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GATE’22 EE: Daily Practices Quiz 23-Aug-2021

GATE 2022 is being organized by IIT, Kharagpur. Through this exam, the qualified candidates get admission into master’s & direct doctoral programs in the field of Engineering/ Architecture/ Technology/ Arts & Science, etc. Qualification in GATE is mandatory for seeking admission to various institutions.

GATE’22 EE: Daily Practices Quiz 23-Aug-2021

Each question carries 2 marks
Negative marking: 2/3 mark

Total Questions: 05
Total marks: 10
Time: 12 min.

Q1. A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 Ω and armature current is 10A. of the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be
(a) 1.79 Ω
(b) 2.1 Ω
(c) 18.9 Ω
(d) 3.1 Ω

Q2. A field excitation of 20 A in a certain alternator result in an armature current of 400A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is….
(a) 1V
(b) 10V
(c) 100V
(d) 1000V

Q3. The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KCL equations and the number of independent KVL equations, respectively, are
(a) 2 and 5
(b) 5 and 2
(c) 3 and 4
(d) 4 and 3

Q4. In the Bode – plot of a unity feedback control system, the value of phase of 𝐺(𝑗𝜔) at the gain cross over frequency is – 125⁰. The phase margin of the system is
(a) – 125⁰
(b)– 55⁰
(c) 55⁰
(d)125⁰

Q5. In the case of metals, the valence and conduction bands have
(a) no overlap, energy gap =0
(b) no overlap, energy gap is small
(c) overlap, energy gap =0
(d) no overlap, energy gap is large

SOLUTIONS

S1. Ans.(a)
Sol. ACQ: I_a1=10 A,r=1 Ω & ϕ_2=0.9 ϕ_1
As torque is constant.
∴I_a1 ϕ_1=I_a2 ϕ_2
⇒I_a2=10/0.9=11.11 A
Let R Ω resistance to be added to maintain the same speed and torque.
∵N α E_b/ϕ⇒N_1/N_2 =E_b1/E_b2 ×ϕ_2/ϕ_1
⇒1=((220-10×1))/(220-11.11(1+R))×(0.9 ϕ_1)/ϕ_1
∴1+R=2.79
⇒R=1.79 Ω.

S2. Ans.(d)
Sol. Internal resistance=2000/400=5 Ω
The magnitude of the internal voltage drops within the machine at a load current of 200A=200×5=1000 V.

S3. Ans.(d)
Sol. ACQ: B=7, N=5
∴ The number of independent KCL equations=N-1=5-1=4
The number of independent KVL equations=B-(N-1) =7-4=3

S4. Ans.(c)
Sol. The Phase angle of 𝐺(𝑗𝜔) at ω = ω_gc = -125⁰
Where; ω_gc = gain cross-over frequency
ϕ_gc = ∠G(jω_gc) = -125⁰
The phase margin = 180⁰ + ϕ_gc
= 180 − 125
= 55⁰

S5. Ans.(c)
Sol. The materials can be classified by the energy gap between their valence band and the conduction band. The valence band is the band consisting of the valence electron, and the conduction band remains empty. Conduction takes place when an electron jumps from valence band to conduction band and the gap between these two bands is forbidden energy gap. Wider the gap between the valence and conduction bands, higher the energy it requires for shifting an electron from valence band to the conduction band.
In the case of conductors, this energy gap is absent or in other words conduction band, and valence band overlaps each other. Thus, electron requires minimum energy to jump from valence band. The typical examples of conductors are Silver, Copper, and Aluminium.
In insulators, this gap is vast. Therefore, it requires a significant amount of energy to shift an electron from valence to conduction band. Thus, insulators are poor conductors of electricity. Mica and Ceramic are the well-known examples of insulation material.
Semiconductors, on the other hand, have an energy gap which is in between that of conductors and insulators. This gap is typically more or less 1 eV, and thus, one electron requires energy more than conductors but less than insulators for shifting valence band to conduction band.

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