01․ A cumulatively compounded dc motor delivers rated load power at rated speed. If the series field is short-circuited, then
Series field resistance is negligible so after short circuiting it there will not be any affect on armature current i.e. armature current remains constant. Due to short circuited series field total flux will also reduce i.e. speed will increase as Nα1/Φ.
02․ A 240 V dc series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is 0.3 Ω. The value of resistance which must be added to obtain rated torque at 1000 rpm is
We know,
03․ A DC series motor drawing an armature current of Ia is operating under saturated magnetic conditions. The torque developed in the motor is proportional to
We know Torque TαΦIa
As motor is operating under saturation condition so torque will be independent of flux i.e. TαIa.
04․ A dc shunt motor runs at a no-load speed of 1140 rpm. At full load, armature reaction weakens the main flux by 5% whereas the armature circuit voltage drops by 10%. The motor full-load speed in rpm is
We know, 
05․ A dc shunt motor has external resistance Ra and Rf in the armature and field circuits respectively. Armature current at starting can be reduced by keeping
Armature current at starting can be reduced by keeping Ra maximum and Rf minimum.
06․ Consider the following statements regarding the speed control of dc motors :
1. ward-leonard method is suitable forconstant-torque drives
2. Ward-leonard method is suitable for constant-power drives
3. field-control method facilitates speed control below base speed
4. Armature-resistance control method is more efficient than ward-leanard method
5. Field-control method is suitable for constant-torque drives
6. Armature-resistance control method is suitable for constant-torque drives
From these, the correct answer is
Armature-resistance control and ward-leonard methods are suitable for constant-torque drives and ward-leonard method is also suitable for constant-power drives.
07․ Consider the following statements:
To control the speed of a dc shunt motor above the base speed over a reasonably wide range, the motor must
1. have compensating winding
2. have interpole winding
3. be started using a 3-point starter
4. be started using a 4-point starter
Of these statements
To control the speed of a DC shunt motor above the base speed over a reasonably wide range, the motor must go through field weakening method. So, to avoid excessive armature reaction due to field weakening motor must be provided with compensating winding and inter poles. To avoid unwanted pull back of starter field must be connected separately with the supply i.e. 4 pin starter should be used.
08․ φA DC shunt machine generates 250 V on open circuit at 1000 RPM. Armature resistance is 0.5 Ω, field resistance 250 Ω. When running as a motor takes 4 A at 250 V. Calculate the speed when running as a motor taking 40 A at 250 V. Armature reaction weakens the field by 4%.
As Generator : O.C. = 250 V, 1000 RPM, flux = φ1
As motor on load : Il = 40 A at 250 V, Ish = 1 A and Ia= 3 A.
Motor on Load : Il=40 A at 250 V, φ2 = 0.96*φ1 and Ia = 39 A.
Eb2 = 230.5 V
From,
N2 = 960 RPM.
N2 = 960 RPM.09․ A DC shunt machine generates 250 V on open circuit at 1000 RPM. Armature resistance is 0.5 Ω, field resistance 250 Ω. When running as a motor takes 4 A at 250 V. Calculate the efficiency when running as a motor taking 40 A at 250 V. Armature reaction weakens the field by 4%.
Total fixed loss including field copper loss = 1000 W
Copper loss in armature on load = 760.5 W
Total losses = 1760.5 W
Total losses = 250×40⇒10000 W
∴ efficiency ζ = [1-(losses/input)]×100⇒82.5%
10․ For a DC machine having P no of pair of pole pairs, Z no of conductors having length L of resistivity ρ and cross section S are arranged. what is the armature resistance?
Resistance of each conductor = ρL/S
and Z/2a conductors in series
Resistance of each armature path = PLZ/2aS
There are 2a Such resistances are in parallel in the armature
∴ Armature resistance = PZL/ 2aS×1/2a⇒ PZL/ 4Sa²
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