Question 7.4: A four-pole 220 V dc shunt generator supplies a load of 3 kW...
A four-pole 220 V dc shunt generator supplies a load of 3 kW at 220 V. The resistance of the armature winding is 0.1 Ω and that of the field winding is 110 Ω. Calculate the total armature current, the current flowing through armature conductors, and the EMF induced. Assume that the armature winding is wave wound.
Output power = 3 kW = 3000 W
Power = output voltage, V × Output current, I_{L}
I_{L}=\frac{3000}{220}=13.6 AFrom the figure it can be seen that
\begin{aligned} I_{f} &=\frac{V}{R_{f}}=\frac{220}{110}=2 A \\ I_{a} &=I_{L}+I_{f} \\ &=13.6+2=15.6 A \end{aligned}The armature winding is wave wound. The number of parallel paths is 2. That is, all the armature conductors are connected in such a way that half the armature current flows through each path. Thus, current flowing though each armature conductor will be I_{a}/2 i.e., \frac{15.6}{2}=7.8 Amps. E is the EMF induced in the armature. A voltage drop of I_{a} R_{a} takes place in the armature winding when it is supplying current. The remaining voltage, V is available across the load terminals.
Thus,
E-I_{a} R_{a}=Vor, E=V+I_{a} R_{a} = 220 + 15.6 × 0.1 = 218.44 V
