SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

${\mathrm{N}}_{\mathrm{a}}=300,{\mathrm{I}}_{\mathrm{a}}=1.5\mathrm{A},{\mathrm{N}}_{\mathrm{b}}=600$, ${\mathrm{I}}_{\mathrm{b}}=4\mathrm{A},{\mathrm{N}}_{\mathrm{c}}=600,{\mathrm{I}}_{\mathrm{c}}=3\mathrm{A}$
Let us assume direction of flux in upward direction be +ve and downward direction be -ve
For Coil A, ${\mathrm{MMF}}_{\mathrm{a}}={\mathrm{N}}_{\mathrm{a}}{\mathrm{I}}_{\mathrm{a}}=300\times 1.5=450\mathrm{AT}(+)$
For coil B, ${\mathrm{MMF}}_{\mathrm{b}}={\mathrm{N}}_{\mathrm{b}}{\mathrm{I}}_{\mathrm{b}}=600\times 4=2400\mathrm{AT}(+)$
For coil C,${\mathrm{MMF}}_{\mathrm{c}}={\mathrm{N}}_{\mathrm{c}}{\mathrm{I}}_{\mathrm{c}}=600\times 3=1800\mathrm{AT}(-)$
Net $\mathrm{MMF}=450+2400-1800=1050\mathrm{AT}$
Note: The direction of flux for coil a and coil b are in same direction and in opposite direction for coil c.