UPRVUNL’21 EE: Daily Practices Quiz 19-July-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 08 min.

Q1. Which of the following relation is correct for current carrying capacity of fuse element?
(a) I ∝ d
(b) I² ∝ d²
(c) I² ∝ d³
(d) I² ∝ d
Where d = diameter of fuse element

Q2. A single-phase motor runs normally at 183 rad/sec on 60 Hz power supply. How
many poles must it have?
(a) 10
(b) 8
(c) 12
(d) 4

Q3. ………. Motors never use belt-connected loads.
(a) Series
(b) Shunt
(c) Cumulatively compounded
(d) Differentially compounded

Q4. An autotransformer having voltage transformation ratio 0.8 supplies a load of 10
kW. The power transferred inductively from the primary to the secondary is
(a) 10 kW
(b) 8 kW
(c) zero
(d) 2 kW

Q5. In an induction type meter, maximum torque is produced when the phase angle between two fluxes is….
(a) 0⁰
(b) 45⁰
(c) 60⁰
(d) 90⁰

Q6. Two resistors of resistances 8 Ω and 12 Ω are connected in parallel. The power dissipated in 8 Ω resistor is 9 W. The power dissipated is 12 Ω resistor is
(a) 12 W
(b) 24 W
(c) 6 W
(d) 16 W

SOLUTIONS

S1. Ans.(c)
Sol. Heat produced per sec = Heat lost per sec (convection, conduction, radiation)
⇒ I²R = K × d × l
⇒ I²(ρl/A) = K × d × l
⇒ I²(ρl/(π(d/2)^2 )) = K × d × l
⇒ (I^2/d^2 ) × constant = K × d × l
⇒ ▭(I^2∝d^3 ) also called fuse law.

S2. Ans.(d)
Sol. ω=4π×f/p rad/sec ⇒ P=(4π×60)/183=4.11
Therefore, the motor must have 4 poles.

S3. Ans.(a)
Sol. If the belt is accidentally broken, the motor will run on on-load. The flux drops to practically zero and the motor speed may become dangerously high. For this reason, a d.c. series motor is used only where load is directly connected to the shaft or geared to the shaft.

S4. Ans.(d)
Sol. Power transferred inductively
= (1-0.8) × 10 = 2KW
In auto-transformer, as winding is common in both circuits, most of the energy is transferred by means of electrical conduction and a small part is transferred through induction.

S5. Ans.(d)
Sol. In induction type meter,T_a=KI_se I_sh sinϕ
Where,I_se= series coil current
I_sh= shunt coil current
If 𝟇=90⁰, then torque will be maximum.

S6. Ans.(c)
Sol. P=V^2/R or V=√(PR.) Since voltage is same in a parallel circuit, P_1 R_1=P_2 R_2 or 9×8 = P_2×12
∴P_2=6 W

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