Question

A 250V ,dc shunt motor takes a line current of 20A . resistance of shunt field winding is 200 ohm amature has resistance of 0.3 ohm .find the armature and back emf

Answer 1

Given :

The line current for shunt motor = $I_L$ = 20 A

The resistance for shunt filed winding = $R_s_h$ = 200 ohm

The resistance for armature winding = $R_a$ = 0.3 ohm

Supply voltage = V = 200 Volt

To Find :

( a ) The armature current

( b )  The back emf

Solution :

Let The armature current = $I_a$

For DC shunt motor

Line current = armature current + shunt current

i.e $I_L$ = $I_a + I_s_h$

Or, 20 A = $I_a + I_s_h$           ...........1

∵ Shunt current = $I_s_h$ = $\dfrac{supply voltage}{shunt resistance}$

                                  = $\dfrac{V}{R_s_h}$

                                  = $\dfrac{200 v}{200 ohm}$

i.e  Shunt current = $I_s_h$ = 1 Amp

( a )

From eq 1

$I_a = 20 A - I_s_h$

Or, $I_a$ = 20 A - 1 A

         = 19 A

So, Armature current = $I_a$ = 19 amp

( b )

For Motor

  ∵  EMF = V - $I_a$ × $R_a$

              = 250 - 19 × 0.3

              = 250 - 5.7

              = 244.3 volt

i.e EMF = 244.3 volt

Hence,

( a ) Armature current is 19 amp

( b ) Back EMF is 244.3 volt            Answer

Answer 2

Answer:

Explanation:

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