Download (b) (c) (d)

EEE GATE MODEL PAPER
Apt:
1. What least value must be assigned to * so that the number 197*5462 is divisible by 9?
(a) 1 (b) 3 (c) 4 (d) 2
2.Two numbers are in the ratio of 15:11. If their H.C.F is 13, find the numbers.
(a) 194 and 140 (b) 195 and 143
(c) 185 & 150 (d) 190 & 152
2
2
x y
x 6
3. If
= , find the value of
2
2
y 5
x y
(a)
60
(b)
11
60
(c)
12
61
10
(d)
61
11
4.If in gate exam, a student scores 2 marks for every correct answer and loses 0.5 marks for
every wrong answer. A student attempts all the 100 questions and scores 120
marks. The number of questions he answered correctly was
a. 50
b. 45
c. 60
d. 68
5.A cube is painted blue on all faces is cut into 125 cubes of equal size. How many cubes
are not painted on any face?
a. 8
b. 27
c. 18
d. 16
6.2 men or 4 women can do a work in 22 days. How long will 4 men and 3 women take to
complete the work?
a. 4 days
b. 8 days
c. 16 days
d. 26 days
7. NOTES : SCORE ::
(a) symbols : rebus (b) insignia : military (C) proportions : recipe
(D) program : computer
ANTONYM:
8. BANALITY:
(a) accurate portrayal
(b) impromptu statement
(c) original expression
(d) succinct interpretation
9. During a period of protracted illness, the sick can become infirm, _________ both the
to work and many of specific skills they once possessed.
(a) regaining
(b) denying
(c) pursing
strength
(d) losing
10. Henry went to the conference _____________________about government contracts.
(a) to learn
(b) with the purpose of learning
(c) in order to have the opportunity to learn
(d) in order to be in a position to learn
EXPLANATION:
1-d
2-b
3-a
6.(b) 2*4*22 /4*4 +2*3 = 8
4.(d) C + W =100, 2C – 0.5W=120, thus C = 68
5.(b) 5^3 = 125, thus the number of cut =4, no. of cubes unpainted =3^3 =27
7.Answer is (A)
Hint: (PART AND THE WHOLE) A (musical) score consists of notes; a rebus (meaning
representation of words in the form of pictures or symbols) consists of symbols.(A)
8.Answer is (C)
9. Answer is (D)
Hint: The use of the description “(that) they once possessed” with reference to a skill means that
the people have lost that skill now. The word infirm also means loss of strength by them. Among
the choices, it is losing that is most appropriate for filling in the blank.
10.Ans: (a)
MATHS:
4 2 1 3
11.Given matrix [𝐴] =[6 3 4 7], the rank of the matrix is
2 1 0 1
a)4
b)3
c)2
d)1
12.For what value of a and b , the following simultaneous equation have an infinite number of
solutions ?
X+Y+Z=5, X+3Y+3Z=9, AND X+2Y+az=b
a) 2 & 7
b)3 & 8
c) 8 & 3
d) 7 & 2
13.Which of the following integral is unbounded
𝜋
1
a)∫04 tan 𝑋 𝑑𝑥
b)∫0
14.T he area in first quadrant under curve Y=
𝜋
𝜋
b) 4 -2 𝑡𝑎𝑛−33
a)2
∞
1
𝑑𝑥
𝑋 2 +1
c)∫0 𝑋𝑒 −𝑥 𝑑𝑥
1
is
𝑋 2 +6𝑋+10
𝜋
c) ) 2 - 𝑡𝑎𝑛−33
𝑑𝑋
1
d)∫0
1
dx
1 −𝑋
𝜋
d) 2 + 𝑡𝑎𝑛−33
6
15.The solution of X𝑑𝑌 + Y =𝑋 4 with the condition y(1) =5
𝑋4
1
a) y= 5 +𝑋
4𝑋 4
5
b) y=
4
+5𝑋
𝑋4
c) y= 5 +1
16.The value of the contour integral ∮|𝑧−𝑖|=2
1
𝑧2
+4
𝑋5
d) y= 5 +1
dZ in positive sense
(z-i)=2
√𝜋
2
a)
𝜋
2
b)-
c)−
√𝜋
2
d)
𝜋
2
17. An examination consists of two papers. Paper 1, Paper 2. The probability of failing in Paper 1 is
0.3 and that in Paper 2 is 0.2. given that a student has failed in Paper 2, the probability of failing in
paper 1 is 0.6. the probability of a student failing in both the paper is
(a) 0.5
(b) 0.18
(c) 0.12
(d) 0.06
18. Let x be a random variable following normal distribution with mean +1 and variance 4. Let Y be
another normal variable with -1 and variable unknown. If P(X≤ −1) = 𝑃(𝑌 ≥ 2), the standard
deviation of Y is
(a) 3
(b) 2
(c) √2
(d) 1
19. A root of the equation 𝑥 3 − 𝑥 − 11 = 0 correct to four decimals using bisection method, is
(a) 2.3737
(b) 2.3838
(c) 2.3736
(d) None of these
20. Laplace transform for the function f(x)=cosh(ax) is
𝑎
𝑠
(a) 𝑠2 −𝑎2
(b) 𝑠2 −𝑎2
𝑎
(c) 𝑠2 +𝑎2
𝑎
(d) 𝑠2 +𝑎2
EXPLANATION:
11.(b)
12.(c)
13.(c)
Linear, because the order of convergence of bisection method is ½
14.(c)
15.(d)
16.(a)
17.(d)
It is an improper integral of second king and its integral also (unbounded) at one point in
[0,1].
18.(c)
∞
 ∫0
𝑑𝑥
(𝑥+3)2 +1
−1
 [tan (𝑥 + 3)]∞
0
𝜋
−1
 2 −tan 3
19.(c)
Using Echelon method
Applying R→
𝑅3
2
𝑎𝑛𝑑 𝑅1 ↔ 𝑅3 ,
(𝑅2 → 𝑅2 − 6𝑅1 , 𝑅3 → 𝑅3 − 4𝑅1 )
(𝑅3 → 𝑅3 −
𝑅2
4
)
Now, by Echelon method,
So, rank of [A] = number of non-zero rows = 2
20.(a)
If we put a=2 and b=7, then rank of [A:B]= Rank of A.
NETWORKS:
21.For a given voltage, four heating coils will produce maximum heat, when connected
(a) All in parallel
(b) all in series (c) with two parallel pairs in series
(d) one pair in parallel with the other two in series
22.The energy stored in the magnetic field at a solenoid 30cm long and 3 cm diameter wound
with 1000 turns of wire carrying a current of 10 A is
(a) 0.015J
(b) 0.15J
(c) 0.5J
(d)1.15J
23.Current I(t) = 2𝑒 −2(𝑡−3) U(t-3) A, for t>0 and voltage V(t) is given as 2𝛿(𝑡 − 3), then the
circuit components is /are
(a) R and C
(b) R and L
(c) L and C
(d) Only R
24.Given two coupled inductors 𝐿1 𝑎𝑛𝑑 𝐿2 their mutual inductance M satisfies
(a) M=√𝐿21 + 𝐿22
(b) M>
(𝐿1 +𝐿2 )
2
(c) M> √𝐿1 𝐿2
(d) M≤ √𝐿1 𝐿2
25.The current through an electrical conductor is 1 A when the temperature of the conductor
0℃ and 0.7A when the temperature is 100℃. The current when the temperature of the
conductor is 1200℃ must be
(a) 0.08A
(b) 0.16A
(c) 0.32A
(d) 0.64A
26.The resonant frequency of the given series circuit is
(a)
(b)
(c)
(d)
1
2𝜋√3
1
Hz
Hz
4𝜋√3
1
4𝜋√2
1
M=1H
Hz
2H
Hz
𝜋√2
2H
2F
27.A circuit which resonates at 1MHz has a Q of 100. Bandwidth between half power point is
(a) 10 kHz
(b) 100 kHz
(c) 10 Hz
(d) 100 Hz
2Ω
2Ω
28.Find 𝐼𝑁 and 𝑅𝑁 in the figure given below
(a)
(b)
(c)
(d)
3A, 10/3 Ω
10 A, 4Ω
1.5A, 6Ω
1.5 A, 4Ω
+
15V
4Ω
𝐼𝑁 , 𝑅𝑁
29.How many 200 W/220 V incandescent lamps connected in series would consume the same
total power as a single 100 W/220 V incandescent lamp?
(a) Not possible
(b) 4
(c) 3
(d) 2
EXPLANATION:
21.(a)
𝑅𝑒𝑞 = 𝑅 || R || R || R
= R/4
Maximum heat produced
P=
𝑉2
𝑅𝑒𝑞
=
𝑉2
𝑅
4
=
4𝑉 2
𝑅
( )
So, for a given parallel network 𝑅𝑒𝑞 is minimum and hence, maximum heat produced
because
1
P𝛼 𝑅
𝑒𝑞
22.(b)
L=
𝑁 2 𝜇0 𝐴
𝑙
After substituting and simplified we get,
=
9𝜋2 𝑥 10−5
0.3
Energy E = ½ L𝐼 2
=½x
9𝜋2 𝑥 10−5
0.3
x (10)2
= 0.148 or 0.15 J
23.(b)
Given, I(t) = 2𝑒 −2(𝑡−3) 𝑈(𝑡 − 3)
Taking Laplace transform of I(t)
I(s) =
2𝑒 −3𝑠
𝑠+2
sI(s) + 2I(s) = 2𝑒 −3𝑠
Taking inverse Laplace transform, we have
2I(t) +
𝑑𝐼(𝑡)
𝑑𝑡
= 2𝛿(−3)
Given V(t) = 2𝛿(−3)
2I(t) +
𝑑𝐼(𝑡)
𝑑𝑡
= 𝑉(𝑡)
V(t) = RI(t) + L
𝑑𝐼(𝑡)
𝑑𝑡
R = 2Ω , L = 1 H
So, the circuit components are R and L.
24. (d)
M = K√𝐿1 𝐿2
𝑀
K = √𝐿
1 𝐿2
K≤ 1

𝑀
≤1
√𝐿1 𝐿2
 𝑀 ≤ √𝐿1 𝐿2
25.(b)Let 𝛼 be the temperature coefficient, then
 𝑅𝑡 = 𝑅0 (1 + 𝛼𝑡)
When 𝑅𝑡 is resistance at t℃ and 𝑅0 is resistance at 0℃.
Assuming that for a given potential difference,
I𝛼

1
𝑅
1
𝑅0 (1+𝛼𝑡)
=
0.7
𝑅0
 𝛼 = 0.0043 𝑝𝑒𝑟 ℃
Current I at 1200℃ is given by

1
𝐼
=
𝑅0 (1+ 𝛼𝑡)
𝑅0
 1 + 𝛼. 1200
 I = 0.16A
26.(b)
Given,
 𝐿𝑒𝑞 = 𝐿1 + 𝐿2 + 2𝑀
 2+2+2x1
At resonance,
1
 𝜔𝐿 = 𝜔𝐶
 𝜔=
 𝐼=
1
√𝐿𝑒𝑞 𝐶
1
4𝜋√3
=
1
√12
Hz
27.(a)
𝑓
Q = ∆𝑓
 ∆𝑓 =
𝑓
=
𝑄
106
100
 10 𝑘𝐻𝑧
28.(a)
 𝑅𝑒𝑞 = 𝑅𝑁 = (2||4) + 2
 𝑅𝑁 =
10
3
(15⁄1)
1
1
2)+ ( ⁄2)+ ( ⁄4)
𝑉
𝐼𝑁 = 21 = 3𝐴
 𝑉1 = (1⁄
 𝐼𝑠𝑐 =
Ω
= 6V
29.(d)
In series power = 1/P
1

=
𝑃
1

200
1
+
𝑃1
1
𝑃2
1
+ 200
 P = 200/2
= 100W
2 lamps should be connected in series
SIGNALS AND SYS:
30.U[n] + u[-n] is equal to
(a) 2
(b) 1+𝛿[𝑛]
31.The Laplace transform of signal sin 5t is
(a)
5
(b)
𝑠 2 +5
(c) 2+ 𝛿[n]
𝑠
(c)
𝑠 2 +5
5
𝑒 −3(2−𝑗𝜔)
(b)
2−𝑗𝜔
𝑒 −3(2+𝑗𝜔)
2+𝑗𝜔
EXPLANATION:
30.(b)
31.(c)
∞ 𝑒 𝑗5𝑡−𝑒 −𝑗5𝑡
X(s) = ∫0
=
2𝑗
𝑒 −𝑠𝑡 𝑑𝑡
5
𝑠 3 +25
32.(d)
−𝑛
X(z) = ∑∞
= 𝑧 −𝑘 , 𝑧 ≠ 0
𝑛=−∞ 𝑥[𝑛]𝑧
33.(a)
∞
X(j𝜔) = ∫−∞ 𝑒 −4|𝑡| 𝑒 −𝑗𝜔𝑡 𝑑𝑡
 𝑒 −4|𝑡| = 𝑒 −4𝑡 , 𝑡 > 0
 𝑒 4|𝑡| = 𝑒 4𝑡 , 𝑡 < 0
∞
 ∫−∞ 𝑒 −4𝑡 𝑒 −𝑗𝜔𝑡 + ∫0 𝑒 −4𝑡 𝑒 −𝑗𝜔𝑡 𝑑𝑡

8
16+𝜔2
(c)
(d)
𝑠 2 +25
32.The z-transform of 𝛿[𝑛 + 𝑘], 𝑘 > 0 𝑖𝑠
(a) 𝑧 −𝑘 , 𝑧 = 0
(b) 𝑧 𝑘 , 𝑧 ≠ 0
(c) 𝑧 −𝑘 , 𝑎𝑙𝑙 𝑧
33.The Fourier transform of signal 𝑒 −2𝑡 𝑢(𝑡 − 3) is
(a)
(d) 1
𝑒 3(2−𝑗𝜔)
2−𝑗𝜔
𝑠
𝑠 2 +25
(d) 𝑧 𝑘 , 𝑎𝑙𝑙 𝑧
(d)
𝑒 3(2+𝑗𝜔)
2+𝑗𝜔
Control system:
34.The transmittance from X to Y in the signal flow graph shown in the figure is
(a) 4/3
1
(b) 1
(c) 2/3
-2
X
1
Y
(d) 1/3
35.State transition matrix property is
(a) ∅(𝑡 − 𝑡0 ) = ∅(𝑡). ∅(𝑡0 )
(b) ∅(𝑡 − 𝑡0 ) = ∅(𝑡). ∅−1 (𝑡0 )
(𝑐) ∅(𝑡 − 𝑡0 ) = ∅−1 (𝑡). ∅(𝑡0 )
(d) ∅(𝑡 − 𝑡0 ) = ∅−1 (𝑡). ∅−1 (𝑡0)
7
3
1
36.The step response of a given system is y=1-3 𝑒 −𝑡 + 2 𝑒 −2𝑡 − 6 𝑒 −4𝑡 the transform
function of the system is0
(a) (s+8)/(s+1)(s+2)(s+4)
(b) (s+6)/(s+1)(s+2)(s+4)
(c) s/(s+1)(s+2)(s+4)
(d) none of the above
37.A linear system is described by the following state equation:
0 1
X(t) = AX(t) + BU(t), A=[
] The state-transaction matrix of the system is
−1 0
cos 𝑡 sin 𝑡
]
− sin 𝑡 cos 𝑡
(c) [ – cos 𝑡 − sin 𝑡]
− sin 𝑡 cos 𝑡
sin 𝑡 ]
(b) [ – cos 𝑡
− sin 𝑡 − cos 𝑡
cos 𝑡 − sin 𝑡
(d) [
]
sin 𝑡 cos 𝑡
(a) [
38.The transform function of certain system is given as T(s) = 1/(s+2)(s+200). The 2%
settling time for unit step is
(a) 4 s
(b) 2 s
(c) 8 s
(d) 1 s
Explanation:
34.(c)
Y=2x+(-2y)
3y = 2x
y/x = 2/3
35.(b)
36.(a)
Since, the derivative of step is impulse.
The impulse response of the system is P(t)
7
2
P(t) = dy/dt = 3 𝑒 −𝑡 − 3𝑒 −2𝑡 + 3 𝑒 −4𝑡
P(s) =
=
7⁄3
−
𝑠+1
𝑠+8
3
𝑠+2
(𝑠+1)(𝑠+2)(𝑠+4)
+
2⁄3
𝑠+4
37. (a)
 ∅(𝑡) = 𝐿−1 [𝑠𝐼 − 𝐴]−1
𝑠 0
0 1 −1
 𝐿−1 [[
]−[
]]
0 𝑠
−1 0
𝑠 −1 −1
 𝐿−1 [
]
−1 𝑠
 𝑎𝑝𝑝𝑙𝑦 𝐷 = 𝑠 2 + 1
cos 𝑡 sin 𝑡
[
]
− sin 𝑡 cos 𝑡
38.(b)
Pole at 200 does not affect more hence, dominant pole is at 𝜔 = 2
So, H(s) = 1/s+2
Analog ckts:
39.The circuit inside the box in figure shown below contains only resistor and diodes. The
terminal voltage 𝑉0 is connected to some point in the circuit inside the box. The largest
and smallest possible values of 𝑉0 most nearly to, are respectively.
+15 V
(a) 15V, 6V
Circuit containing
(b) 24 V, zero
𝑉0
Diode and resistor
-9V
(c) 24 v, 6V
(d) 15V, -9V
40.In a CB amplifier, the maximum efficiency could be
(a) 25%
(b) 85%
(c) 99%
(d) 50%
41. The expression for the output voltage 𝑉0 in terms of the input
Voltages 𝑉1 and 𝑉2 in the circuit shown below, assuming the
Operational amplifier to be ideal is
𝑉0 = 𝐴1 𝑉1 + 𝐴2 𝑉2. The values of 𝐴1 𝑎𝑛𝑑 𝐴2 100 k
Would be respectively
10 k
(a) -9 and 10
𝑉2
-(b) 9.9 and -10
𝑉0
+
+
𝑉1
(c) 9 and -10
11 k
99 k
(d) -9.9 and 10
+5 V
42. In the circuit of figure, 𝑉𝐵 = −1 𝑉.
Find the value of β
(a) 103.4
600 kΩ
(b) 135.5
(c) 108.5
(d) 102.4
6 kΩ
-5 V
𝑉
43. A diode whose terminal characteristics are related as 𝐼𝐷 = 𝐼𝑆 ( ), where 𝐼𝑆 is the reverse
𝑉𝑇
saturation current and 𝑉𝑇 is the thermal voltage (V=25mV), is biased at 𝐼𝐷 = 2𝑚𝐴. its
dynamic resistance is
(a) 25Ω
(b) 12.5Ω
(c) 50Ω
(d) 100Ω
Explanation:
39.(d)
The output voltage cannot exceed the positive power supply voltage and cannot be
lower than the negative power supply voltage.
40.(c)
41.(b)
42.(c)
 𝑉𝐵 = −𝐼𝐵 𝑅𝐵
 𝐼𝐵 = 1.67𝜇𝐴
 𝐼𝐸 = −1.7𝑉

𝐼𝐸
𝐼𝐵
= (𝛽 + 1) =
0.271𝑚𝐴
β = 108.5
2𝜇
43.(b)

𝜕𝑉
𝜕𝐼𝐷
=
25𝑥10−3
2𝑥10−3
= 12.5 Ω
Digital:
44. A logical expression in the Sum Of Product (SOP) is suitable for implementation using
(a) AND gates
(b) NOR gates
(c) NAND gates
(d) EX-OR
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
45. The logic function f = (𝑥. 𝑦̅) + (𝑥̅ . 𝑦) is the same as
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
(a) f=(x+y)(𝑥̅ + 𝑦̅)
(b) f = (𝑥̅
+ 𝑦̅) + (𝑥 + 𝑦)
̅̅̅̅̅̅̅
(c) f= (𝑥, 𝑦). (𝑥̅ , 𝑦̅)
(d) None of the above
46.If the operating frequency of 8085 is2 MHz, find the time required to execute MOV A, M.
(a) 3µs
(b) 3.5µs
(c) 4µs
(d) 4.5µs
Explanation:
44.(c)
SOP is suitable for implementation using OR gates or NAND gates
45.(d)
Putting x=1, y=1
In given functions and all other options, then check result
46.(b)
Clock frequency f = 2 MHz
T-state = clock period (1/f) = 0.5µs
T-states for MOV A, M = 7
Time for instruction: (7T) x 0.5 = 3.5µs
Ps:
47.The formula for specific speed 𝑁𝑠 of turbine is
𝑁(𝑃)2
(a) 𝑁𝑠 =
(b) 𝑁𝑠 =
𝐻 3/2
𝑁(𝑃)1/2
𝐻 3/4
(c) 𝑁𝑠 =
𝑁(𝑃)1/2
𝐻 5/4
(d)𝑁𝑠 =
𝑁√𝑃
𝐻
48.A 100 km long transmission line is loaded at 110 kV. If the loss of line is 5 MW and the
load is 150 MVA, the resistance of the line is
(a) 0.806 Ω/phase (b) 8.06 Ω/phase
(c) 0.0806Ω/phase
(d) 80.6Ω/phase
49.When a line to ground fault occurs, the current in a faulted phase is 100A. The zero
sequence current in this case will be
(a) Zero
(b) 33.3 A
(c) 66.6 A
(d) 100A
50.If the inductance and capacitance of a system are 1.0 H and 0.01 µF respectively and the
instantaneous value of current interrupted is 10 A, the voltage across the breaker contacts will
be
(a) 50 kV
(b) 100 kV
(c) 60 kV
(d) 57 kV
51.A Mho relay is a
(a) Voltage restrained directional relay
(b) Voltage controlled over current relay
(c) Directional restrained over current relay
(d) Directional restrained over voltage relay
52.Shunt capacitors are used to raise the power factor of the load of 100 kW from 0.8 lag to
unity, supply being 3-phase at 10kV. In star bank, the capacitance will be
(a) 0.8µF
(b) 2.4 µF
(c) 1.6µF
(d) 3.2µF
Explanation:
47.(c)
 𝑁𝑠 =
𝑁 √𝑃
𝐻 1.25
=
𝑁(𝑃)1/2
𝐻 5/4
48.(d)
Since, 𝑃𝐿 = 𝐼𝐿2 𝑅 = 𝑉𝐿 𝐼𝐿
 𝐼𝐿 =
𝑃𝐿

𝑉𝐿
150𝑋106
110𝑋103
=
150
110
3
𝑥 103
 1.363 x 10 A
 R = 80.6 Ω/ phase
49.(b)
 𝐼𝑎0 =
𝐼𝑓
3
= 33.3𝐴
50.(b)
100
51.(a)
A mho relay is a voltage restrained direction relay
52.(b)
3
kVAR = 4 𝑥 100000 = 75000
 𝐶𝑠 =
75000
2𝜋𝑥50𝑥(10000)2
 2.4 𝜇𝐹
Machines:
53.A 400 V/100 V, 10 kVA two-winding transformer is reconnected as an autotransformer across a suitable voltage source. The maximum rating of such an
arrangement could be
(a) 50 kVA
(b) 15 kVA
(c) 12.5 kVA
(d) 8.75 kVA
54.A 4-pole DC generator is running at 1500 rpm. The frequency of current in the
armature winding will be
(a) 25 Hz
(b) 50 Hz
(c) 100 Hz
(d) 200 Hz
55.Slip test is performed to determine
(a) Slip
(b) Direct axis reactance and quadrature axis
(c) Positive sequence reactance and negative sequence reactance
(d) Sub-transient reactance
56.A synchronous motor with negligible armature resistance runs at load angle of 20° at
the rated frequency. If supply frequency is increased by 10%, keeping other parameter
constant, the new load angle will be
(a) 12°
(b) 18°
(c) 20°
(d) 22°
57.A 440V shunt motor has an armature resistance of 0.5Ω and shunt field resistance of
650Ω. If the no load current is 3 A, then current in the armature will be
(a) 2.32 A
(b) 3A
(c) 0.68 A
(d) 880A
58.Two transformers operating in parallel will share the load according to their
(a) Leakage reactance
(b) pu impedance
(c) Efficiency
(d) rating
Explain:
53.(a)
Primary current 𝐼1 =
10𝑥103
Secondary current 𝐼2 =
= 25 𝐴
400
10𝑥103
100
= 100 𝐴
Total current I = 𝐼1 + 𝐼2 = 125𝐴
Maximum kVA rating =
125𝑥400
1000
= 50𝑘𝑉𝐴
54.(b)
Frequency of current in the armature winding

55.(b)
56.(c)
𝑃𝑁𝑠
120
=
4𝑥1500
120
= 50 𝐻𝑧
57.(a)
 𝐼𝑠ℎ =
440
= 0.677 𝐴
650
 𝐼𝑎 = 𝐼𝐿 − 𝐼𝑠ℎ = 3 − 0.677
 2.32A
58.(d)
Measurements:
59.A resistance of 10 kΩ with 5% tolerance is connected in series with 5 kΩ resistor of
10% tolerance. What is the tolerance limit for the series network?
(a) 5%
(b) 6.67%
(c) 10%
(d) 8.33%
60.A slide-wire is used for measurement of current in the circuit. The voltage drop across
standard resistor of 0.2Ω is balance at 83cm. Find the magnitude of current, if the
standard cell emf of 1.53 V is balanced at 42 m.
(a) 10 A
(b) 12.56 A
(c) 13.04 A
(d) 14.95 A
61.A 50Hz voltage is measured with a moving iron voltmeter and a rectifier type AC
voltmeter connected in parallel. If the meter readings are 𝑉1 𝑎𝑛𝑑 𝑉2 respectively and the
meters are free from calibration errors, then the form factor of the AC voltage may be
estimated as
(a)
𝑉1
(b)
𝑉2
1.11𝑉1
(c)
𝑉2
√2𝑉1
𝑉2
(d)
𝜋𝑉1
𝑉2
Explanation:
59.(b)
Error in 10 kΩ resistance = 10x
5
100
10
= 0.5 𝑘
Error in 5 kΩ resistance = 5x100 = 5𝑘
Total measurement resistance = 10+0.5+5+0.5 = 16 kΩ
Original resistance = 10+5 = 15 kΩ
Error =
16−15
15
1
𝑥100 = 15 𝑥100 = 6.67%
60.(d)
Voltage drop per unit length = 1.53/42 = 0.036 V/cm
Voltage drop across 83 cm length = 0.036x83=2.99V
Current through resistor
I= 2.99/0.2 = 14.95 A
61.(b)
From factor of the wave = RMS value/ Mean value
Moving iron instrument will show rms value. Rectifier voltmeter is calibrated to read
rms value of sinusoidal voltage i.e., with form factor of 1.11
𝑉
2
Mean value of the applied voltage = 1.11
From factor =
Pe:
𝑉1
𝑉2 /1.11
=
1.11 𝑉1
𝑉2
62. A fully controlled three-phase converter has to supply an output voltage of 270 V. The
minimum secondary line voltage of the input transformer is
(a) 490V
(b) 280V
(c) 350V
(d) none of these
63. A single-phase inverter has square wave output voltage. What is the percentage of the
fifth harmonic component in relation to the fundamental component?
(a) 40%
(b) 30%
(c) 20%
(d) 10%
64. A three-phase diode bridge rectifier is fed from a 400V rms, 50 Hz, three-phase AC
source. If the load is purely resistive, the peak instantaneous output voltage is equal to
(a) 400V
(b) 400√2𝑉
2
(c) 400 √3 𝑉
(d)
400
√3
𝑉
65. A step – down chopper is operated in the continuous conduction mode in steady state
with a constant duty ratio D. If 𝑉0 is the magnitude of the DC output voltage and if 𝑉𝑠 is the
𝑉
magnitude of the DC input voltage, the ratio 𝑉0 is given by
𝑠
(a) D
(b) 1-D
(c) 1/1-D
(d) D/1-D
Explanation:
62.(a)
 𝑉𝑚𝑒𝑎𝑛 =
 𝑉𝑙𝑖𝑛𝑒 =
√3
𝑉
𝜋 𝑙𝑖𝑛𝑒
270 𝑥 𝜋
√3
= 490 𝑉
63.(c)
 𝑉05 =
𝑉01
5
 𝑉05 = 20% 𝑜𝑓 𝑉01
64.(b)
Since, load is purely resistive, therefore peak instantaneous voltage.
 𝑉0 = √2𝑉𝑟𝑚𝑠 = 400√2 𝑉
65.(a)