voltage V
_{C}(t) is

(a) 2.25 cos (5t + 150^{0})

(b) 2.25 cos (5t – 150^{0}) V

(c) 2.25 cos (5t + 140.71^{0}) V

(d) 2.25 cos (5t – 140.71^{0}) V

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

(a) 2 sin (2t + 5.77^{0}) A

(b) cos (2t – 84.23^{0}) A

(c) 2sin (2t – 5.77^{0}) A

(d) cos (2t + 84.23^{0}) A

- The s-domain equivalent of the following circuit is

(c) both (a) and (b)

(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 **

- (a)

V_{O} (s) = 1/j_{o}C/R + 1/J_{O}C V_{I}(s) = 1/1 + j_{o}RC V_{I}(s)

given, V_{I}(t) = 2 sin 10^{3}t, RC = 1 ms

hence, 0 = 10^{3} rad/s

in phasor form,

V_{I }= 2<0^{0}

Putting values

V_{O}(s) 1/1 + j x 10^{3} x 10^{-3} v_{i} (s)

= 1/1 + j V_{I} (S)

In phasor form,

V_{O} = 1/2 <-45^{0} x 2 <0^{0}

1< – 45^{0}

hence, V_{A} (t) = sin (t – 45^{0}) = sin (10^{3}t – 45^{0})

- (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

- (c)

Z_{N} = Source impedance

R + J_{O}L = 1 + J x 1 x 1 = 1 + j

for maximum power transfer to load,

Z_{L} = Z_{S} = 1 – J = R + 1/J_{O}C

The load Z_{L} is 1 resistor in series with 1 F capacitor.

- (d)

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

- (d)

4/1 + 1 + 12/1 + 2

V_{1} = 1 = 6 V

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

- (b)

After replacing all the sources by their internal impedance.

R_{TH} = (3||6) + (8||8) = 6

- (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.

I_{1} = 20 – 6 – 8/2 + 4 + 2 = 0.75 mA

- (c)

After killing the source, R_{TH} = 6

- (b)

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

- (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

- (c)

applying KVL in circuit

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

differentiating and re-arranging,

2d^{2}I(t)/dt^{2} = 2dI(t)/dt + I(t) = cos t

- (a)

Resistance of current source is in parallel with it.

let voltage of node A is V_{A}.

applying kCL at node A

V_{1} – V_{A}/5 + V_{1}/5 = V_{A}/5

V_{1} = 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_{A}I = V_{A}. V_{1}/5

V_{A}^{2} /5 = 20 x 20/5 = 80 W

- (c)
- (d)

incidence matrix is given by

- (a)
- (d)

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

- (b)

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

- (b)

re-drawn circuit under steady state is given below.

V_{C} = 4 x 5/4 + 1 = 4 V

- (a)

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

- (c)

**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

I_{1} = 16/3 x 1/2 = 8/3 A

**Case 2** when |V_{O}| = 5 V source is treated, the equivalent circuit becomes

I_{2} = V/R_{EQ} = 5/1 + 1||1

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

I_{1} = I_{2} x 1/1 + 1

= 10/3 x 1/2

I_{1 }= 5/3 A

The net current in 1 resistance

= I_{1} + I_{1} = 8/3 + 5/3 = 13/3 A

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

- (c)

applying KCL at node A,

voltage of node A is e_{o} with reference to ground.

12 – e_{o}/4 = e_{0}/4 + e_{o}/4

3e_{o}/4 = 12/4

e_{o} = 4 V

- (a)

Z_{A} = Z_{AB} . Z_{AC}/Z_{AB} + Z_{BC} + Z_{AC}

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

3Z^{2}/3 3Z = Z/3

Similary, Z_{B} = Z/3

Z_{C} = Z/3

- (c)

Admittance matrix equation

[I_{1} I_{2}] = [Y_{11} Y_{12} Y_{21} Y_{22}] = [V_{1} V_{2}]

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

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

The parameter Y_{12} = I_{1}/V_{2}|_{V1 = 0}

V_{2} = – 20 I_{1}

Y_{12} – 1/20 = – 0.05 V

74 (d)

- (d)

Applying KVL in path FDCB,

01 – E – 5 = 10

E = 16 V

- (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

V_{AB} = 1/1 + 2 . 10 = 10/3 V (say V_{1})

**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

- (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.

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

- (a)

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

- (b)

given H(s) = X/Y = s + 1/s^{2} + s + 1 ……………(i)

X (t) = cos t = 1<0^{0}

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 + /j^{2}0^{2} + j + 1 = j + 1/-1 + j + 1

j + 1/j = 2 <45^{0}/<90^{0} = j_{2} <-45^{0}

1<0^{0}/Y = 2 <- 45^{0}

Y = 1/2 <45^{0}

- (b)

for the given circuit,

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

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

I(s) = V_{2}(S)/Z(S) = 10<0^{0}/5 + s + 1

I(s) = 2 <0^{0} (j + 1)

= 2 <0^{0}. 2 <45^{0} = 2 2<45^{0}

- (d)

given that

V_{2} = 2 cos 2t = 2 <0^{0}

I_{L}(s) = V_{2}/I_{S} = 4 V_{2}/S

I_{C}(s) = 1/3s_{/}2 = 3s. v_{2}/2

total current flowing through 1 resistance

= I_{2} (s) + I_{C} (s)

= V_{2} [4/S + 3s/2

= V_{2} [8 + 3s/2/2s]

voltage drop across resistance

= 1.V_{2} = cos 2t = 2 <0^{0}

0 = 2

= v_{2 }[8 – 3 x 4/2j x 2]

= V_{2} (-4/4J)

= – V_{2} <90^{0}

By using triangle law

V_{S} = 2/2 <45^{0}

- (c)

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} <90^{0} ………….(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 <45^{0} = V_{S} /2 2 = <- 45^{0} = …………..(2)

From eq. (1) and (2) it is clear that the phase difference between the I_{C} and I_{L} = [90 – (-45^{0})] = 135^{0}

- (c)
- (a)

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)

- (c)

by applying KCL,

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

I = 1 + J = 2<45^{0}

I (T) = 2 cos (2 x 50 + 45^{0})

I (t) = 2 cos (100 + 45^{0}) A

- (d)

by applying KCL at super node in following figure,

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

I = 8 + 6J

I = 10<53.1^{0}

I (t) = 10 cos (100t + 53.1^{0})

- (a)

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

- (d)

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

= 3 + J7.5 = 8.08 <68.2^{0}

I = 20 <0/8.08 <68.2^{0} = 2.48 < – 68.2^{0} A

I (t) = 2.48 cos (300t – 68.2^{0}) A

- (d)

Y = 1/2 + J(1 mF) 10^{3} = 0.5 + j = 1.12<63.43^{0} V

V_{C} = (1 <0)/1.12 <63.43^{0} = 0.89 <- 63. 43^{0} V

- (a)

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

= 5 – J5 = 5 5 <- 45^{0}

V_{C} = (1<0) (5 <-90^{0})/5/2 <-45^{0}

= 1/2 <-45^{0} V

V_{C} (t) = 1/2 cos (2t – 45^{0}) V

- (a)

Z(s) = 1/s + (1) (1 + 2s)/1 + 1 + 2s = s^{2} + 1.5s + 1/s(1 + s)

- (c)

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

Z(s) = s(s + 1)/2s^{2} + s + 1

- (b)

1/Z(s) = 1/2 + s/s^{2} + 1 = s^{2} + 2s + 1/2 (s^{2} + 1)

Z(s) = 2(s^{2} + 1)/(s + 1)^{2}

- (d)

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

= s(5s + 6)/3s^{2} + 7s + 6

- (a)

by applying KVL in the circuit

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

10 – 4i = 0

I = 2.5 A

- (d)

0 = 2 x 10 x 10^{3} = 2 x 10^{4}

Y = j(1uf) (2 x 10^{4}) + -j/(160uf) (2 x 10^{4}) = 1/36

= 0.0278 – j0.0366

Z = 1/Y = 13.16 + J17.33

- (d)

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

= 9 + J11

Z = 14.21 <50.71^{0}

V_{C} = 8 <0 (4 < – 90^{0})/14.21 <50.71^{0}

= 2.25 <140.71^{0} V

V_{C} (t) = 2.25 cos (5t – 140.71^{0}) v

- (b)

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 – 84.23^{0}) A

- (a)

the circuit is as shown below.

- (c)

the circuit is as shown below.