Networks Signals and Systems Subject Wise
Question 1 |
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see (b) in the figure), what is the current through the short circuit at Port 1?

0.5 A | |
1 A | |
2 A | |
2.5 A |

Question 2 |
for 0 ≤ t < T
where R0 = 1 Ω, and C = 1 F. We are also given that T = 3 R0C and the source voltage is Vs = 1 V. If the current at time
t = 0 is 1 A, then the current I(t), in amperes, at time t = T/2 is __________ (rounded off to 2 decimal places).
Fill in the Blank Type Question |

Question 3 |
Sin(1000 t) + cos(1000 t) | |
2 sin(1000 t) + 2 cos(1000 t) | |
3 sin(1000 t) + cos(1000 t) | |
Sin(1000 t) + 3 cos(1000 t) |

Question 4 |
The ratio


Fill in the Blank Type Question |

V1=

V2=




Question 5 |

where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value 2 Amperes is _______.

Fill in the Blank Type Question |

Question 6 |

The parameter for the given two-port network (in ohms, correct to two decimal places) is ____________.
Fill in the Blank Type Question |

Question 7 |
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Question 8 |
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Question 9 |
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Question 10 |



The magnitude of the current (in amperes, accurate to two decimal places) through the source is ____________.
Fill in the Blank Type Question |

Total resistance RT is resultant of following combination,

We have values for R1 and R3,


So,


Question 11 |


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Question 12 |
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Question 13 |

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->Hence average value of V0 i.e. output of half wave rectifier is given by,

Question 14 |
The current i(t) (in ampere) at t = 0.5 seconds is ________

Fill in the Blank Type Question |

Question 15 |

T-1 =T | |
T2 =T | |
Determinant (T) = 0 | |
Determinant (T) = 1 |
if AD – BC = 1 i.e. Determinant(T) = 1
Note : If network is reciprocal, then Determinant (T) = 1
Question 16 |


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then any voltage or any current of any element should have also 3 terms based on this option B. and D. are eliminated.



Question 17 |

Fill in the Blank Type Question |



VC=I*XC=25V
Amplitude of 25 Volt VC = 25 Volt
Hence, the correct answer is 25
Question 18 |
5 V, 25 V | |
10 V, 30 V | |
15 V. 35 V | |
0 V, 20 V |

Current flowing through both the parallel 4 Ω will be I.
So,





V1 = 5 volts
V2 =25 volts
Question 19 |

Fill in the Blank Type Question |

Question 20 |

The value of IL that maximizes the power absorbed by the constant current load is
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So we can say under MPT Vs / 2 will appear on the load
so

Question 21 |

If time t is in seconds, the capacitor voltage Vc (in volts) for t>0 is given by
4(1 —exp (—t/0.5)) | |
10 — 6 exp (—t/0.5) | |
4(1 — exp (-0.6)) | |
10 — 6 exp (—t/0.6) |


Question 22 |

Fill in the Blank Type Question |
IR = I
IL = QI < – 90°
IC = QI < 90°
For parallel RLC circuit

Question 23 |

Where the entries are in



Then the value of


Fill in the Blank Type Question |

Question 24 |


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Question 25 |
Fill in the Blank Type Question |

Applying KVL,

Thus, current through

Question 26 |

Fill in the Blank Type Question |


Question 27 |
Cos (200t) + 2sin (500t) | |
2 cos (200t) + 4sin (500t) | |
Sin (200t) + 2cos (500t) | |
2 sin (200t) + 4cos (500t) |

since there are 2 frequency term output will also have 2 frequency term.



So w.r.t. 4sin500t output must be 4sin500t without any change in amplitude and phase, this is satisfied by only option B.
Question 28 |

Fill in the Blank Type Question |

Question 29 |

Fill in the Blank Type Question |


at t = 0+ when is open circuited, the capacitors will have an ideal voltage source of values 4V and 6V so the circuit will be

So the current through 2Ω resistor at t = 0+ should be 4/(2+2) = 1A

Question 30 |
Fill in the Blank Type Question |


By using KCL at node ‘a’.

KCL at b
-

Question 31 |


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And from the figure We get

Question 32 |

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Question 33 |
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Question 34 |

Fill in the Blank Type Question |

If current I = 0,
Voltage at node A = input voltage
Applying KCL at node A,

or

Solving equation, we get

Question 35 |

Fill in the Blank Type Question |
Transfer function



For steady state response,

But Td(s) is negative, so response is also negative, i.e. -0.5
Question 36 |

1. A parallel strip containing the jΩ axis
2. A parallel strip not containing the jΩ axis
3. The entire s-plane
4. A half plane containing the jΩ axis
A | |
B | |
C | |
D |
and

So, it is a finite duration signal and for the finite duration signal ROC is always entire s-plane.
Question 37 |

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Question 38 |
150 | |
100 | |
50 | |
200 |

Question 39 |

10 | |
25 | |
5 | |
30 |

Question 40 |

10 | |
25 | |
5 | |
30 |

Question 41 |
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Question 42 |
2.333 | |
1.333 | |
1.255 | |
3.333 |

Question 43 |

Then the value of β is _________
2 | |
-2 | |
1 | |
-1 |

Now, we have ROC


Coefficient of s is zero. From above equation, we conclude the result as

Question 44 |
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Z1 = First section Output impedance
Z2 = Second section Input impedance
For maximum power transfer, upto 1st section is

Question 45 |
Consider the configuration shown in the figure which is a portion of a larger electrical network
For and currents i1=2A, i4=-1A, i5=-4A which one of the following is TRUE?
I6=5A | |
I3=-4A | |
Data is sufficient to conclude that the supposed currents are impossible | |
Data is insufficient to identify the current ![]() |

Question 46 |
A Y-network has resistances of 10 Ω each in two of its arms, while the third arm has a resistance of 11 Ω in the equivalent D − network, the lowest value in Ω among the three resistances is ______________.
25.09 | |
27.09 | |
28.09 | |
29.09 |


i.e, lowest value among three resistances is

Question 47 |
(18 + j 1.5) kVA | |
(18 - j 1.5) kVA | |
(20 + j 1.5) kVA | |
(20 - j 1.5) kVA |

Question 48 |

0.324 | |
0.368 | |
0.408 | |
0.442 |

Question 49 |

5 | |
10 | |
15 | |
20 |

Differentiate with respect to time,


For critically damped response,

This is the required value of C to have critically damped response i(t)
Question 50 |
in series with a current source | |
in parallel with a voltage source | |
in series with a voltage source | |
in parallel with a current source |

Question 51 |
A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are 1 H, 1 F, and 1Ω, respectively, the operating angular frequency (in rad/s) is ________.
0.25 | |
0.30 | |
0.40 | |
0.45 |
The operating frequencyat which current leads the supply.
again magnitude of current is half the value at resonance
By substituting R, L & C values,
So, operating frequency
Question 52 |

1.24 | |
1.28 | |
1.32 | |
1.36 |

The y-parameter are added


Question 53 |

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Question 54 |

Which one of the following represents the current i(t)?
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Question 55 |

0.2 Amp | |
0.4 Amp | |
0.5 Amp | |
0.8 Amp |

Question 56 |
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Question 57 |

2r/s | |
3 r/s | |
4r/s | |
5r/s |


Equating imaginary term to zero i.e.,


Question 58 |
5 | |
10 | |
15 | |
20 |

Question 59 |

Voltage controlled voltage source | |
Voltage controlled current source | |
Current controlled current source | |
Current controlled voltage source |

The dependent source represents a current controlled current source
Question 60 |
2.8 | |
3.4 | |
5 | |
4.4 |

Question 61 |
21.25 | |
25.50 | |
31.25 | |
35.0 |

Question 62 |

The value of Re/R is ________
2.0 | |
2.618 | |
4.0 | |
4.618 |


Then equivalent resistance is

→ For the given network, we can split in to R is in series with



Question 63 |
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Question 64 |


3 | |
4 | |
5 | |
7 |

Question 65 |


125/100 and 80/100 | |
100/100 and 80/100 | |
100/100 and 100/100 | |
80/100 and 80/100 |
Given transformer



Attenuation factor will be 0.8
So,

or,

at

at

Question 66 |
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective O factor at the same operating frequency is
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Question 67 |
in µC stored in the effective capacitance across the terminals are respectively,

2.8 and 36 | |
7 and 119 | |
2.8 and 32 | |
7 and 80 |

Question 68 |

The current Is in Amps in the voltage source and voltage Vs in Volts across the current source respectively, are
13, –20 | |
8, –10 | |
–8, 20 | |
–13, 20 |

Question 69 |
2 | |
3.33 | |
10 | |
12 |

Question 70 |


44.2 | |
50 | |
62.5 | |
125 |







Question 71 |
50 Ω | |
100 Ω | |
5 kΩ | |
10.1 k Ω |

To find thevenin impedance across node 1 and 2. Connect a 1 V source and find the current through voltage source.
Then

By applying KCL at node B and A
iAB + 99ib = ITh
ib = iA + IAB
⇒ ib – iA + 99ib = ITh
⇒ 100ib – iA = iTh …(1)
By applying KVL in outer loop
10 × 103ib = 1
ib = 10-4 A
And 10 × 103ib = -100iA
⇒ iA = -100iA
From equation (i)
100iA + 100ib = ITh
⇒ ITh = 200ib
⇒ 200 × 104 = 0.02

Question 72 |
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0 A |

Question 73 |

0.8 Ω | |
1.4 Ω | |
2 Ω | |
2.8 Ω |

Current through 3 V source is

So power delivered to circuit B by circuit A is
P = i2R + i1 × 3

for P to be maximum



49(2+R) – 98R – 21(2+R) = 0
98 + 42 = 49R + 21R

Question 74 |
–5V | |
2V | |
3V | |
6V |


= 3A …(1)
KCL at node B, we have
i = i2 + i1
i2 + i1 = 3A ..(2)
KCL at node E, we have
i1 = i3 + i4 …(3)
KCL at node D we have
i5 = i2 + i3 + i4
i5 = i2 + i1
i5 = 3A
KCL at node F, we have
i6 + 2 + i5 = 0
i6 = -2-i5
So, VC – VD = 1 × i6 = -5V
Question 75 |
(i) 1 Ω connected at port B draws a current of 3 A
(ii) 2.5 Ω connected at port B draws a current of 2 A

With 10 V dc connected at port A. the current drawn by 7 Ω connected at port B is
3/7 A | |
5/7 A | |
1A | |
9/7 A |
IL = VTh / (RTh +RL)
(i) RL=1 Ω, IL =3A,
Implies 3 = VTh / (RTh +1),
(ii) For RL=2.5 Ω, IL = 2A.
Implies 2= VTh / (RTh +2.5)
Divide above equations (i) and (ii),
3/2 = (RTh + 2.5)/ (RTh + 1),
3 RTh + 3 = 2 RTh + 5.
RTh = 2 Ω.
Substitute RTh in the above equation, VTh = 3(2+1) = 9V.
For RL=7 Ω, IL= VTh(2+RL) = 9/(2+7) = 1A
Question 76 |
Y21 = Y11 and Y22 = Y21 | |
Y21 = Y12 and Y11 = Y22 | |
Y21 = Y22 and Y11 = Y22 | |
Y11 = Y22 and Y21 = Y22 |

Question 77 |
input impedance and forward current gain | |
reverse voltage gain and output admittance | |
input impedance and reverse voltage gain | |
output impedance and forward current gain |

V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2

∴ h11 is input impedance

∴ h21 is forward current gain
∴ Option A
Question 78 |
V1 = 6V2 – 4I2
I1 = 7V2 – 2I2
A, B, C and D parameters are
6,-4 Ω, 7 mho and -2 | |
6, 4 Ω, 7 mho and 2 | |
-6, 4 Ω, -7 mho and 2 | |
6, 4 Ω, -7 mho and -2 |

Question 79 |
30.5 μF | |
41.5 μF | |
64.0 μF | |
76.8 μF |

PR = 400 watt
VR = 160 V


Question 80 |
19.6 A and 51.5o | |
27.4 A and -51.5o | |
19.6 A and -51.5o | |
27.4 A and 51.5o |

Question 81 |
172.7 V | |
184.5 V | |
196.3 V | |
208.1 V |

Question 82 |
2.5 V | |
3.4 V | |
4.3 V | |
5.2 V |
C = 680μF = 28V
Voutput average = 28V
RL = 200Ω
Charge stored while charging = charge discharged while discharging
∴


∴

∴ Option B
Question 83 |

In(9) sec | |
ln(9)0.125 sec | |
ln(9)0.25 sec | |
9 sec |
Taking laplace transform of given circuit then finding the voltage across capacitor . then after using of concept of maxima/minima , we get the time when Vc(t) will be maximum i.e. t=ln(9)0.125 sec
Refer the Topic Wise Question for Time Domain and Frequency Analysis of Linear circuits Networks
Question 84 |

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for parallel connection Y-parameters will be added , from the given figure we can observe that given circuit is symmetrical and reciprocal.
We can split the given circuit in two equivalent circuit which are connected in parallel.

Y-parameters for this circuit will be :







2nd circuit consist of only capacitors., it is also a reciprocal and symmetrical circuit, Y-parameters for this circuit will be









On adding these two parameters , we get eqivalent Y-parameter.
Refer the Topic Wise Question for Two Port Networks Networks
Question 85 |


5 ppm/0C | |
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9 ppm/0C | |
4.5 ppm/0C |




When



where







Given ,


On putting these values ,

Refer the Topic Wise Question for Basic Electrical Engineering Networks
Question 86 |
-3 ppm/0C | |
3 ppm/0C | |
-6 ppm/0C | |
6 ppm/0C |
Formula for resonance frequency in LC tank circuit =

Differentiate .r.t temperature(T), only Capacitance will vary because here temperature coefficient of



And C=C[1+


From (1) and (2)



Now as we know for any change in temperature(


=

=

This equation shows that temperature coefficient of


Given,


Therefore,


= -3ppm/

Refer the Topic Wise Question for Steady State equations and Analysis Networks
Question 87 |


0.0148 A and 0.055 A | |
0.0148 A and -0.055 A | |
0.0296 A and 0.11 A | |
0.0296 A and -0.11 A |
For 0 < t < 1 time constant(

, switch is connected to “1”. Therefore circuit becomes,


=500

= 2.5

So the capacitor slowly charges to 20V with following equation

At t=



= 20[1-0.37]
=12.6 V
i.e At

I at


After 1 time constant (





Therefore, I =

=

= 0.0548A

Refer the Topic Wise Question for Time Domain and Frequency Analysis of Linear circuits Networks
Question 88 |

Reciprocal and symmetric | |
Rciprocal but not-symmetric | |
Symmetric but non-rciprocal | |
Neither symmetric nor reciprocal |
Normally we know dependent source make a circuit non-reciprocal. As for symmetrical circuit , there should be mirror image in circuit , but in figure we can clearly see that there is no any line of symmetry. So given circuit is Non-reciprocal and not symmetric.
To check we have to calculate Z-parameter.

When







When











Refer the Topic Wise Question for Two Port Networks Networks
Question 89 |

2/3 A | |
3/2 A | |
1/3 A | |
None of the above |
In the given circuit , we can use superposition principle. While using the voltage source current source is open circuited, and while using the current source voltage source is short circuited.
And calculate

Refer the Topic Wise Question for Network Equations and Solution Methods Networks
Question 90 |

16.67 A/sec and -4.16 x 104 A/sec | |
16.67 A/sec and -3.33 x 104 A/sec | |
25 A/sec and -4.16 x 104 A/sec | |
25 A/sec and -3.33 x 104 A/sec |
In steady state at t=


Inductor short circuit ,

Capacitor open circuit,


=


=

Inductor does not allow sudden change in current and capacitor does not allow sudden change in voltage.
So


The equivalent circuit at t=


Where ,

Also,


=

For inductor,


=

= 16.67 A/sec
KVL for capacitor branch,
100V=

On differentiating w.r.t ‘t’
0=20


Also,


Also., C



Put (2) in (1)
0= 20

Given, C=2






Refer the Topic Wise Question for Steady State equations and Analysis Networks
Question 91 |
Constant Current, Topping, Float | |
Topping, Constant Current, Float | |
Float, Topping, Constant Current | |
Float, Constant Current, Topping |
The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation. Lead acid batteries should be charged in three stages, which are [1] constant-current charge, [2] topping charge and [3] float charge. The constant-current charge applies the bulk of the charge and takes up roughly half of the required charge time; the topping charge continues at a lower charge current and provides saturation, and the float charge compensates for the loss caused by self-discharge.
Refer the Topic Wise Question for Basic Electrical Engineering Networks
Question 92 |

K2 x s/(s + 5) | |
(s + 5)/(s x K2) | |
K2 x s/(s + 6) | |
(s 6)/(s x K2) |
Given=> Z1= K1 [(s+2)/(s+5)]
Now, We have to determine what is Z2. Now according to general concept, Z2 can be written as: Ls||(R1+R2)
This can also be written as: (R1+R2)Ls/Ls+R1+R2.................(a)
Now given options are having 'K' term in them. It means we have to bring Z1 in the scenario.
So, given=> Z1= K1 (s+2)/(s+5)
Now in order to get R1, R2 and Ls in terms of K, we have to reduce Z1 and bring it in a form so that it can be compared with the Z1 given in the question.
Now looking at the figure we can write:
Z1= R1||(R2+Ls)=> R1(R2+Ls)/R1+R2+Ls
So, we have to bring it in a particular form.
Now=> in question's Z1 term=> 's+' term is present. It means that we have to somehow take Ls common from here. So taking Lcommon so that 's' remains inside the bracket.
So on taking 'L' term out, and simplifying we will get: Z1= (R2/L +s)R1 / (R1+R2)/L +s
Now this can easily be compared with Z1 given in question. On comparing, we get:
R1=K1, R2/L=2
So, now putting these values in (a) we get:
Z2= (R1+R2)Ls/Ls+R1+R2
=> L(R1+R2)(s)/ R1+R2/L +s
Now, R1+R2/L=5(From comparing), R1=K1, and R1/L=3.
So, on putting this, we will get Z2= sK/(s+5)
Refer the Topic Wise Question for Two Port Networks Networks