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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

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