GATE 2015 EE – SET 2 – Question 48

Solution :

$$V_{a1}$$ = 100V , $$N_{1}$$ = 1000 rpm , $$I_{a1} $$= 10A ,$$R_{a}$$ = 1Ω

$$\tau _{L} \alpha N^{2}$$ , $$V_{a2}$$= ? ,$$ N_{2}$$ = 500 rpm , $$E_{b} = V – I_{a}R_{a}$$

$$E_{b1}$$= $$V_{a1}-I_{a1}{R_{a}}$$
$$E_{b1}$$ = 100-(10)(1)=90V at $$N_{1}$$

$$E_{b1}$$$$ \alpha N$$$$ \Rightarrow$$ seperately excited DC motor $$\Rightarrow$$$$ \phi$$ to be constant.

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

$$E_{b2}$$ = $$\frac{500}{1000}$$(90) = 45 V

$$I_{a}$$$$\alpha$$$$ \tau $$$$\Rightarrow $$$$\phi$$ is constant.

$$I_{a}$$$$\alpha$$ $$N^{2}$$ $$\Rightarrow$$ $$\frac{I_{a_{1}}}{I_{a_{2}}}$$ = $$\frac{N_{1}^{2}}{N_{2}^{2}}$$

$$I_{a2}$$=$$(\frac{N_{2}}{N_{1}})^{2} I_{a1}$$ =

$$ (\frac{500}{1000})^{2}$$10 = 2.5 A

$$E_{b}$$ = $$V_{a} – I_{a}R_{a}$$

$$V_{a_{2}}$$ =$$E_{b_{2}} + I_{a2}R_{a}$$

$$V_{a_{2}}$$ = 45 + (2.5)(1) = 47.5 v