SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

E.M.F. generated by a generator is ${\mathrm{E}}_{\mathrm{a}}=\frac{\mathrm{\varphi }\mathrm{ZNP}}{60\mathrm{A}}$
As the number of parallel paths is not given take $\mathrm{A}=1$
${\mathrm{E}}_{\mathrm{a}}=\frac{\mathrm{\varphi }\mathrm{ZNP}}{60}$ .....(1)
${\mathrm{\omega }}_{\mathrm{m}}=$ speed in $\mathrm{rad}∕\sec$
${\mathrm{\omega }}_{\mathrm{m}}=\frac{2\mathrm{\pi }\mathrm{N}}{60}$
$\mathrm{N}=\frac{30{\mathrm{\omega }}_{\mathrm{m}}}{\mathrm{\pi }}$ ......(2)
$\mathrm{Z}=2\times {\mathrm{N}}_{\mathrm{c}}$ .......(3)
${\mathrm{N}}_{\mathrm{c}}=$ number of coil turns
From equations (1), (2) and (3)
${\mathrm{E}}_{\mathrm{a}}=\frac{\mathrm{\pi }\times 2{\mathrm{N}}_{\mathrm{c}}\times 30{\mathrm{w}}_{\mathrm{m}}\times \mathrm{P}}{60\times \mathrm{\pi }}$
${\mathrm{E}}_{\mathrm{a}}=\frac{\mathrm{\pi }{\mathrm{\omega }}_{\mathrm{m}}{\mathrm{N}}_{\mathrm{c}}\mathrm{P}}{\mathrm{\pi }}$