Quiz: Electrical Engineering
Exam: SSC JE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. In the equivalent circuit of an induction motor, the mechanical load is represented by
(where r is actual rotor resistance, (s)=slip)
(a) r_2/s
(b) r_2 s/((1-s))
(c) r_2 s/((1+s))
(d) r_2 ((1-s))/s
Q2. In a three-phase induction motor, the maximum torque…………….
(a) Does not depend on r_2
(b) Is proportional to rotor resistance r_2
(c) Is proportional to 1/r_2
(d) Is proportional to r_2^2
Q3. An electric motor in which rotor and stator fields rotate simultaneously is called……. motor.
(a) Induction
(b) Synchronous
(c) DC
(d) Universal
Q4. Deep bar rotors are used in induction motor to increase
(a) Starting current
(b) Full load current
(c) Starting torque
(d) Full load torque
Q5. Two watt-meters, connected to measure power in a three-phase balanced circuit, read 100 W each. The reactive power absorbed by the circuit is
(a) 100 Var
(b) 200 Var
(c) 300 Var
(d) 0 Var
Q6. In the circuit show in fig. the current I_1 and I_2 respectively are
(a) 4A and 8A
(b) 2A and 4A
(c) 4A and 2A
(d) 6/5A and 12/5A
Q7. The equivalent resistance in ohms in the circuit shown is
(a) 1/6
(b) 6
(c) 3/4
(d) 4/3
Q8. If r is the radius of the conductor and R is the radius of the sheath of the cable, the cable operates stably from the point of view of dielectric strength if:
(a) r/R>1
(b)r/R<1 (c) r/R<0.368 (d) r/R>2.718
Q9. Laminated cores, in electrical machines, are used to reduce
(a) copper loss
(b) eddy current loss
(c) hysteresis loss
(d) all of the above
Q10. If coper loss of transformer at 7/8th full load is 4900 W, then its full load copper loss
would be
(a)5600 W
(b) 6400 W
(c) 373 W
(d) 429 W
SOLUTIONS
S1. Ans.(d)
Sol. R_mechanical=r_2/s(1-s)
S2. Ans.(a)
Sol. T_max=(kE_2^2)/(2X_2 )
So, maximum torque does not depend on r_2.
S3. Ans.(b)
Sol. For synchronous motor, N_s=N_r
S4. Ans.(c)
Sol. Deep bars rotors are used in induction motors to increase the starting torque, deep bar rotor provide more rotor resistance at the running condition and low rotor resistance at the running condition.
S5. Ans.(d)
Sol. Total reactive power=√3 (W_1-W2)=√3 (100-100)=0 Var
S6. Ans.(b)
Sol. Applying current division rule –
I_1=2/(2+4)×6
=2A
I_2=4/(2+4)×6
=4 A
S7. Ans.(c)
Sol. G_eff=(G_1×G_2)/(G_1+G_2 )=(2×4)/(2+4)=8/6
Req = 1/G_eff =6/8=3/4 Ω.
S8. Ans.(c)
Sol. The cable operates stably from the point of view of dielectric strength if:
R/r>e or 2.718
∴r/R<1/e or 0.368
S9. Ans.(b)
Sol. To reduce eddy current loss laminated cores are used.
S10. Ans.(b)
Sol. Copper loss at fractional load=x² full load copper loss
⇒4900=(7/8)^2×P_(F.L.)
⇒P_(F.L).=6400 W
