1、电动势与磁通的关系 φ= pp sin at e,=-n. do ON,, cos at=√2E1sn(Ot-90°) e aton, m cos at=2E2 sin( at-90') ON10 cos ot=√2 E sin(Ot-90°) E1 oN, Dm-444N, n Q=2m=2×3.14×50 ON, 4.44 fN,O O Io=4.44 N,Dφ=Φm sinωt 1、电动势与磁通的关系 cos 2 sin( 90 ) 1 1 1 1  = − = − N t = E t − dt d e N   m    cos 2 sin( 90 ) 2 2 2 2  = − = − N t = E t − dt d e N   m    cos 2 sin( 90 ) 1 1 1 1 1 1  = − = − N t = E t − dt d e N   m        m m fN N E    1 1 1 4.44 2 = = m m fN N E    2 2 2 4.44 2 = = m m fN N E       1 1 1 1 1 4.44 2 = =  = 2f = 23.1450