Quiz: Electrical Engineering
Exam: UPSSSC-JE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. Holes are drilled on the opposite sides of the spindles of an energy meter to
(a)Avoid creep on load
(b)Balance the disc
(c)Dissipate heat generated due to eddy currents
(d)Increases the deflection torque
Q2. The input of an A.C. circuit having power factor of 0.8 lagging is 40 kVA. The power
drawn by the circuit is
(a) 12 kW
(b) 22 kW
(c) 32 kW
(d) 64 kW
Q3. Which of the following expressions is true for apparent power in an A.C. circuit?
(a) VI cos ∅
(b) Vav.×Iav.
(c) V(r.m.s.) ×I(r.m.s.)
(d) Vpeak×Ipeak
Q4. The efficiency for maximum power transfer to the load is
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Q5. In a parallel R-L-C circuit Susceptance is equal to
(a) 1/X
(b) 1/R
(c) R/Z^2
(d) X/Z^2
Q6.A circuit component that opposes the change in circuit voltage is
(a) Resistance
(b) Capacitance
(c) Inductance
(d) All the above
Q7. A series resonant circuit implies
(a) Zero power factor and maximum current
(b) Unity power factor and maximum current
(c) Unity power factor and minimum current
(d) Zero power factor and minimum current
Q8. A 6 pole, 50 Hz, 3-phase induction motor is running at 950 rpm and has rotor cu loss of 5 Kw. Its rotor input is……….KW.
(a)100
(b)10
(c)95
(d)5.5
Q9. Reactance relay is normally preferred for protection against
(a)Over load currents only
(b)Phase faults only
(c)Earth faults only
(d)High voltage protection only
Q10. Which of the following relay is used for the protection of medium transmission line?
(a)Impedance Relay
(b)Mho’s Relay
(c)Differential Relay
(d)None of the above
SOLUTIONS
S1. Ans.(a)
Sol. The slow but continuous rotation of the disc when only the pressure coils are excited but no current is flowing in the circuit is called “creeping”.
To overcome this creeping effect on no-load, two holes are drilled in the disc on a diameter, i.e. on opposite sides of the spindle. This causes sufficient distortion of the field to prevent rotation when one of the holes comes under one of the poles of the shunt magnet.
S2. Ans.(c)
Sol. P.F.= P/S
Or
P = P.F. x S = 0.8 x 40 = 32 kW
S3. Ans.(c)
Sol. The product of voltage and current if and only if the phase angle differences between current and voltage are ignored. In an AC circuit, the product of the r.m.s voltage and the r.m.s current is called apparent power.
S4. Ans.(b)
Sol. For maximum power transfer to load
Z_L=Z_s or R_L=R_s
I=V/(R_s+R_L )=V/(2R_s ) (Since R_L=R_s)
P_in=V.I=(R_s+R_L ).I
= I^2 (R_s+R_L )=I^2 (2R_s)
P_out=P_load=I^2 R_L=I^2 R_s
η = P_out/P_in
= (I^2 (R_s))/(I^2 (2R_s ) )=1/2=0.5
% η=50%
So, the efficiency for maximum power transfer.
S5. Ans.(a)
Sol. For Parallel RLC Circuit
1/Z=1/R+J(1/X_L +1/X_C )
Y = 1/R+J(1/X)
For here Susceptance = 1/X
S6. Ans.(b)
Sol. A capacitance is a circuit component that opposes the change in circuit.
S7. Ans.(b)
Sol. At resonance, we know that
ω=ω_0=1/√LC& X_L=X_c then
Impedance Z = √(R^2+(X_L-X_c )^2 )
Here L = Inductance
C = Capacitance
⇒ Z = R so there is unity power factor and Current is maximum.
S8. Ans.(a)
Sol. rotor speed (N_r)=950 rpm
And synchronous speed of the motor, N_s= 120f/P=(120×50)/6=1000 rpm
∴slip(s)=(N_s-N_r)/N_s =(1000-950)/1000=0.05
Rotor cu loss is given as, P_L=sP_r
Where, P_r=rotor input
∴P_r=P_l/s=(5×10^3)/0.05=100 KW.
S9. Ans.(c)
Sol. A reactance relay is an overcurrent relay with directional limitations. So, it is normally used against earth faults only.
S10. Ans.(a)
Sol. Relay used for different configuration of transmission lines……
Short transmission line: reactance relay
Medium transmission line: impedance relay
Long transmission line: mho relay
