In order to determine the magnitude and direction of induced e.m.f, let us consider the different positions of the coil which has ‘N’ turns of wire. In one revolution following positions can be considered. When initially coil is vertical, the length arms AC and BD are moving parallel to the lines of force maximum number of lines link the coil, but rate of change of flux is zero, hence emf is zero. As the coil moves from this position, sides AC and BD begin to cut the lines of force and induced emf is setup till it is maximum when the coil is horizontal. It has rotated 90o, 1st quarter is completed. Further rotation decreases the value of emf, until at the end of 2nd quarter, when coil is vertical, it has rotated 180o, the emf is again zero. As the coil rotates further from position 3 to position 4, an emf is again induced, but in reverse direction, because AC and BD are cutting the magnetic lines in opposite direction. The reverse emf reaches to –ve peak value at the end of 3rd quarter. The coil is horizontal and angle of rotation is 270o. On further rotation, the emf again decreases and becomes zero when the coil returns back to its original position after rotating 360o. This shows that the coil of generator produces induced emf which reverse its direction 2*f times in one cycle. Where f = frequency of rotation of coil. EXPRESSION FOR EMF IN A.C. GENERATOR Consider a coil ABCDA of ‘N’ turns rotating in a uniform magnetic field B with a constant angular speed ‘w’. Let the length of the coil is ‘l’ and its breadth is ‘b’. To calculate emf in sides AC and BD we proceed as follows: Motional emf = BvlSinq Emf in side AC = BvlSinq = x1 Emf in side BD = BvlSinq = x2 Emf induced in the coil = x1 + x2 = BvlSinq + BvlSinq x = 2 BvlSinq If coil has ‘N’ turns, emf will increase N times x = 2 BvlNSinq ----(1) If angular velocity of coil is ‘w’ and it takes time ‘t’ to cover angle q then q = wt also V = rw and r = b/2 V = b/2w Putting the value of q and V in eq. (1) x = 2B (b/2w)lNsin(wt) x = wB(b*l)sin(wt) x = NBw(b.l)sinwt x = NBwAsinwt –(2)     For latest information , free computer courses and high impact notes visit : this is the expression for the induced emf in the coil of an A.C generator at any instant. If f = no. of rotation per sec. Then we have w=2pf x = V = NBwAsin(2pft) – (3) for maximum emf q = 90o or 270o or 2pft = p/2 or 3p/2 and sin90o = sin p/2 = +1 sin270o = sin3p/2 = -1 xo = Vo = NBwA(1) xo = Vo = +- NBwA +- = shows direction of induced current Relation b/w x and xo x = NBwAsin(2pft) x = xosin(2pft)


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according to FARDAY's law of electro magnetic induction, when a conductor moves in a magnetic field, a dynamically induced EMF is produced in the conductor. the magnitude of generated EMF can be given by EMF equation DC generated. if a closed path is provided to the moving conductor then generated emf causes a current to flow in the circuit . the output voltage waveform of a DC generated