armature conductors and running at 500 r.p m. supplies a load of emf 260V determine the flux in the generator Ex02: A 4-pole d.c. generator runs at 750 r.p.m. and the armature is

To determine the flux in the generator, we can use the formula for the generated emf (E) in a DC generator:<br /><br />\[ E = \frac{N \cdot \phi \cdot Z}{120} \]<br /><br />Where:<br />- \( E \) is the generated emf (260V)<br />- \( N \) is the speed of the armature in revolutions per minute (500 r.p.m.)<br />- \( \phi \) is the flux per pole<br />- \( Z \) is the number of armature conductors<br /><br />We need to rearrange the formula to solve for the flux (\( \phi \)):<br /><br />\[ \phi = \frac{E \cdot 120}{N \cdot Z} \]<br /><br />Given:<br />- \( E = 260 \) V<br />- \( N = 500 \) r.p.m.<br />- \( Z = 4 \) (assuming a standard number of conductors for simplicity)<br /><br />Now, plug in the values:<br /><br />\[ \phi = \frac{260 \cdot 120}{500 \cdot 4} \]<br /><br />\[ \phi = \frac{31200}{2000} \]<br /><br />\[ \phi = 15.6 \text{ Wb} \]<br /><br />Therefore, the flux in the generator is 15.6 Webers.