250+ TOP MCQs on Graphical Representation of External Characteristics and Answers

DC Machines Multiple Choice Questions on “Graphical Representation of External Characteristics”.

1. No load point of DC generator is __________
a) Intersection of OCC and Rf line
b) Point on the Y axis at rated field current
c) Point on the X axis at rated terminal voltage
d) Can’t find through graphical interpretation
Answer: a
Clarification: Intersection of OCC with field resistance line gives the no-load point. All the value so obtained from x and y axis respectively gives the terminal voltage and field current of a DC generator at no-load.

2. Rf line intersects with OCC in ________
a) 1st quadrant
b) 2nd quadrant
c) 3rd quadrant
d) 4th quadrant
Answer: a
Clarification: Rf line is a straight line passing through the origin and having constant slope, so rising always in a positive direction. OCC starts from some positive value on Y axis and increases till maximum point, afterwards it starts becoming constant, where generally it intersects with Rf line.

3. How armature resistance effect is shown graphically?
a) By adding IaRa product horizontally with Rf line
b) By subtracting IaRa product vertically with Rf line
c) By adding IaRa product vertically with Rf line
d) By subtracting IaRa product horizontally with Rf line
Answer: c
Clarification: For representing the voltage drop in armature resistance, we add product IaRa vertically with Rf line at minimum 2 points and draw line parallel to Rf to get line with V + IaRa = constant.

4. For determining IaRa maximum _____________
a) Distance between Rf line and v+ IaRa line is taken
b) Distance between OCC and v+ IaRa line is taken
c) Maximum Distance between OCC and v+ IaRa line is taken
d) Can’t calculated graphically
Answer: c
Clarification: OCC when starts from some residual voltage on y axis, goes on increasing till some maximum value and then starts reducing slightly. The bulk is formed where maximum distance between OCC and v+ IaRa line is taken, to get effective frop in armature.

5. How demagnetization effect of armature reaction is shown graphically?
a) By shifting origin towards +ve y axis
b) By shifting origin towards +ve x axis
c) By shifting origin towards -ve y axis
d) By shifting origin towards -ve x axis
Answer: d
Clarification: The demagnetization caused by armature reaction can be quantified by equivalent field current Ifd which can be taken as proportional to the armature current Ia. So, by shifting the origin towards -ve x axis by Ifd we can show armature reaction graphically.

6. External characteristics of DC shunt motor lies in ______________
a) 1st quadrant
b) 2nd quadrant
c) 1st and 2nd quadrant
d) 4th quadrant
Answer: c
Clarification: When external characteristics are plotted on graphs, and if effect of armature reaction is not considered graph lies in 1st quadrant only. When effect of armature reaction is considered we shift the origin thus, characteristics lies in 2nd quadrant as well.

7. From magnetization characteristic at If = 7.1 A, Ea = 225 V at 1000 rpm. What will be the terminal voltage at speed 950 rpm?
a) 225 V
b) 235 V
c) 214 V
d) 220 V
Answer: c
Clarification: Speed is directly proportional to the back emf of a machine. So, Ea = 225 V at 1000 rpm
Ea (950 rpm) = 225 x 950 / 1000, this will give speed of given DC machine at 950 rpm. Upon calculations we get, Ea = 213.7 V= 214 V approx.

8. For a given compound DC machine, Net field current obtained from characteristic is equal to 7.5 A, where shunt field current is equal to 5 A, armature current is 505 A, demagnetizing current equal to 0.95 A and shunt field winding of 1000 turns at rated speed of 1000 rpm. What will be the series field turns?
a) 7
b) 8
c) 5
d) 9
Answer: a
Clarification: From the excitation balance equation, If + [Nse/Nf] Ia – Ifd =If (net).
5 + 505(Nse /1000) – 0.95 = If (net).
Calculating for Nse by substituting If (net), we get Nse = 6.8 that is 7 turns.

DC Machines, .