Quiz: Electrical Engineering 14 May 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1.A keeper is used for
(a)restoring of lost flux
(b)amplification of flux
(c)providing a closed path for magnetic flux
(d)changing the direction of magnetic line of forces

Q2. A load is connected to a circuit. At the terminals to which the load is connected, Rth = 20 Ω and Vth = 80 V. The maximum power supplied to the load is
(a) 360 W
(b) 80 W
(c) 10 W
(d) 120 W

Q3. When a network is loaded by a resistance equal in value to its Norton resistance, the Norton current is 8 A. The current through the load will be
(a) 8 A
(b) 4
(c) 16 A
(d) None of these

Q4. A 140 V source has a series internal resistance of 1 Ω. The maximum power that can be delivered to a load is
(a) 2100 W
(b) 4900 W
(c) 3600 W
(d) 7800 W

Q5. A 8 A source has a shunt internal resistance of 100 Ω. The maximum power that can be delivered to a load is
(a) 925 W
(b) 1250 W
(c) 1600 W
(d) None of above

Q6. Materials which lack permanent magnetic dipole are known as

(a) paramagnetic
(b) Diamagnetic
(c) Ferromagnetic
(d) Ferrimagnetic

Q7. Increase in percentage of carbon in steel reduces
(a) resistivity
(b) coercive force
(c) Permeability
(d) Retentivity

Q8. Which motor can conveniently operate at lagging as well as leading power factor?
(a) Squirrel cage induction motor
(b) Wound rotor induction motor
(c) Synchronous motor
(d) DC shunt motor

Q9. The insulating material employed for motor windings are classified according to
(a) Motor kW output rating
(b) level of temperature rise
(c) Controller size
(d) Overload protection available

Q10. The difference between the indicated value and the true value of a quantity
(a) Gross error
(b) Absolute error
(c) Dynamic error
(d) Relative error

Solution

S1. Ans.(c)
Keeper is used for providing a closed path for magnetic flux.

S2. Ans.(b)
Sol. Under maximum power transfer, R_L-R_Th and voltage across R_L=V_Th/2.
∴P_(max.)=(V_Th/2)²/R_Th =(80/2)²/20=80 W

S3. Ans.(b)
Sol. Under maximum power transfer, R_L=R_N. Therefore, the Norton current will divide equally between R_L and R_N so that current through R_L is I_N/2
i.e. I_N= 8/2 = 4 A

S4. Ans.(b)
Sol. Here, E_Th = 140 V and R_Th = 1 Ω. For maximum power in the load, R_L=R_Th = 1 Ω.
Therefore, voltage across R_L=E_Th/2 = 140/2 = 70 V.
∴P_max.=(70)²/R_L =(70)^2/1=4900 W

S5. Ans.(c)
Sol. Here, I_N = 8 A and R_N = 100 Ω. For maximum power in the load, R_L=R_N = 100 Ω.
Therefore, current through load = I_N/2 = 8/2 = 4 A
∴ P_max = (4)² × R_L = (4)² × 100 = 1600 W

S6. Ans.(b)

Sol. Materials which lack permanent magnetic dipole are known as Diamagnetic.

S7. Ans.(d)
Sol.
Increase in percentage of carbon in steel reduces retentivity.

S8. Ans.(c)
Sol. Synchronous motor

S9. Ans.(b)
Sol. level of temperature rise

S10. Ans.(b)
Sol. The difference between the indicated value and the true value of a quantity is absolute error.

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