Quiz: Electrical Engineering

Exam: NLC

Topic: Miscellaneous

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

Q1. An 8-pole duplex lap winding will have ……. parallel paths.

(a) 8

(b) 4

(c) 32

(d) 12

Q2. Hysteresis loss in a d.c. machine is directly proportional to ………….

(a) speed

(b) (speed)²

(c) (speed)^1.6

(d) none of the above

Q3. If W_C is the constant loss and R_a is the armature resistance of a d.c. generator, then load current I_L corresponding to maximum efficiency is

(a) I_L=√(R_a/W_c )

(b) I_L=W_c/√(R_a )

(c) I_L=R_a/√(W_c )

(d) I_L=√(W_c/R_a )

Q4. A 4-pole, lap-wound d.,c. shunt generator has an armature winding consisting of 110 turns each of 0.004 Ω. The armature resistance is

(a) 0.5 Ω

(b) 1 Ω

(c) 0.25 Ω

(d) 0.0275 Ω

Q5. The current drawn by a 220 V d.c. motor of armature resistance 0.75 Ω and back e.m.f. 180 V is

(a) 106 A

(b) 53.33 A

(c) 533.33 A

(d) 5.33 A

Q6. With the increase in speed of a d.c. motor,

(a) both back e.m.f. and line current fall

(b) both back e.m.f. and line current increase

(c) back e.m.f. falls and line current increases

(d) back e.m.f. increases and line current decreases

Q7. In a d.c. series motor, if armature current is reduced to 60% of its original value, the torque of the motor will be reduced by

(a) 64%

(b) 36%

(c) 50%

(d) 70%

Q8. A d.c. shunt motor runs at 500 r.p.m. at 220 V. A resistance of 10.5 Ω is added in series with the armature for speed control. The armature resistance is 0.5 Ω. The current to stall the motor will be

(a) 60 A

(b) 20 A

(c) 70 A

(d) 80 A

Q9. A d.c. series motor is running at rated speed without any additional resistance in series. If an additional resistance is placed in series, the speed of motor

(a) increases

(b) decreases

(c) remains same

(d) none of the above

Q10. A transformer is an efficient device because it ………………..

(a) is a static device

(b) uses inductive coupling

(c) uses capacitive coupling

(d) uses electric coupling

Silution

S1. Ans (d)

Sol. With a n-plex lap winding, the number of parallel paths is P n where

P=number of poles

N=2, 3 etc. and stands for duplex, triplex etc.

Hence, 6-pole duplex lap winding will have 6×2= 12 parallel paths

S2. Ans (a)

Sol. Hysteresis power loss, P_h∝B_max^1.6 f

Now, f=NP/120 so that f ∝ N (speed). Therefore, P_h∝N

S3. Ans (d)

Sol. The efficiency of a d.c. generator is maximum when

Variable loss = Constant loss

Or I_L^2 R_a=W_c or I_L=√(W_c/R_a )

S4. Ans (d)

Sol. Total resistance of 110 turns = 110 ×0.004=0.44 Ω

Since there are 4 parallel paths in armature,

Resistance of each path=0.44/4=0.11 Ω

Now, there are four such resistances in parallel each of value 0.11 Ω .

∴ Armature resistance, r_a=0.11/4=0.0275 Ω

S5. Ans (b)

Sol. Current drawn = (V-E_b)/R_a =(220-180)/0.75=53.33 A

S6. Ans (d)

Sol. For a d.c. motor, line current = (V-E_b)/R_a .

S7. Ans (a)

Sol. In a d.c. series motor, the torque developed T∝I_a^2. For the first case, T_1∝I_a^2; For the second case, T_2∝(0.6 I_a )^2.

∴ T_2/T_1 =((0.6 I_a)²)/(I_a^2 )=(0.36 I_a^2)/(I_a^2 )=0.36

∴ 1-T_2/T_1 =1-0.36=0.64 or (T_1-T_2)/T_1 =0.64 or 64%

S8. Ans (b)

Sol. The stalling torque is the maximum torque (i.e. value of load torque plus frictional torque) which the motor will just fail to provide. At stalling torque, E_b=0.

∴ Stalling current = (V-E_b)/((R_a+R))=(220-0)/((0.5+10.5))=220/11=20 A

S9. Ans (b)

Sol. When additional resistance is placed in series, the voltage available across the armature decreases. Hence, the speed of the motor falls.

S10. Ans (a)

Sol. Since transformer is a static device, the friction and windage losses are absent.