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GATE 2015 EE - SET 1 Question 48

Solution :

 E_{b}\alpha N   , \frac{E_{b1}}{E_{b2}} = \frac{N_{1}}{N_{2}}

N_{1} = 1000 rpm , E_{b1} = 200V ,R_{a} = 1Ω

N_{1} = 500 rpm , T_{3} = 0.5 T_{2} , N_{3} = 520 rpm

\frac{E_{b1}}{E_{b2}} = \frac{N_{1}}{N_{2}}

\frac{200}{E_{b2}} = \frac{1000}{500}

E_{b2}  = 100 V

\frac{E_{b1}}{E_{b3}} = \frac{N_{1}}{N_{3}}

\frac{200}{E_{b3}} = \frac{1000}{520}

E_{b3} = 104 V

E_{b}=V- I_{a}R_{a}
E_{b2} = V - I_{a2}R_{a}
100 = V _ I_{a2}(1)
100 = V - I_{a2} ----------(1)

E_{b3} = V - I_{a3}R_{a}

104 = V - I_{a3} --------(2)

T α I

\frac{T_{2}}{T_{3}} = \frac{I_{a2}}{I_{a3}}

\frac{T_{2}}{0.5T_{2}} = \frac{I_{a2}}{I_{a3}}

I_{a3}=0.5I_{a2} ----------(3)

I_{a3} = \frac{I_{a2}}{2}

substitute (3) into (2)

104 = V - I_{a3}

104 = V - 0.5  I_{a2} --------------(4)

Solving (1) and (4)

100 = V - I_{a2}

104 = V - 0.5 I_{a2}

I_{a2} = 8A.

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