For Questions 17 to 19 refer to the data given
below:
DC supply of 200 V is applied to a series RL network where R = 20 Ω and
L = 0.5 H
(17) At the instant of switching, the rate of change of current will
be
(A) 400 A/sec
(B) 300 A/sec
(C) 200 A/sec
(D) 100 A/sec
ANSWER: 400 A/sec
(18) At t = L/R, the rate of change of current will be nearly
(A) 450 A/sec
(B) 350 A/sec
(C) 250 A/sec
(D) 150 A/sec
ANSWER: 150 A/sec
(19) The steady value of current will be
(A) 1 A
(B) 10 A
(C) 100 A
(D) 20 A
ANSWER: 10 A
(20) Let R and L are in parallel connection. ω = 1000, Φ = +45° and
|Z| = 50√2.
The value of R and L will be
(A) R =1Ω, L = 10H
(B) R =10Ω, L = 1H
(C) R =100Ω, L = 0.1H
(D) R =1000Ω, L = 0.01H
ANSWER: R =100Ω, L = 0.1H
For Questions 21 to 23 refer to the figure given below
(21) The value of current supplied by the battery is
(A) 11.7 A
(B) 11.7 mA
(C) 1.17 A
(D) 1.17 mA
ANSWER: 11.7 mA
(22) The voltage developed across 20 K resistor will be
(A) 140 V
(B) 115 V
(C) 120 V
(D) 100 V
ANSWER: 120 V
(23) The current through the cross branch 10 K resistor will be
(A) 34 mA
(B) 3.4 mA
(C) 340 mA
(D) 40 mA
ANSWER: 3.4 mA
For Questions 24 to 26 refer to the data given below:
Inherent resistance of an inductor is 50O and a time constant of 1.25
seconds. The inductor is now rewound with the same weight of wire of
half the radius.
(24) Inductance value will be
(A) 1000 H
(B) 100 H
(C) 10 H
(D) 1 H
ANSWER: 1000 H
(25) The new constant will be
(A) More than 1.25 sec
(B) Less than 1.25 sec
(C) 1.25 sec
(D) Information provided is not sufficient
ANSWER: 1.25 sec
(26) When the inductor is connected to 600 V dc source, the stored
energy will be nearly
(A) 200 J
(B) 281 J
(C) 140 J
(D) 100 J
ANSWER: 281 J
(27) Two inductors when connected in parallel, the equivalent
inductance is 4 H. When they are connected in series, the equivalent
inductance is 20 H. The value of individual inductance should be
(A) 10 H and 10 H
(B) 8 H and 12 H
(C) 6.5 H and 12 H
(D) 4 H and 16 H
ANSWER: 4 H and 16 H
(28) What will be the voltage needed across pq so that the voltage
drop across the 15 ohm resistor is 45 volts.
(A) 150 V
(B) 200 V
(C) 100 V
(D) 50 V
ANSWER: 150 V
(29) The open circuit I-V characteristic is
(A) A horizontal line through the origin
(B) A horizontal line above the origin
(C) A vertical line away from origin
(D) A vertical line through origin
ANSWER: A vertical line through origin
(30) The condition of a linear resistor, 0 < R < infinity
represents
(A) Voltage controlled resistor
(B) Current controlled resistor
(C) Neither voltage controlled nor current controlled resistor
(D) Both voltage controlled and current controlled resistor
ANSWER: Both voltage controlled and current controlled
resistor
(31) A non-linear resistor is represented as
(A) V + 10 i = 0
(B) 3V + i = 10
(C) V = i²
(D) All of the above
ANSWER: V = i²
(32) The super-position theorem is applicable to
(A) Linear and non linear responses only
(B) Linear responses only
(C) Linear, non linear and time variant responses
(D) None of the above
ANSWER: Linear responses only
(33) When one resistor of a delta-connected circuit is open, power
will be
(A) Reduced by 1/3
(B) Zero
(C) Reduced to 1/3
(D) Unaltered
ANSWER: Reduced by 1/3
(34) In 3 phase star connected network, line voltage is same as
(A) 3 phase voltage
(B) √3phase voltage
(C) 1/√3 phase voltage
(D) Phase voltage
ANSWER: 1/√3 phase voltage
(35) Consider the Norton equivalent circuit. When a 1 ohm resistor
is connected across the input terminals, equivalent resistor,
equivalent voltage and current will be
(A) 2/3 ohm, -10 V, -6 A
(B) 2/3 ohm, 10 V, -6 A
(C) 2/3 ohm, -5 V, 6 A
(D) 2/3 ohm, 5 V, 6 A
ANSWER: 2/3 ohm, 10 V, -6 A
(36) Millman’s theorem yields
(A) Equivalent impedance
(B) Equivalent resistance
(C) Equivalent voltage source
(D) Equivalent voltage or current source
ANSWER: Equivalent voltage or current source
For Questions 37 to 39 refer to the figure shown below:
(37) The impedance Za of the star connection will be
(A) (1 + j 5.1) Ω
(B) (0.3 + j 7.4) Ω
(C) (1.5 + j 8.1) Ω
(D) (2 + j 10.5) Ω
ANSWER: (0.3 + j 7.4) Ω
(38) The impedance Zb of the star connection will be
(A) (13 + j 13.5) Ω
(B) (0.3 + j 7.4) Ω
(C) (1 + j 7) Ω
(D) (18 + j 3.5) Ω
ANSWER: (13 + j 13.5) Ω
(39) The impedance Zc of the star connection will be
(A) (8 + j 12) Ω
(B) (1 + j 1.8) Ω
(C) (24 + j 13.8) Ω
(D) (8 + j 1.8) Ω
ANSWER: (8 + j 1.8) Ω
(40) To solve a network, the number of independent equations is
equal to
(A) The number of chords
(B) The number of branches
(C) Sum of the number of chords and branches
(D) Sum of the number of chords, branches and nodes
ANSWER: The number of chords
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