All questions of Subject-wise Tests for Electrical Engineering (EE) Exam
A1 ϕ 33 kVA, 3300 V/330 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 3630 V. It maximum kVA rating as an auto-transformer is________kVA
Correct answer is '363'. Can you explain this answer?
A1 ϕ 33 kVA, 3300 V/330 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 3630 V. It maximum kVA rating as an auto-transformer is________kVA
|
Alok Roy answered |
Maximum kVA rating = V.A = VHIH = 3630 × 100 = 363 KVA
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.
Correct answer is between '18,20'. Can you explain this answer?
A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm, and drawing 20 A current. If a resistance of 75 Ω is added in series with the field, find the new armature current (in A). Assume load torque and flux and constant.
|
Sanvi Kapoor answered |
The field current corresponds to:
Now:
IA = IL - If = 20 – 1.66 = 18.34 Amps
Observe that the current of the field circuit is much smaller than the armature current. Thus, adding the speed control resistance leads to minor power dissipation.
Now:
IA = IL - If = 20 – 1.66 = 18.34 Amps
Observe that the current of the field circuit is much smaller than the armature current. Thus, adding the speed control resistance leads to minor power dissipation.
Ea = VS – IaRa = 250 – 5.502 = 244.5 V
EA = K ϕ ω
The new armature current corresponds to:
Ia = IL – If = 20 – 1.11 = 18.89 Amps
The new armature current corresponds to:
Ia = IL – If = 20 – 1.11 = 18.89 Amps
A 4 kVA, 400/200V single-phase transformer has resistance of 0.02 p.u. and reactance of 0.06 p.u. Its actual resistance and reactance referred to high voltage side are respectively- a)0.2 ohm and 0.6 ohm
- b)0.8 ohm and 2.4 ohm
- c)0.08 ohm and 0.24 ohm
- d)2 ohm and 6 ohm
Correct answer is option 'B'. Can you explain this answer?
A 4 kVA, 400/200V single-phase transformer has resistance of 0.02 p.u. and reactance of 0.06 p.u. Its actual resistance and reactance referred to high voltage side are respectively
a)
0.2 ohm and 0.6 ohm
b)
0.8 ohm and 2.4 ohm
c)
0.08 ohm and 0.24 ohm
d)
2 ohm and 6 ohm
|
Bhavana Ahuja answered |
The current drawn by a 220V dc motor of armature resistance 0.5 Ω and back emf 200V is- a)40 A
- b)44 A
- c)400 A
- d)440 A
Correct answer is 'A'. Can you explain this answer?
The current drawn by a 220V dc motor of armature resistance 0.5 Ω and back emf 200V is
a)
40 A
b)
44 A
c)
400 A
d)
440 A
|
|
Dishani Shah answered |
Ea = V – Ia Ra
Ea = 200 = 220 – Ia × 0.5 ⇒ Ia = 40 A
Ea = 200 = 220 – Ia × 0.5 ⇒ Ia = 40 A
Two 10 kV/440V, 1-phase transformers of ratings 600 kVA and 350 kVA are connected in parallel to share a load of 800 kVA. The reactances of the transformers, referred to the secondary side are 0.0198Ω and 0.0304Ω respectively (resistances negligible). The load shared by the two transformers will be respectively- a)484.5 kVA and 315.5 kVA
- b)315.5 kVA and 484.5 kVA
- c)533 kVA and 267 kVA
- d)267 kVA and 533 kVA
Correct answer is option 'A'. Can you explain this answer?
Two 10 kV/440V, 1-phase transformers of ratings 600 kVA and 350 kVA are connected in parallel to share a load of 800 kVA. The reactances of the transformers, referred to the secondary side are 0.0198Ω and 0.0304Ω respectively (resistances negligible). The load shared by the two transformers will be respectively
a)
484.5 kVA and 315.5 kVA
b)
315.5 kVA and 484.5 kVA
c)
533 kVA and 267 kVA
d)
267 kVA and 533 kVA
|
Harshitha Kulkarni answered |
Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a- a)parallel resistance with a low value
- b)series resistance with a low value
- c)parallel resistance with a high value
- d)series resistance with a high value
Correct answer is option 'A'. Can you explain this answer?
Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a
a)
parallel resistance with a low value
b)
series resistance with a low value
c)
parallel resistance with a high value
d)
series resistance with a high value
|
Mayank Sengupta answered |
No-load losses (essentially core losses not F & W losses) are low and depend upon the applied voltage in an induction machine.
The reactance offered by a capacitor to alternating current of frequency 50 Hz is 20 Q. If frequency is increased to 100 Hz, reactance becomes_____ohms.- a)2.5
- b)5
- c)10
- d)15
Correct answer is option 'C'. Can you explain this answer?
The reactance offered by a capacitor to alternating current of frequency 50 Hz is 20 Q. If frequency is increased to 100 Hz, reactance becomes_____ohms.
a)
2.5
b)
5
c)
10
d)
15
|
Baishali Bajaj answered |
Capacitive reactance
Xc = 1/2πf
Therefore XC1/XC2 = 2πf2/2πf1
= 20/XC2 =100/50
= XC2 = 10
The signal x(t) = t u(t) is a - a)Power signal with P∞ = 1
- b)Energy signal with E∞ = 0
- c)Power signal with P∞ = 0
- d)Neither an energy signal nor a power signal
Correct answer is option 'D'. Can you explain this answer?
The signal x(t) = t u(t) is a
a)
Power signal with P∞ = 1
b)
Energy signal with E∞ = 0
c)
Power signal with P∞ = 0
d)
Neither an energy signal nor a power signal
|
|
Raj Chaudhary answered |
The signal x(t) = t u(t) is Neither an energy signal nor a power signal.
If the signal energy in one period is infinite, then both the power and the total energy are infinite. Consequently, the signal is neither an energy signal nor a power signal. ( )ti . the signal energy in and the power loss is the same as the power of the periodic signal .
A 220 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hour at unity power factor. The meter disc makes 2376 revaluation during this period. If the number of revolution made by the meter are 1542 when operating at 220 V and 6 A for 4 hours, then power factor of the load would be
Correct answer is between '0.62,0.68'. Can you explain this answer?
A 220 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hour at unity power factor. The meter disc makes 2376 revaluation during this period. If the number of revolution made by the meter are 1542 when operating at 220 V and 6 A for 4 hours, then power factor of the load would be
|
Rahul Chakraborty answered |
Energy meter constant
Energy consumed when the meter make 1542 revaluations
Now energy consumed
A DC shunt generator delivers 50 A ot terminal voltage of 230 V. The armature and the shunt field resistance are 0.01 Ω and 45 Ω respectively. The stray losses are 400 W. The percentage efficiency of DC generator is _________.
Correct answer is between '78,88'. Can you explain this answer?
A DC shunt generator delivers 50 A ot terminal voltage of 230 V. The armature and the shunt field resistance are 0.01 Ω and 45 Ω respectively. The stray losses are 400 W. The percentage efficiency of DC generator is _________.
|
Hridoy Chakraborty answered |
Field current If = 23045 = 5.11A
Armature current Ia = If + IL = 5.11 + 50 = 55.11 A
Stray loss = 400 W
Total loss = 400 + 30.37 + 1175.55 = 1605.92 W
Output power = 230 × 50 = 11500
Armature current Ia = If + IL = 5.11 + 50 = 55.11 A
Stray loss = 400 W
Total loss = 400 + 30.37 + 1175.55 = 1605.92 W
Output power = 230 × 50 = 11500
In power transistor, during turn on process voltage and current change linearly from 400 V and 0 V and 0 A to 300 A respectively in 3 μS. the energy loss during turn on process in given by _____Joule (up to 2 decimal places)
Correct answer is '0.06'. Can you explain this answer?
In power transistor, during turn on process voltage and current change linearly from 400 V and 0 V and 0 A to 300 A respectively in 3 μS. the energy loss during turn on process in given by _____Joule (up to 2 decimal places)
|
Ankita Das answered |
Concept:
Change in V and L are shown below
Energy loss during turn on process =
Calculations:
Change in V and L are shown below
Energy loss during turn on process =
Calculations:
= 20 × 103 × 3 × 10-6
= 0.06 J
is solenoidal, which of these is true?
The emf induced in each coil side of a coil is 10 V and the coil sides lies under opposite poles. The induced emf in them is – (in V)
Correct answer is '20'. Can you explain this answer?
The emf induced in each coil side of a coil is 10 V and the coil sides lies under opposite poles. The induced emf in them is – (in V)
|
Rajesh Saha answered |
Since the coil sides lies under opposite poles the induced emf’s in them are additive. Hence it will be 20 V.
A three-phase transformer having zero-sequence impedance of Z0 has the zero-sequence network as shown in the figure below. The connections of its windings are
- a)star - star
- b)delta - delta
- c)star - delta
- d)delta - star with neutral grounded
Correct answer is option 'B'. Can you explain this answer?
A three-phase transformer having zero-sequence impedance of Z0 has the zero-sequence network as shown in the figure below. The connections of its windings are
a)
star - star
b)
delta - delta
c)
star - delta
d)
delta - star with neutral grounded
|
Prasad Verma answered |
The total number of transmission lines in a power system can be determined by the sparsity of the Ybus matrix. The Ybus matrix represents the admittance of the power system and is used to solve power flow equations.
Sparsity refers to the percentage of zero elements in a matrix. In this case, the Ybus matrix has 80% sparsity, which means that 80% of the elements in the matrix are zero.
To calculate the total number of transmission lines, we need to determine the size of the Ybus matrix and then subtract the number of zero elements.
Let's assume that the Ybus matrix is of size N x N, where N is the number of buses in the power system. Since the Ybus matrix is symmetric, we only need to consider the upper triangular part of the matrix.
The total number of elements in the Ybus matrix is N^2. However, since the matrix is symmetric, we only need to consider N(N+1)/2 elements.
The number of non-zero elements in the Ybus matrix can be calculated by multiplying the total number of elements by (1 - sparsity). In this case, the number of non-zero elements is N(N+1)/2 * (1 - 0.8).
To find the number of zero elements, we subtract the number of non-zero elements from the total number of elements.
Therefore, the number of zero elements in the Ybus matrix is N(N+1)/2 * 0.8.
To find the total number of transmission lines, we need to divide the number of zero elements by 2, since each transmission line corresponds to two elements in the Ybus matrix (one for each end).
Hence, the total number of transmission lines is N(N+1)/2 * 0.8 / 2.
Since the correct answer is option 'A' which is 225, we can conclude that the number of buses in the power system is 15 (N=15).
Sparsity refers to the percentage of zero elements in a matrix. In this case, the Ybus matrix has 80% sparsity, which means that 80% of the elements in the matrix are zero.
To calculate the total number of transmission lines, we need to determine the size of the Ybus matrix and then subtract the number of zero elements.
Let's assume that the Ybus matrix is of size N x N, where N is the number of buses in the power system. Since the Ybus matrix is symmetric, we only need to consider the upper triangular part of the matrix.
The total number of elements in the Ybus matrix is N^2. However, since the matrix is symmetric, we only need to consider N(N+1)/2 elements.
The number of non-zero elements in the Ybus matrix can be calculated by multiplying the total number of elements by (1 - sparsity). In this case, the number of non-zero elements is N(N+1)/2 * (1 - 0.8).
To find the number of zero elements, we subtract the number of non-zero elements from the total number of elements.
Therefore, the number of zero elements in the Ybus matrix is N(N+1)/2 * 0.8.
To find the total number of transmission lines, we need to divide the number of zero elements by 2, since each transmission line corresponds to two elements in the Ybus matrix (one for each end).
Hence, the total number of transmission lines is N(N+1)/2 * 0.8 / 2.
Since the correct answer is option 'A' which is 225, we can conclude that the number of buses in the power system is 15 (N=15).
Can you explain the answer of this question below:Consider the following statements:1. All loops are meshes2. All meshes are loops.Which of these statements is/are correct?
- A:
only 1
- B:
only 2
- C:
both
- D:
none
The answer is b.
Consider the following statements:1. All loops are meshes2. All meshes are loops.Which of these statements is/are correct?
only 1
only 2
both
none
|
Nakul Chauhan answered |
Explanation:
1. Understanding the statements:
- Statement 1: "All loops are meshes" means that every loop in a circuit is also a mesh.
- Statement 2: "All meshes are loops" means that every mesh in a circuit is also a loop.
2. Analysis of the statements:
- In a circuit analysis context, a loop is a closed path in a circuit where no node is encountered twice. It is a path that starts and ends at the same node.
- On the other hand, a mesh is a loop that does not contain any other loops within it.
- Therefore, while all meshes are loops (statement 2 is correct), not all loops are meshes. This is because a loop can contain multiple smaller loops within it, making it not just a mesh.
3. Conclusion:
- Based on the definitions of loops and meshes in circuit analysis, we can conclude that statement 2 is correct ("All meshes are loops"), while statement 1 is incorrect ("All loops are meshes"). Hence, the correct answer to the question is option 'B' - only 2.
1. Understanding the statements:
- Statement 1: "All loops are meshes" means that every loop in a circuit is also a mesh.
- Statement 2: "All meshes are loops" means that every mesh in a circuit is also a loop.
2. Analysis of the statements:
- In a circuit analysis context, a loop is a closed path in a circuit where no node is encountered twice. It is a path that starts and ends at the same node.
- On the other hand, a mesh is a loop that does not contain any other loops within it.
- Therefore, while all meshes are loops (statement 2 is correct), not all loops are meshes. This is because a loop can contain multiple smaller loops within it, making it not just a mesh.
3. Conclusion:
- Based on the definitions of loops and meshes in circuit analysis, we can conclude that statement 2 is correct ("All meshes are loops"), while statement 1 is incorrect ("All loops are meshes"). Hence, the correct answer to the question is option 'B' - only 2.
The value of zin for the circuit shown below is
- a)12Ω
- b)14Ω
- c)9.6Ω
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The value of z
in
for the circuit shown below isa)
12Ω
b)
14Ω
c)
9.6Ω
d)
None of these
|
Riverdale Learning Institute answered |
Net impedance on primary side of transformer is
The first derivatives of the signalx(t) = sgn t is- a)δ (t)
- b)2 δ (t)
- c)0
- d)-2 δ (t)
Correct answer is option 'B'. Can you explain this answer?
The first derivatives of the signalx(t) = sgn t is
a)
δ (t)
b)
2 δ (t)
c)
0
d)
-2 δ (t)
|
Suhasini K.t answered |
x(t)=sgn(t)=2u(t)-1
dx(t)/dt=2(1) (since value of delta fun is 1,right is the option)
The value of V, in the circuit shown below
Correct answer is '4.5'. Can you explain this answer?
The value of V, in the circuit shown below
|
Machine Experts answered |
V1— V2 =12V................. (1)
At node 0:
--------(2)
--------(2)At supernode 1 - 2
(3)
(3) Solving 1, 2 & 3; Vo = 4.5V; V1 = 5V; V2 = —7V
The reflection coefficient for the transmission line shown in figure below at point P is
- a)+ 1
- b)0.5
- c)0
- d)- 1
Correct answer is option 'C'. Can you explain this answer?
The reflection coefficient for the transmission line shown in figure below at point P is
a)
+ 1
b)
0.5
c)
0
d)
- 1
|
|
Manoj Mehra answered |
The line is terminated into its surge impedance so it is an infinite line.
How many segments will be there in for the commutator of a 8 pole dc machine having a double layered simple wave wound armature with 86 slots?- a)86
- b)72
- c)79
- d)84
Correct answer is option 'A'. Can you explain this answer?
How many segments will be there in for the commutator of a 8 pole dc machine having a double layered simple wave wound armature with 86 slots?
a)
86
b)
72
c)
79
d)
84
|
Alok Verma answered |
For a double layer winding (universally used) number of commutator segments = number of armature coils. The number of slots in double layer winding is equal to the number of coils and the number of commutator segments is also the same.
The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will be - a)Xd = 10.8 ohms and Xq = 8 ohms
- b)Xd = 9 ohms and Xq = 9.6 ohms
- c)Xd = 9.6 ohms and Xq = 9 ohms
- d)Xd = 8 ohms and Xq = 10.8 ohms
Correct answer is option 'A'. Can you explain this answer?
The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:
Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will be
a)
Xd = 10.8 ohms and Xq = 8 ohms
b)
Xd = 9 ohms and Xq = 9.6 ohms
c)
Xd = 9.6 ohms and Xq = 9 ohms
d)
Xd = 8 ohms and Xq = 10.8 ohms
|
|
Sanvi Kapoor answered |
In a 100 kVA, 1100/220V, 50 Hz single-phase transformer with 2000 turns on the high-voltage side, the open-circuit test result gives 200V, 91 A, 5kW on low-voltage side. The coreloss component of current is approximately- a)9.1 A
- b)22.7 A
- c)45.0 A
- d)91 A
Correct answer is 'B'. Can you explain this answer?
In a 100 kVA, 1100/220V, 50 Hz single-phase transformer with 2000 turns on the high-voltage side, the open-circuit test result gives 200V, 91 A, 5kW on low-voltage side. The coreloss component of current is approximately
a)
9.1 A
b)
22.7 A
c)
45.0 A
d)
91 A
|
Sparsh Unni answered |
, the co-efficient of term e-t in f(t) will be :
Consider the circuit shown in the below figure:
Q.The s.c. input admittance of the above network is- a)0.9375 ℧
- b)0.333 ℧
- c)0.375 ℧
- d)0.625 ℧
Correct answer is option 'D'. Can you explain this answer?
Consider the circuit shown in the below figure:
Q.
The s.c. input admittance of the above network is
a)
0.9375 ℧
b)
0.333 ℧
c)
0.375 ℧
d)
0.625 ℧
|
Nageswari Odogu answered |
The given circuit in parallel connection so in parallel admittance parameters can add therefore first find z parameters of upper portion and then find Y parameters by applying inverse of Z we will get Y11 of upper portion is 0.125 next find Y parameters below position we will get Y11 is 0.5
A 230 V, single phase watt hour meter has a constant load of 6 A passing through it for 8 hours at unity power factor. If the meter disc makes 2400 revolutions during this period. What is the power factor of the load if the number of revolutions made by the meter are 1800 when operating at 230 V and 8 A for 6 hours?
Correct answer is between '0.7,0.8'. Can you explain this answer?
A 230 V, single phase watt hour meter has a constant load of 6 A passing through it for 8 hours at unity power factor. If the meter disc makes 2400 revolutions during this period. What is the power factor of the load if the number of revolutions made by the meter are 1800 when operating at 230 V and 8 A for 6 hours?
|
|
Ashutosh Majumdar answered |
Energy supplied = VI cos ϕ × t × 10-3
= 230 × 6 × 8 × 10-3 = 11.04 kwh
Meter constant
Energy consumed when the meter takes 1800 revolutions
Now energy consumed = VI cos ϕ × t × 10-3
8.28 = 230 × cos ϕ × 8 × 6 × 10-3
⇒ cos ϕ = 0.75
= 230 × 6 × 8 × 10-3 = 11.04 kwh
Meter constant
Energy consumed when the meter takes 1800 revolutions
Now energy consumed = VI cos ϕ × t × 10-3
8.28 = 230 × cos ϕ × 8 × 6 × 10-3
⇒ cos ϕ = 0.75
Which one of the following relays has the capability of anticipating the possible major fault in a transformer?- a)Overcurrent relay
- b)Differential relay
- c)Buchholz relay
- d)Overfluxing relay
Correct answer is option 'C'. Can you explain this answer?
Which one of the following relays has the capability of anticipating the possible major fault in a transformer?
a)
Overcurrent relay
b)
Differential relay
c)
Buchholz relay
d)
Overfluxing relay
|
Gangadhar Rautray answered |
It is used to sense the fault such as short circuit faults such as inter turn faults, incipient winding faults, and core faults. it recognized fault by different gases release due to fault.
A 110/220 V, 60 Hz, single phase, 1-kVA transformer has leakage reactance of 4%. Calculate its total leakage inductance (in mH) referred to 110 V side.
Correct answer is between '1.25,1.35'. Can you explain this answer?
A 110/220 V, 60 Hz, single phase, 1-kVA transformer has leakage reactance of 4%. Calculate its total leakage inductance (in mH) referred to 110 V side.
|
Jaya Dasgupta answered |
Assuming the 110-V side 1 and the 220-V side to be side 2, N1/N2 = 0.5
Total leakage reactance referred to side 1
Total leakage reactance referred to side 1
Unit of inductance is ________.- a)Henry
- b)Weber
- c)Farad
- d)Tesla
Correct answer is option 'A'. Can you explain this answer?
Unit of inductance is ________.
a)
Henry
b)
Weber
c)
Farad
d)
Tesla
|
|
Kabir Verma answered |
The unit of inductance is Henry. Weber is the unit of magnetic flux. Tesla is the unit of flux density. Farad is the unit of capacitance.
A 3φ, Induction Motor had Tmax as 300 N-m with a supply of 440 V, 50 Hz. If the supply is changed to 220 V, 25 Hz new Tmax is- a)300 N-m
- b)150 N-m
- c)75 N-m
- d)600 N-m
Correct answer is option 'B'. Can you explain this answer?
A 3φ, Induction Motor had Tmax as 300 N-m with a supply of 440 V, 50 Hz. If the supply is changed to 220 V, 25 Hz new Tmax is
a)
300 N-m
b)
150 N-m
c)
75 N-m
d)
600 N-m
|
Shivam Das answered |
Sorry, but I'm not sure what you mean by "A 3". Can you please provide more information or clarify your question?
The switch of below circuit was open for long and at t = 0 it is closed. What is the final steady state voltage across the capacitor and the time constant of the circuit?
- a)0 V and 0.1 sec
- b)20 V and 0.2 sec
- c)10 V and 0.2 sec
- d)10 V and 0.1 sec
Correct answer is option 'D'. Can you explain this answer?
The switch of below circuit was open for long and at t = 0 it is closed. What is the final steady state voltage across the capacitor and the time constant of the circuit?
a)
0 V and 0.1 sec
b)
20 V and 0.2 sec
c)
10 V and 0.2 sec
d)
10 V and 0.1 sec
|
Constructing Careers answered |
Final value = 20/2 = 10 Volt
Time constant τ= 5 × 103 × 20 × 10-6
= 0.1 sec
Time constant τ= 5 × 103 × 20 × 10-6
= 0.1 sec
A DC shunt motor of 200 V, 10.5 A, 2000 rpm has an armature resistance of 0.5 Ω and field resistance of 400 Ω. It drives a load whose torque is constant at rated motor torque. What is the value of armature current if the source voltage drops to 175 V?- a)9.7 A
- b)12.4 A
- c)11.4 A
- d)10.7 A
Correct answer is option 'C'. Can you explain this answer?
A DC shunt motor of 200 V, 10.5 A, 2000 rpm has an armature resistance of 0.5 Ω and field resistance of 400 Ω. It drives a load whose torque is constant at rated motor torque. What is the value of armature current if the source voltage drops to 175 V?
a)
9.7 A
b)
12.4 A
c)
11.4 A
d)
10.7 A
|
Raghav Nambiar answered |
Given V1 = 200 V
Armature resistance (Ra) = 0.5 Ω
Field winding resistance (Rsh) = 400 Ω
Field current = 200/400 = 0.5 A
Load current = 10.5 A
Armature current = 10.5 – 0.5 = 10 A
Given that load torque is constant.
Armature resistance (Ra) = 0.5 Ω
Field winding resistance (Rsh) = 400 Ω
Field current = 200/400 = 0.5 A
Load current = 10.5 A
Armature current = 10.5 – 0.5 = 10 A
Given that load torque is constant.
A separately excited DC generator when running at 1500 rpm, supplies 200 A at 150 V. what will be the load current if speed drops to 1000 rpm. The resistance of armature is 0.25 Ω and voltage drop due to brush contact is 2 V- a)130.33 A
- b)132.67 A
- c)135.50 A
- d)138.75 A
Correct answer is option 'B'. Can you explain this answer?
A separately excited DC generator when running at 1500 rpm, supplies 200 A at 150 V. what will be the load current if speed drops to 1000 rpm. The resistance of armature is 0.25 Ω and voltage drop due to brush contact is 2 V
a)
130.33 A
b)
132.67 A
c)
135.50 A
d)
138.75 A
|
|
Harshad Singh answered |
Load resistance = 150/200 = 0.75Ω
E1 = 150 + (200 × 0.25) + 2
= 202 V
N1 = 1500 rpm, N2 = 1000 rpm
= 134.67 V
V = 134.67 – 0.25 I – 2
I = Load current
V = 132.67 – 0.25 I
E1 = 150 + (200 × 0.25) + 2
= 202 V
N1 = 1500 rpm, N2 = 1000 rpm
= 134.67 V
V = 134.67 – 0.25 I – 2
I = Load current
V = 132.67 – 0.25 I
Consider the circuit shown in figure
Q.
- a)(s+ 1)1
- b)
- c)s(s +1)1
- d)(s + 1).s
Correct answer is option 'B'. Can you explain this answer?
Consider the circuit shown in figure
Q.
a)
(s+ 1)1
b)
c)
s(s +1)1
d)
(s + 1).s
|
Yash Patel answered |
Using current division rule
Number of twigs in a tree are? n- number of nodes- a)n
- b)n+1
- c)n-1
- d)n-2
Correct answer is 'C'. Can you explain this answer?
Number of twigs in a tree are? n- number of nodes
a)
n
b)
n+1
c)
n-1
d)
n-2
|
Neha Choudhury answered |
Twig is a branch in a tree. Number of twigs in a tree are n-1. If there are 4 nodes in a tree then number of possible twigs are 3.
The operating characteristic of a distance relay in the R-X plane is shown in the figure below, It represents operating characteristic of a
- a)reactance relay
- b)directional impedance relay
- c)impedance relay
- d)mho relay
Correct answer is option 'D'. Can you explain this answer?
The operating characteristic of a distance relay in the R-X plane is shown in the figure below, It represents operating characteristic of a
a)
reactance relay
b)
directional impedance relay
c)
impedance relay
d)
mho relay
|
Prasant Verma answered |
First of all Consider the universal Torque equation of mho relay...then draw it on paper (there will be sinuosoidal term)
The control system shown in the given fig has an internal rate Feedback loop. The closed-loop system for open and closed condition of the switch will be respectively-
- a)Unstable and stable
- b)Unstable and Unstable
- c)Stable and Unstable
- d)Stable and Stable
Correct answer is option 'A'. Can you explain this answer?
The control system shown in the given fig has an internal rate Feedback loop. The closed-loop system for open and closed condition of the switch will be respectively-
a)
Unstable and stable
b)
Unstable and Unstable
c)
Stable and Unstable
d)
Stable and Stable
|
Cstoppers Instructors answered |
A two-winding transformer is used as an auto-transformer. The kVA rating of the auto-transformer compared to the twowinding transformer will be- a)3 times
- b)2 times
- c)1.5 times
- d)same
Correct answer is option 'B'. Can you explain this answer?
A two-winding transformer is used as an auto-transformer. The kVA rating of the auto-transformer compared to the twowinding transformer will be
a)
3 times
b)
2 times
c)
1.5 times
d)
same
|
Vedika Singh answered |
Answer would depend upon the voltage ratio
and type of auto-transformeri.e.., set-up or step-down.
For step-up transformer
Where a < 1
Let, N1 = N2 (Ideal two winding transformer) For step-up transformer
A single phase transformer takes a no load current of 1.3 A when high voltage winding is kept open. If the iron loss component of no load current is 0.5 A, the magnetizing component of no load current will be____ (in A)
Correct answer is between '1.1,1.3'. Can you explain this answer?
A single phase transformer takes a no load current of 1.3 A when high voltage winding is kept open. If the iron loss component of no load current is 0.5 A, the magnetizing component of no load current will be____ (in A)
|
|
Ameya Gupta answered |
No load current I0 = 1.3 A
Iron loss component IW = 0.5 A
Magnetizing component of no load current
Iron loss component IW = 0.5 A
Magnetizing component of no load current
A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.- a)7.2 KVAR
- b)12.47 KVAR
- c)16.63 KW
- d)9.6 KW
Correct answer is option 'A'. Can you explain this answer?
A wattmeter reads 5.54 kW when its current coil is connected in the red phase and its voltage coil is connected between the neutral and red phase of a symmetrical three phase system supplying a balanced load of 30 A at 400 V. What will be the reading of the instrument if the current coil remains unchanged and voltage coil be connected between blue and yellow phases. The phase sequence is RYB.
a)
7.2 KVAR
b)
12.47 KVAR
c)
16.63 KW
d)
9.6 KW
|
|
Anoushka Kumar answered |
Given data:
Current, I = 30 A
Voltage, V = 400 V
Power, P = 5.54 kW
To find: Reading of the instrument when voltage coil is connected between blue and yellow phases.
Solution:
Let's first calculate the power factor of the load using the given data.
Apparent power, S = VI = 30 x 400 = 12000 VA
Power factor, cosφ = P/S = 5.54/12 = 0.46
Sinφ = √(1 - cos²φ) = √(1 - 0.46²) = 0.89
Now let's calculate the phase angle between the voltage and current using the power factor.
Tanφ = sinφ/cosφ = 0.89/0.46 = 1.93
φ = tan⁻¹(1.93) = 63.65°
We know that the power measured by a wattmeter is given by P = VIcosφ. Since the current coil remains unchanged, the reading of the instrument will only depend on the voltage and power factor.
When the voltage coil is connected between the neutral and red phase, the phase angle between the voltage and current is zero, so the power factor is cosφ = 1. In this case, P = VIcosφ = 30 x 400 x 1 = 12000 W.
When the voltage coil is connected between the blue and yellow phases, the phase angle between the voltage and current is φ = 63.65°, so the power factor is cosφ = 0.46. In this case, P = VIcosφ = 30 x 400 x 0.46 = 5520 W.
The difference between the two readings is the reactive power of the load, which is given by Q = √(S² - P²) = √(12000² - 5520²) = 10368 VAR.
Since the load is balanced and the phase sequence is RYB, the reactive power is the same for all three phases. Therefore, the reading of the instrument when the voltage coil is connected between the blue and yellow phases is Q/3 = 10368/3 = 3456 VAR = 3.456 kVAR.
Therefore, the correct answer is option (A) 7.2 kVAR. (Note that this answer is incorrect and the correct answer should be 3.456 kVAR).
Current, I = 30 A
Voltage, V = 400 V
Power, P = 5.54 kW
To find: Reading of the instrument when voltage coil is connected between blue and yellow phases.
Solution:
Let's first calculate the power factor of the load using the given data.
Apparent power, S = VI = 30 x 400 = 12000 VA
Power factor, cosφ = P/S = 5.54/12 = 0.46
Sinφ = √(1 - cos²φ) = √(1 - 0.46²) = 0.89
Now let's calculate the phase angle between the voltage and current using the power factor.
Tanφ = sinφ/cosφ = 0.89/0.46 = 1.93
φ = tan⁻¹(1.93) = 63.65°
We know that the power measured by a wattmeter is given by P = VIcosφ. Since the current coil remains unchanged, the reading of the instrument will only depend on the voltage and power factor.
When the voltage coil is connected between the neutral and red phase, the phase angle between the voltage and current is zero, so the power factor is cosφ = 1. In this case, P = VIcosφ = 30 x 400 x 1 = 12000 W.
When the voltage coil is connected between the blue and yellow phases, the phase angle between the voltage and current is φ = 63.65°, so the power factor is cosφ = 0.46. In this case, P = VIcosφ = 30 x 400 x 0.46 = 5520 W.
The difference between the two readings is the reactive power of the load, which is given by Q = √(S² - P²) = √(12000² - 5520²) = 10368 VAR.
Since the load is balanced and the phase sequence is RYB, the reactive power is the same for all three phases. Therefore, the reading of the instrument when the voltage coil is connected between the blue and yellow phases is Q/3 = 10368/3 = 3456 VAR = 3.456 kVAR.
Therefore, the correct answer is option (A) 7.2 kVAR. (Note that this answer is incorrect and the correct answer should be 3.456 kVAR).
If all the sequence voltages at the fault point in a power system are equal, then the fault is a- a)three-phase fault
- b)line to ground fault
- c)line to line fault
- d)double line to ground fault
Correct answer is option 'D'. Can you explain this answer?
If all the sequence voltages at the fault point in a power system are equal, then the fault is a
a)
three-phase fault
b)
line to ground fault
c)
line to line fault
d)
double line to ground fault
|
Pallavi Deshpande answered |
For a solid fault i.e, Zf = 0, Va1 = Va2 = Va0 for L-L-G fault.
The network shown below is
- a)Over damped
- b)Under damped
- c)Critically damped
- d)Un-damped
Correct answer is option 'B'. Can you explain this answer?
The network shown below is
a)
Over damped
b)
Under damped
c)
Critically damped
d)
Un-damped
|
Naroj Boda answered |
- Roots are complex conjugates
The current drawn by a 120V d.c. motor with back e.m.f. of 110V and armature resistance of 0.4 ohm is- a)4 A
- b)25 A
- c)274 A
- d)300 A
Correct answer is option 'B'. Can you explain this answer?
The current drawn by a 120V d.c. motor with back e.m.f. of 110V and armature resistance of 0.4 ohm is
a)
4 A
b)
25 A
c)
274 A
d)
300 A
|
Arya Datta answered |
Eb = Vt – Ia Ra
∴ 110 = 120 – Ia × 0.4 ⇒ Ia = 25A
∴ 110 = 120 – Ia × 0.4 ⇒ Ia = 25A
Chapter doubts & questions for Subject-wise Tests - GATE Electrical Engineering (EE) Mock Test Series 2026 2025 is part of Electrical Engineering (EE) exam preparation. The chapters have been prepared according to the Electrical Engineering (EE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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