A four-pole DC shunt generator with a wave wound armature having 390 conductors has to supply a load of 500 lamps each of 100 W at 250 V. Allowing 10 V for the voltage drop in the connecting leads between the generator and the load and brush drop of 2 V . Calculate the speed at which the generator should be driven. The flux per pole is 30 m Wb and the value of R_{a}=0.05 \Omega and R_{s h}=65 \Omega
The conventional circuit diagram of the DC shunt generator is shown in Fig. 4.64.
\text { Total load }=500 \times 100 W
I_{L}=\frac{500 \times 100}{250}=200 A
Voltage drip in leads, V_{L}=10 V
Voltage across shunt field winding,
V_{s h} =V+V_{L}=250+10=260 V
I_{s h} =V_{s h} / R_{s h}=260 / 65=4 A
I_{a} =I_{L}+I_{s h}=200+4=204 A
\text { Armature drop }=I_{a} R_{a} =204 \times 0.05=10 \cdot 2 V
Total brush drop, \quad 2 v_{b}=2 V
Generated emf,
E_{g} =V+I_{a} R_{a}+V_{L}+2 v_{b}
=250+10 \cdot 2+10+2=272 \cdot 2 V
\text { Now, }
E_{g}=\frac{P \phi N Z}{60 A} \text { or } 272 \cdot 2=\frac{4 \times 30 \times 10^{-3} \times N \times 390}{60 \times 2}
\text { or } N=\frac{272 \cdot 2 \times 60 \times 2}{4 \times 30 \times 10^{-3} \times 390}=698 \text { rpm }
