# 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

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

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

$$I_{a3}$$=0.5$$I_{a2}$$ ———-(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.