# GATE 2015 EE – SET 2 – Question 23

Solution :

P = 4 , V= 230V , Power = 5kW , N = 1200 rpm

Induced EMF , E = $$\frac{\phi NZ}{60}*\frac{P}{A}$$

For lap connected , P = A = 4 ,

For wave connected , A = 2

E$$\alpha$$ $$\frac{1}{A}$$ = $$\frac{E_{lap}}{E_{wave}}$$=$$\frac{A_{wave}}{A_{lap}}$$=$$\frac{2}{4}$$= $$\frac{1}{2}$$

$$\frac{E_{lap}}{230}$$ = $$\frac{1}{2}$$

$$E_{lap}$$ = $$\frac{230}{2}$$=115 V

When A doubles  , I also doubles ,  So I_{wave} = 2I_{wave}

$$P_{lap}$$= $$E_{lap}*I_{lap}$$= $$\frac{E_{wave}}{2}* 2I_{wave}$$= $$E_{wave} * I_{wave}$$

P_{lap} = P_{wave} = 5KW