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for a parallel rlc circuit the incorrect statement among the following is , for parallel rlc circuit which one of the following statements is not correct ? answers given below after questions.

Unit Exercise – 1

1 Mark Questions

  1. For the two-port network shown below, the short-circuit admittance parameter matrix is

(a) [4  -2/-2 4]

(b) [1  -0.5 /-0.5  1]

(c) [1  0.5 /0.5  1]

(d) [4  2/2 4]

  1. For parallel RLC circuit, which one of the following statements is not correct?

(a) the bandwidth of the circuit decreases increased

(b) the bandwidth of the circuit remains same if L is increased

(c) at resonance, input impedance is a real quantity

(d) at resonance, the magnitude of input impedance attains its minimum value

  1. The nodal method of circuit analysis is based on

(a) KVL and ohm’s law

(b) KCL and ohm’s law

(c) KCL and KVL

(d) KCL, KVL and ohm’s law

  1. The equivalent capacitance for the network shown in the figure is

(a) c/4

(b) 5c/13

(c) 5c/2

(d) 3c

  1. Twelve 1 resistors are used as edge to form a cube. the resistance between two diagonally opposite corners of the cube is

(a) 5/6

(b) 6/5

(c) 1

(d) none of these

  1. Twelve 1 H inductors are used as edge to form a cube. the inductance between two diagonally opposite corners of cube is

(a) 5/6 H

(b) 10/6 H

(c) 2 H

(d) 3/2 H

  1. The equivalent resistance as seen between the terminals (a, b) is

(a) 2

(b) 4

(c) 1

(d) not possible

  1. A fully charged mobile phone with a 12 V battery is good for a 10 min. talk-time. Assume that during the talk-time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. how much energy does the battery deliver during this talk-time?

(a) 220 j

(b) 12 kj

(c) 13.2 kj

(d) 14.4 kj

  1. In th interconnection of ideal sources shown in the figure, it is known that the 60 v source is absorbing power.

Which of the following can be the value of the current source?

(a) 10 A

(b) 13 A

(c) 15 A

(d) 18 A

  1. If the transfer function of the following network is VO(s)/V1(s) = 1/2 + sCR

The value of the load resistance R is

(a) R/4

(b) R/2

(c) R

(d) 2R

  1. Given that F(s) is the one-sided laplace transform of f(t), the laplace transform of is

(a) sf(s) – f(0)

(b) 1/s F(s)

(c) F d

(d) 1/s [F(s) – f(0)]

  1. The network element and V-I characteristics are shown in fig. (a) and fig. (b). the element is

(a) non-linear, active and bilateral

(b) linear, passive and bilateral

(c) non-linear, passive and non-bilateral

(d) non-linear, active and non-bilateral

  1. The current having the waveform shown in figure is flowing in a resistance of 1. the average power is

(a) 1000 W

(b) 1000/2 W

(c) 1000/3 W

(d) 1000/4 W

  1. The current l(t) in the conductor A is as shown in figure. the charge that is transported by this current through surface SO of the conductor during the time interval 0 < t < 6 s is

(a) 7.5 C

(b) 8.5 C

(c) 5.5 C

(d) zero

  1. Current through and voltage across a device are given in figure. energy (approximate) absorbed by the device in the interval 0 < t < 2 s is

(a) 8.5 j

(b) 10 j

(c) 11.4 j

(d) 5 j

  1. Given the current through R is l = 1 A. the value of R is

(a) 0

(b) 2

(c) 4

(d) 8

  1. The equivalent inductance of the network is

(a) L1 + L2 – 2 M

(b) L1 + L2 + 2 M

(c) L1 + L2 – M

(d) L1 + L2

  1. Which one of the following networks is the Y-equivalent of circuit shown in figure?
  2. Consider the circuit graph shown in figure. each branch of circuit graph represents a circuit element. the value of voltage V1 is

(a) -30 V

(b) 25 V

(c) -20 V

(d) 15 V

  1. In the following circuit, the switch S is closed at t = 0. the rate of change of current dl/dt (0+) is given

(a) 0

(b) RSIS/L

(c) (R + RS)IS/L

(d) oo

  1. In the following graph, the number of trees (p) and the number of cut-sets (Q) are

(a) p = 2, q = 2

(b) p = 2, q = 6

(c) p = 4, q = 6

(d) p = 4, q = 10

  1. The pole- zero plot given below corresponds to a

(a) low-pass filter

(b) high-pass filter

(c) band-pass filter

(d) notch filter

  1. Consider the following graphs :

Non-planar graph(s) is/are

(a) 1 and 3

(b) 4 only

(c) 3 only

(d) 3 and 4

  1. Consider the following circuits :

The planar circuits are

(a) 1 and 2

(b) 2 and 3

(c) 3 and 4

(d) 4 and 1

  1. A tree of the graph shown below is

(a) a d e h

(b) ac f h

(c) a f h g

(d) ae f g

  1. For the graph shown below correct set is

Node branch           twigs             link

(a)  4          6                   4                     2

(b) 4           6                  3                    3

(c) 5           6                  4                   2

(d) 5           5                4                    1

  1. In the following circuit, the voltage VA is

(a) 4.33 V

(b) 4.09 V

(c) 8.67 V

(d) 8.18 V

  1. In the following circuit, the voltage V2 is

(a) 0.5 V

(b) 1.0 V

(c) 1.5 V

(d) 2.0 V

  1. In the following circuit, the voltage V1 is

(a) 120 V

(b) -120 V

(c) 90 V

(d) -90 V

  1. In the following circuit, the voltage V1 is

(a) 0.4 VS

(b) 1.5 VS

(c) 0.67 VS

(d) 2.5 VS

  1. In the following circuit the voltage VA is

(a) -11 V

(b) 11 V

(c) 3 V

(d) -3 V

  1. An independent voltage source in series with an impedance ZS = RS = JXS delivers a maximum average power to a load impedance ZL when

(a) ZL = RS + JXS

(b) ZL = RS

(c) ZL = JXS

(d) ZL = RS = JXS

  1. The RC circuit shown in the figure is

(a) a low-pass filter

(b) a high-pass filter

(c) a band-pass filter

(d) a band-reject filter

  1. The condition on R,L and C such that step response y(t) in figure has no oscillations, is

(a) R > 1/2 L/C

(b) R > L/C

(c) R > 2 L/C

(d) R = 1/LC

  1. The ABCD-parameters of an ideal n:1 transformer shown in figure are [n 0 0 x] the value of x will be a

(a) n

(b) 1/n

(c) n2

(d) 1/n2

  1. In a series RLC circuit R = 2 K, L = 1 H and C = 1/400 uf. the resonant frequency is

(a) 2 x 104 hz

(b) 1 x 104 hz

(c) 104 hz

(d) 2 x 104 hz

  1. The maximum power that can be transferred to the load resistor RL from the voltage source in figure is

(a) 1 W

(b) 10 W

(c) 0.25 W

(d) 0.5 W

  1. All the resistances in the figure are 1 each. the value of I will be

(a) 1/15 A

(b) 2/15 A

(c) 4/15 A

(d) 8/15 A

  1. The time constant of the network shown in the figure is

(a) 4RC

(b) 3RC

(c) 4RC

(d) 2RC/3

  1. The energy stored in the magnetic field at a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current of 10 a, is

(a) 0.015 j

(b) 0.15 j

(c) 0.5 j

(d) 1.15 j

  1. In the following circuit, the values of VTH and RTH are

(a) – 2 V, 6/5

(b) 2V, 5/6

(c) 1 V, 5/6

(d) -1 V, 6/5

  1. A simple equivalent circuit of the two-terminal network shown in figure is
  2. In the following circuit, the values of IN and RN are

(a) 3 A, 10/3

(b) 10 A, 4

(c) 1.5 A, 6

(d) 1.5 A, a

  1. In the following circuit, the values of VTH and RTH are

(a) 2 V, 4

(b) 4 V, 4

(c) 4 V, 5

(d) 2 V, 5

  1. In the following circuit, the values of IN and RN are

(a) 4 A, 3

(b) 2 A, 6

(c) 2 A, 9

(d) 4 A, 2

  1. For the following circuit, the values of VTH and RTH are

(a) – 100 V, 75

(b) 155 V, 55

(c) 155 V, 37

(d) 145 V, 75

  1. Consider the network graph shown in figure. which one of the following is not a tree of this graph?
  2. The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in figure, is

(a) L1 + L2 + M

(b) L1 + L2 – M

(c) L1 + L2 + 2M

(d) L1 + L2 – 2M

  1. The circuit shown in figure with R = 1/3, L = 1/4 H, C = 3F has input voltage V(t) = sin 2t. the resulting current I(t) is

(a) 5 sin (2t + 53.10)

(b) 5 sin (2t + – 53.10)

(c) 25 sin (2t + 53.10)

(d) 25 sin (2t – 53.10)

  1. For the RL circuit shown in figure the input voltage v(t) – u(t). the current I(t) is

Answers and solution

Unit Exercise – 1

  1. (a) Short-circuit admittance parameters are given by

Two-port current in terms of voltage

[I1/I2] = [Y11 Y12/Y21 Y22] [V1/V2]

I1 = Y11 V1 + Y12 V2

I2 = Y22 Y1 + Y22 V2

Y11 = I1/V1|V2 = 0

Y12 = I1/V2|V1 = 0

Y21 = I2/V1|V2 = 0

Y22 = I2/V2|V1 = 0

To calcuate Y11 and Y21

Terminals 2 – 2′ make shorten V2 = 0

Now, circuit is

Y11 = I1/V1|V2 = 0 = 1/0.5110 x 5 = 1/0.25 = 4

V1 = – 0.5 I2

I2/V1 = – 1/0.5 I2

I2/V1 = – 1/0.5 = – 2

Y21 = I2/V1 = – 2

To calcuate Y12 and Y22

Terminals 1 – 1′ make shorten as V1 = 0.

Now, circuit is

V2 = 0.5I1

Y12 = I1/V2|V1 = 0 = 1/-0.5 = – 2

V2 = (0.5110.5)I2 = 0.25I2

Y22 = I2/V2|V1  = 0 = 1/0.25 = 4

Hence, [Y11 Y12/Y21 Y22] = [4 -2/-2 4]

Alternate Method

For general network

The admittance parameters

Y11 = YA + YC

Y12 = – YC

Y21 = – YC

Y22 = YC + YB

Convert impedance to admittance in circuit

Y11 = 2 + 2 = 4

Y12 = – 2

Y21 = – 2

Y22 = 2 + 2 = 4

[Y11 Y12/Y21 Y22] = [4 -2 /-2 4]

  1. (d)

parallel resonant circuit

Q = 00/s0

so = o0/Q

  1. OORC = R/0OL = R/R C/L

so = BO = 1/LC x 1L/RL = 1/RC

So = BO = 1/LC x 1/L/RC = 1/RC

O0 = 1/LC

Hence, bandwidth of parallel resonant circuit

BW = 1/RC

i.e., BW decreases when R is increased and remains constant if L is increased.

Impedance curve of parallel resonant circuit At resonance, from figure input impedance is a real quantiy and attains its maximum value.

  1. (b)

Nodal circuit analysis method is based on KCL and ohm’s law.

  1. (b)

Circuit for equivalent capacitance by applying the concept of series parallel combination of the capacitance.

1/CEQ = 1/C + 1/C + 1/5C/3

= 1/C + 1/C + 3/5C = 1/C [5 + 5 + 3/5]

CEQ = 5C/13

  1. (a)

Suppose value of resistance be R each.

Applying KVL by selecting any path between A and B,

VAB = 1/3. R + 1/6 R + 1/3 R

VAB = IR [5/6]

VAB/I = 5R/6

RAB = 5R/6

Where, R = 1 for this problem

so, RAB = 5/6

  1. (a)

Apply same concept as done in earlier problem.

  1. (a)

RAB = 1 + (2||RAB)

RAB = 1 + 2.RAB/2 + RAB

RAB = 2 + RAB + 2RAB/2 + RAB

2RAB + (RAB)2 = 2 + RAB + 2RAB

(RAB)2 – RAB – 2 = 0

R2AB – 2RAB + RAB – 2 = 0

RAB (RAB – 2) + 1 (RAB – 2) = 0

(RAB + 1) (RAB – 2) = 0

RAB = – 1,2

RAB = – 1not possible

RAB = 2

  1. (c)

Energy delivered during talk-time

E = V (t) I (t) dt

I (t) = 2 A = constant = 2 |v(t) dt

2 x shaded area

= 2 x 1/2 x (10 + 12) x 60 x 10

= 13.2 kj

  1. (a)

Given, 60 V source is absorbing power, it means that current flow from +ve to -ve terminal in 60 V source (as indicated I1 in figure)

I + I1 = 12 A ………….(i)

current source must be have the value less than 12 A to satisfies eq . (1).

  1. (c)

Impedance of capacitor XC = 1/sC

VO = ILRI ………………….(i)

From current division rule,

IL = I x XC/RL + XC

VO = I x CRL/RL + XC ………..(ii)

Equivalent impedance of circuit

ZEQ = R + (XC||RL)

= R + XCRL/XCRL

We can write

I = VI/ZEQ

Putting value in eq. (ii)

VO (S) = VI (s)/XCRL x XCRL/XC + RL

VO(s)/VI (s) = XCRL/RXC + RRL + XCRL

Dividing by XCRL,

VO(s)/VI (s) = 1/1 + R/XC + R/RL = 1/1 + RsC + R/RL

R = RL

VO/VI (S) = 1/2 + sCR

  1. (b)

Laplace transform of definite integral

L[f(t) dt] = F(s)/s

laplace transform of indefinite integral

L[ f (t) dt] = F(s)/s + r-1(0+)/s

  1. (d)

The given figure is non-linear because after 1 V the output becomes constant. any element will be passive if the ratio V|I is positive any time, so the given output shown that the element in passive.

Element will be bilateral if and only if the magnitude (i.e., V or I) is same in both the directions. here, the given element is non-bilateral.

  1. (c)

Average power = [1/1 (10t)2 dt ] R

[By using formula average power = 1/T |I2 dt]. R

= 100 [t2 dt] x 10 = 1000 [t3/3]1

1000 x 1/3 = 1000/3 W

  1. (c)

current through the surface is given by

I(t) = dQ/dt where, Q = charge flowing

Q = |I(t) dt

for 0 < t < 6 s,

Q = I (t) dt = Area under the curve I (t) – t

= Area OBC – Area BCDE

= 1/2 x 3 x 5 – [(1/2 x 1 x 1) + (1 x 1) + (1/2 x 1 x 1)]

= 5.5 C

  1. (a)

Energy absorbed by the device is given

E = |P(t) dt

where, p (t) = power absorbed by the device

E = |V(t) I(t) dt

= 10 (1 – e-7t) dt + |0. (e-7(t-1)) dt

= 10 (1 – e-7t) dt

= 10 (t)1 – 10 [e-7t/7]1 = 10 – 10 [-e-7t/7 + 1/7]

= 10 – 10/7 (1 – e-7t) = 8.5 j

  1. (c)

Applying superposition theorem,

Case 1 taken current source, circuit becomes as given in figure.

let the current flowing in R is I1

I1 = 2 x 2/2 + R

= 4/2 + R ……………….(i)

Case ii  taken voltage source, circuit becomes as given in figure.

Let the current flowing is R is I2

I2 = 2/2 + R………………………(ii)

since current in both the cases flowing in the same direction, so net current

I = I1 + I2

1 = 4/2 + R + 2/2 + R (I = 1, given)

2 + R = 6

R = 6 – 2 = 4

  1. (a)

since, in one inductor current is leaving to dot and in other inductor current is entering to dot. so,

LQE = L1 + L2 – 2M

  1. (c)

by using delta to star conversion,

R1 = R12 x R13/R12 + R23 + R31 = J5 x J5/J5 + J5 – J5 = J5

R2  = R12 x R23/R12 + R23 + R31

= J5 x (-j5)/j5 + j5 – j5 = – j5

R3 = R13 x R23/R12 + R23 + R31

= J5 x – j5/j5 + j5 – j5 = – j5

  1. (d)

the circuit is as shown below.

100 = 65 + V2          V2 = 35 V

V3 – 30 = V2              V3 = 65 V

105 + V4 – V3 – 65 = 0   V1 = 25 V

V4 + 15 – 55 + V1 = 0   V1 = 15 V

  1. (b)

changing current source to voltage source,

Applying KVL,

ISRS = – I (t) RS – I (t) R – L dI(t)/dt = 0

t = 0+

I(0+) = 0 because at t = 0 there is no curent  flow through inductor and inductor opposes the instantaneous flow of current in circuit and I(0) = I(0+) = 0

dI(0+) dt = ISRS/L

  1. (c)

A tree is connected subgraph of a convected subgraph containing all the nodes of the graph but containing to loops.

  1. (d)

In pole-zero plot there are two transmission zeros are located on the jo-axis, at the complex conjugate location, then the magnitude response exhibits a zero transmission at 0 – 0.

  1. (b)

other three circuits can be drawn on plane without crossing as shown below.

  1. (a)

the circuits 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch.

  1. (c)

following figure shows that afhg is a tree of given graph.

  1. (b)

in this graph there are 4 nodes and 6 branches.

twig = n- 1 = 4 – 1 = 3

link = b – n + 1 = 6 – 4 + 1 = 3

  1. (c)

using nodal analysis,

VA – 10/4 + VA/2 = 4

VA = 8.67 V

  1. (d)

V2/20 + V2 + 10/30 = 0.5

V2 = 2 V

  1. (d)

-V1/60 + -V1/60 + 6 = 9

V1 = – 90 V

  1. (b)

V1 = 4VS/6R + VS/3R

1/6R + 1/3R + 1/6R

V1 = 1.5 VS

  1. (b)

VA = 2 (3 + 1) + 3(1)

VA = 11 V

  1. (d)

For maximum power transfer, load impedance ZL should be complex conjugate of the series impedance ZS.

ZL = ZS

= R3 – JXS

The maximum power transfer will be

P = V2/4RS

  1. (c)

Impedance of capacitor XC = 1/0C

where, o = 0

XC = 0

Capacitor behaves as open-circuit.

VO = 0

XC  = 0

Capacitor behaves as short-circuit.

VO = 0

Hence, RC is circuit shown is band-pass filter.

  1. (c)

Transfer function of circuit

Y(s)/U(s) = 1/sC/sL + R + 1/sC = 1/S2LC + sRC + 1

Dividing numerator and denominator by LC

1/LC/S2 + R/L s + 1/LC

Characteristic equation

s2 + R/L s + 1/LC = 0

Comparing with general expression,

s2 + 2E0ns + 02n  = 0

on = 1/LC and 2E0N = R/L

E = 1/2 x R/L x LC

E = R/2 C/L

For response Y(t) has no oscillation;

E > 1

R > 2 L/C

  1. (b)

Two-port output expressed as input parameters in ABCD- parameters

[V1/V1] = [A B  C   D][V2/I2]

V1 = AV2 + BI2

I1 = CV2 + DI2 ………….(1)

We know that for ideal transformer,

V1/V2 = I2/I1 = N/1

V1 = nV2

I1 = I2/N

We can write V1 = nV2 + 0I2

Comparing rith eq. (i)

[A B C D] = [n  0  0   1/n]

hence,  X = 1/n

  1. (b)

For series resonant circuit

resonant preparing  fr = 1/2 LC

= 1/2 1/400 x 10-6 x 1 = 104

  1. (c)

For maximum power transfer to load resistor RL, RL must be equal to 100

maximum power V2/4RL = 102/4 x 100 = 0.25 W

  1. (d)

In order to calcuate I in the given figure draw the equivalent model.

1/VEQ = 1/15/8 = 8/15

  1. (a)

the constant

= REQC

REQ = 4R||4R = 2R

C = 2C

= 2R . 2C = 4RC

  1. (b)

we know

energy = u . N2 A/I

4 x 10-7 x (1000)2 x (3/2 x 10-2)2/30 x 10-2

= 0.15 j

  1. (C)

VTH = (2) (3) (1)/3 + 3 = 1 V

RTH = 1||5 = 5/6

  1. (b)

after killing all source equivalent resistance is R. open circuit voltage = V1.

  1. (a)

the circuit is as shown below.

RN = 2||4 + 2 = 10/3

V1 = 15/2/1/2 + 1/2 + 1/4 = 6 V

ISC = IN = V1/2 = 3 A

  1. (b)

VTH = (6) (6)/3 + 3 = 4 V

RTH = (3||6) + 2 = 4

  1. (d)

the short-circuit current across the terminal is

ISC = 6 x 4/4 + 2 = 4 A = IN, RN = 6||3 = 2

  1. (b)

For the calculation of RTH. if we kill the sources the 20 resistance is inactive because 5 A source will be open-circuit.

RTH = 30 + 25 = 55

VTH = 5 + 5 x 30 = 155 V

  1. (b)

A tree is connected subgraph of a connected graph containing all the nodes of the graph but containing no loop.

option (b) has loop. so, it is not a tree.

hence, (b) cannot be tree of the given graph.

  1. (d)

A current entering the dotted terminal of one coil produce an open-circuit voltage with positive voltage refernce at the dotted terminal of the second coil.

LEQ = dI/dt = L1  dI/dt + L2 dI/dt – M dI/dt – M dI/dt

LEQ = L1 + L2 – 2M

  1. (a)

V (t) = sin 2t

in phasor form,

V = 1 <00

Admittance of circuit

Y = YR + YL + YC

= 1/R + 1/f0L + J0C

= 3 + 1/jo3 x 1/4 + j x 2 x 3

= 3 – j2 + j6

= 3 + 4 j

in phasor form,

= 32 + 42 < tan-14/3

Y = 5<53.10

I = V.Y = 1<800 x 5<53.10

= 5<53.10

I = 5 sin (2t + 53.10)

  1. (c)

Applying KVL in loop,

V1(t) = L dI(t)/dt + I(t) R

u(t) = dI (t)/dt + 2I(t)

solution of equation

I(t) = 1/2 (1 – e-2t) u(t)

I(t) = 1/2 = 0.5

At t = 1/2.

I(t) = 1/2 (1 – e-1)   = 0.31