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 R_{0} = 1 Ω, and C = 1 F. We are also given that T = 3 R_{0}C and the source voltage is V_{s} = 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 is ______
Fill in the Blank Type Question 
V_{1}=
V_{2}=
=
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 twoport network (in ohms, correct to two decimal places) is ____________.
Fill in the Blank Type Question 
Question 7 
Fill in the Blank Type Question 
Question 8 
Fill in the Blank Type Question 
Question 9 
Fill in the Blank Type Question 
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 R_{T} is resultant of following combination,
We have values for R_{1} and R_{3},
and
So, Thus, current through 11 V voltage source is,
Question 11 
Question 12 
Fill in the Blank Type Question 
Question 13 
Fill in the Blank Type Question 
>Hence average value of V_{0} i.e. output of half wave rectifier is given by, Answer Range ( 3.15 to 3.21 )
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  
T^{2} =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 
If the source has 3 frequency terms as given
then any voltage or any current of any element should have also 3 terms based on this option B. and D. are eliminated.
If we take option C.. It has 3 frequency term but it also suggest there is a phase change so 4„ but amplitude must be same as input as ak is present which may not be true always.
So option A. is correct, as it suggest frequency term of output and inputs are same with possible change in amplitude and phase, because we have ().
Question 17 
Fill in the Blank Type Question 
V_{C}=I*X_{C}=25V
Amplitude of 25 Volt V_{C} = 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, by KVL
by KVL
V_{1} = 5 volts
V_{2} =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
So we can say under MPT V_{s} / 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 
I_{R} = I
I_{L} = QI < – 90°
I_{C} = QI < 90°
For parallel RLC circuit
Question 23 
Where the entries are in Suppose
Then the value of (in) equals to
Fill in the Blank Type Question 
Question 24 
Question 25 
Fill in the Blank Type Question 
Applying KVL,
Thus, current through
Question 26 
Fill in the Blank Type Question 
Question 27 
The output voltage V_{o}(t) is
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.
If we take 4sin500t first i.e. W = 500 then on the output section, this parallel LC combination have so it is open circuit and V0 = Vi
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 
And from the figure We get
Question 32 
Fill in the Blank Type Question 
Question 33 
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
(unit impulse response)
For steady state response,
But T_{d}(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 splane
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 splane.
Question 37 
Question 38 
150  
100  
50  
200 
Question 39 
10  
25  
5  
30 
Question 40 
10  
25  
5  
30 
Question 41 
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 Taking Laplace for given ROC, we have
Coefficient of s is zero. From above equation, we conclude the result as
Question 44 
Z_{1} = First section Output impedance
Z_{2} = Second section Input impedance
For maximum power transfer, upto 1^{st} section is
Question 45 
Consider the configuration shown in the figure which is a portion of a larger electrical network
For and currents i_{1}=2A, i_{4}=1A, i_{5}=4A which one of the following is TRUE?
I_{6}=5A  
I_{3}=4A  
Data is sufficient to conclude that the supposed currents are impossible  
Data is insufficient to identify the current 
Question 46 
A Ynetwork 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 yparameter are added
Question 53 
Question 54 
Which one of the following represents the current i(t)?
Question 55 
0.2 Amp  
0.4 Amp  
0.5 Amp  
0.8 Amp 
Question 56 
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 
For an infinite ladder network, if all resistance are having same value of R
Then equivalent resistance is
→ For the given network, we can split in to R is in series with
Question 63 
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 q_{1} and q_{2} at the chosen operating frequency. Their respective resistances are R_{1} and R_{2}. When connected in series, their effective O factor at the same operating frequency is
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 I_{s} in Amps in the voltage source and voltage V_{s} 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 
W
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
i_{AB} + 99i_{b} = I_{Th}
i_{b} = i_{A} + I_{AB}
⇒ i_{b} – i_{A} + 99i_{b} = I_{Th}
⇒ 100i_{b} – i_{A} = i_{Th} …(1)
By applying KVL in outer loop
10 × 10^{3}i_{b} = 1
i_{b} = 10^{4} A
And 10 × 10^{3}i_{b} = 100i_{A}
⇒ i_{A} = 100i_{A}
From equation (i)
100i_{A} + 100i_{b} = I_{Th}
⇒ I_{Th} = 200i_{b}
⇒ 200 × 10^{4} = 0.02
Question 72 
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 = i^{2}R + i_{1} × 3
for P to be maximum will be aero
_{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 = i_{2} + i_{1}
i_{2} + i_{1} = 3A ..(2)
KCL at node E, we have
i_{1} = i_{3} + i_{4} …(3)
KCL at node D we have
i_{5} = i_{2} + i_{3} + i_{4}
i_{5} = i_{2} + i_{1}
i_{5} = 3A
KCL at node F, we have
i_{6} + 2 + i_{5} = 0
i_{6} = 2i_{5}
So, V_{C} – V_{D} = 1 × i_{6} = 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 
I_{L = }V_{Th} / (R_{Th} +R_{L})
(i) R_{L}=1 Ω, I_{L} =3A,
Implies 3 = V_{Th} / (R_{Th} +1),
(ii) For R_{L}=2.5 Ω, I_{L} = 2A.
Implies 2= V_{Th} / (R_{Th} +2.5)
Divide above equations (i) and (ii),
3/2 = (R_{Th} + 2.5)/ (R_{Th} + 1),
3 R_{Th} + 3 = 2 R_{Th} + 5.
R_{Th} = 2 Ω.
Substitute R_{Th }in the above equation, V_{Th} = 3(2+1) = 9V.
For R_{L}=7 Ω, I_{L}= V_{Th}(2+R_{L}) = 9/(2+7) = 1A
Question 76 
Y_{21} = Y_{11} and Y_{22} = Y_{21}  
Y_{21} = Y_{12} and Y_{11} = Y_{22}  
Y_{21} = Y_{22} and Y_{11} = Y_{22}  
Y_{11} = Y_{22} and Y_{21} = Y_{22} 
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 
V_{1} = h_{11} I_{1} + h_{12} V_{2}
I_{2} = h_{21} I_{1} + h_{22} V_{2}
∴ h_{11} is input impedance
∴ h_{21} is forward current gain
∴ Option A
Question 78 
V_{1} = 6V_{2} – 4I_{2}
I_{1} = 7V_{2} – 2I_{2}
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 
P_{R} = 400 watt
V_{R} = 160 V
Question 80 
19.6 A and 51.5^{o}  
27.4 A and 51.5^{o}  
19.6 A and 51.5^{o}  
27.4 A and 51.5^{o} 
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
V_{output average} = 28V
R_{L} = 200Ω
Charge stored while charging = charge discharged while discharging
∴ × C = I_{DC} × T_{o}
∴ = 3.4 V
∴ 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 
for parallel connection Yparameters 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.
Yparameters for this circuit will be :
=
=
=
2^{nd} circuit consist of only capacitors., it is also a reciprocal and symmetrical circuit, Yparameters for this circuit will be
=
=
=
=
On adding these two parameters , we get eqivalent Yparameter.
Refer the Topic Wise Question for Two Port Networks Networks
Question 85 
5 ppm/^{0}C  
ppm/^{0}C  
9 ppm/^{0}C  
4.5 ppm/^{0}C 
are temperature coefficient of respectively.
When are connected in series. Then
(1+) ,
where ,
equivalent temperature coefficient
= +
=
Given ,
On putting these values ,
Refer the Topic Wise Question for Basic Electrical Engineering Networks
Question 86 
3 ppm/^{0}C  
3 ppm/^{0}C  
6 ppm/^{0}C  
6 ppm/^{0}C 
Formula for resonance frequency in LC tank circuit =
Differentiate .r.t temperature(T), only Capacitance will vary because here temperature coefficient of (permittivity ) is given.
= …………………. (1)
And C=C[1+], from this …………………. ..(2)
From (1) and (2)
= , on simplifying it we get
Now as we know for any change in temperature(), new frequency will be
=
=
This equation shows that temperature coefficient of will be .
Given,6ppm
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,
=RC
=500
= 2.5 econds
So the capacitor slowly charges to 20V with following equation
At t=,
=
= 20[10.37]
=12.6 V
i.e At , the capacitor voltage is = 12.6 V
I at = =0.0148A
After 1 time constant () i.e. at t=, equivalent circuit becomes
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 notsymmetric  
Symmetric but nonrciprocal  
Neither symmetric nor reciprocal 
Normally we know dependent source make a circuit nonreciprocal. 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 Nonreciprocal and not symmetric.
To check we have to calculate Zparameter.
When
→
+ →
When →
+ → =
→
→not symmetrical
→not reciprocal
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 10^{4} A/sec  
16.67 A/sec and 3.33 x 10^{4} A/sec  
25 A/sec and 4.16 x 10^{4} A/sec  
25 A/sec and 3.33 x 10^{4} A/sec 
In steady state at t=
Inductor short circuit ,
Capacitor open circuit, 0A
=
=
Inductor does not allow sudden change in current and capacitor does not allow sudden change in voltage.
So and
The equivalent circuit at t=is,
Where ,
Also,
=v
For inductor,
=
= 16.67 A/sec
KVL for capacitor branch,
100V=
On differentiating w.r.t ‘t’
0=20 + …………………………..(1)
Also, =
Also., C = → ………….(2)
Put (2) in (1)
0= 20
Given, C=2
=4.16
= =4.16 A/s
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] constantcurrent charge, [2] topping charge and [3] float charge. The constantcurrent 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 selfdischarge.
Refer the Topic Wise Question for Basic Electrical Engineering Networks
Question 92 
K_{2} x s/(s + 5)  
(s + 5)/(s x K_{2})  
K_{2} x s/(s + 6)  
(s 6)/(s x K_{2}) 
Given=> Z1= K1 [(s+2)/(s+5)]
Now, We have to determine what is Z_{2}. Now according to general concept, Z_{2} can be written as: Ls(R_{1}+R_{2})
This can also be written as: (R_{1}+R_{2})Ls/Ls+R1+R2.................(a)
Now given options are having 'K' term in them. It means we have to bring Z_{1} in the scenario.
So, given=> Z_{1}= K_{1} (s+2)/(s+5)
Now in order to get R_{1}, R_{2} and L_{s} in terms of K, we have to reduce Z_{1} and bring it in a form so that it can be compared with the Z_{1} given in the question.
Now looking at the figure we can write:
Z_{1}= R_{1}(R_{2}+L_{s})=> R_{1}(R_{2}+L_{s})/R_{1}+R_{2}+L_{s}
So, we have to bring it in a particular form.
Now=> in question's Z_{1} term=> 's+' term is present. It means that we have to somehow take L_{s} common from here. So taking Lcommon so that 's' remains inside the bracket.
So on taking 'L' term out, and simplifying we will get: Z_{1}= (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= (R_{1}+R_{2})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