# Monthly Archives: June 2017

## 5th International Conference on Advances in Science, Engineering, Technology and Natural Resources BANGKOK,August 4-5, 2017,Bangkok, Thailand.

5th International Conference on Advances in Science, Engineering, Technology and Natural Resources BANGKOK,August 4-5, 2017,Bangkok, Thailand. This conference is a joint conference with RMUTT (Rajamangala University of Technology Thanyaburi) Thailand conference link on RMUTT website is: http://research.en.rmutt.ac.th/?p=3840 Relevant Selective papers will be … Continue reading

## IEC 61850 Europe 2017, 26-28 Sep 017, Amsterdam, Netherlands

IEC 61850 Europe 2017, 26-28 Sep 017, Amsterdam, Netherlands Now firmly established as the European end-user forum for IEC 61850 experts and implementation leaders, this dedicated 3-day conference, exhibition and networking forum provides the information, inspiration, and connections you need … Continue reading

## GATE 2015 EE – SET 2 – Question 47

Solution : $$N_{1}$$ = 100 , $$N_{2}$$=50 ,$$N_{3}$$ = 50 $$I_{2}$$ = 2$$\angle 30^{0}$$ $$I_{3}$$ = 2$$\angle 150^{0}$$ $$I_{1}$$ = ? I $$\alpha$$$$\frac{1}{N}$$ $$\frac{I_{A}}{I_{B}}$$ = $$\frac{N_{B}}{N_{A}}$$ $$I_{A}$$ = $$\frac{N_{B}}{N_{A}} * I_{B}$$ $$I_{1}$$ = $$\frac{N_{2}}{N_{1}}*I_{2} + \frac{N_{3}}{N_{1}}*I_{3}$$ … Continue reading

## GATE 2015 EE – SET 2 – Question 46

Solution : For minimum equivalent inductance , the coils are  connected in parallel, Each inductance = L, Mutual inductance = 0 , Equivalent  inductance = α L $$L_{eq}$$=$$\frac{(L)(L)}{L+L}$$=$$\frac{L^{2}}{2L}$$ = $$\frac{L}{2}$$ = 0.5 L $$\therefore$$$$\alpha$$ = 0.5

## 4th International Conference on Innovations in Engineering, Technology, Computers and Industrial Applications

4th International Conference on Innovations in Engineering, Technology, Computers and Industrial Applications  4th International Conference on Innovations in Engineering, Technology, Computers and Industrial Applications (IETCIA-2017) 3rd to 4th August 2017, Pattaya, Thailand. Abstract/Poster/Full Paper Submission Dates: June 26, 2017 FEE: Student–USD 195; Academician–USD … Continue reading

## GATE 2015 EE – SET 2 – Question 41

Solution : GMR = KR GMR = $$\left [ (D_{11}D_{12}D_{13}) (D_{21}D_{22}D_{23})(D_{31}D_{32}D_{33})\right ]^{\frac{1}{m*n}}$$ GMR = KR GMR =$$\left [ (3R*r*0.7788R)^{3}\right ]^{\frac{1}{3*3}}$$ $$\rightarrow$$$$\left [ (0.7788*9*R^{3})^{3} \right ]^{\frac{1}{9}}$$ $$\Rightarrow$$ 1.914R

## GATE 2015 EE – SET 2 – Question 40

Solution : $$\frac{dc_{1}}{dp_{1}}$$ = 0.2$$P_{1}$$+ 50 $$\frac{dc_{1}}{dp_{1}}$$= 76$$\frac{Rs}{MWh}$$ 0.2$$P_{1}$$+ 50 = 76 $$P_{1}$$= 130 MW $$\frac{dc_{2}}{dP_{2}}$$= 0.24$$P_{2}$$ + 40 $$\frac{dc_{2}}{dP_{2}}$$= 68.8$$\frac{Rs}{MWh}$$ 0.24$$P_{2}$$+ 40 = 68.8 $$P_{2}$$= 120 MW $$P_{1} + P_{2}$$ … Continue reading