Question

(a) What is back emf? Derive the relation for the back emf and the supplied voltage in terms of armature resistance. (6)

(b) An 8kw, 230V, 1200 rpm d.c shunt motor has Ra = 0.7W. The field current is adjusted until, on no-load with a supply of 250V, the motor runs at 1250 rpm and draws armature current of 1.6 amps. A load torque is then applied to the motor shaft which causes it to raise to 40 A and the speed falls to 1150 rpm.Determine the reduction in the flux per pole due to the armature reaction. (10)

Answer 1

Part (a)

Back electromotive force (back EMF, Eb) is the voltage generated in the armature of a DC motor when it rotates and cuts the magnetic flux. By Fleming's Right-Hand Rule, this induced emf opposes the applied voltage V, as per Lenz's law. Back EMF acts as a self-regulating mechanism that limits the armature current when the motor is running.

Consider a shunt motor:

$$V\:=\:Applied\:voltage$$

$$I\:=\:Current\:flowing\:through\:the\:circuit$$

$$R_{a}\:=\:Armature\:resistance$$

$$R_{sh}\:=\:Shunt\:field\:resistance$$

$$I_{sh}\:=\:Shunt\:field\:current$$

$$E_{b}\:=\:Back\:EMF$$

$$Net\:voltage\:across\:Armature\:=\:V-E_{b}$$

$$Current\:=\:\frac{V}{R}$$

$$Therefore,\:I_{a}\:=\:\frac{V-E_{B}}{R_{a}}$$

$$I_{a}R_{a}\:=\:V-E_{b}$$

$$E_{b}\:=\:V-I_{a}R_{a}$$

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