The useful flux of a four-pole, lap wound D C generator is 0.07 W b . Its armature carries 220 turns and each turn has 0.004 ohm resistance, if its armature current is 50 A, running at a speed of 900 r p m, determine its terminal voltage.
Question 4.40: The useful flux of a four-pole, lap wound D C generator is 0...
\text { Here, } P=4 ; A=4(\text { lap }-\text { wound }) ; \phi=0.07 Wb ; N=900 rpm ; I_{a}=50 A
No. of armature conductors, Z=2 \times No. of turns =2 \times 220=440
Induced emf, E=\frac{P Z \phi N}{60 A}=\frac{4 \times 400 \times 0 \cdot 07 \times 900}{60 \times 4}=462 V
No. of turns connected in each parallel path =\frac{220}{4}=55
Resistance of each parallel path =55 \times 0.004=0.22 ohm
Four resistances, each of 0.22 ohm, are connected in parallel as shown in Fig. 4.63(a).
\therefore Armature resistance, R_{a}=\frac{0 \cdot 22}{4}=0 \cdot 055 \Omega
Terminal voltage, V=E-I_{a} R_{a} (generator action)
=462-50 \times 0.055 =459 \cdot 25 V
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