Question

A dc shunt motor develop torque of 20 n-m when the armature current is 40a if it is changed as commutative compound motor by adding a series field which increases 20% of flux the armature current is 60a then the torque develope is

Answer 1

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Answer 2

Given :

For a DC shunt motor

Toque develop = 20 n-m

Armature current = 40 amp

For commutative compound motor

%age flux increase = 20%

The armature current = $I_a$ = 60 amp

To Find :

The torque develop in commutative compound motor

Solution :

Let The torque develop  = T n-m

∵ Torque = k × flux × armature current

where k = constant

 For DC shunt motor

Torque = k × flux × armature current

20 n-m = k × ∅ × $I_a$

Or,   20 n-m = k × ∅ × 40

So,    k × ∅ = $\dfrac{20}{40}$

i.e    k × ∅ = 0.5

Or,        ∅ = 0.5

Again

For commutative compound motor

As the flux increase by 20% in commutative compound motor

So,     ∅' = 0.5 + 20% of 0.5

             = 0.5 + 0.2 × 0.5

             = 0.5 + 0.1 = 0.6

So, Flux for  commutative compound motor = ∅' = 0.6 wb

So, Toque develop =  ∅' × $I_a$'

Or,  T = 0.6 × 60

∴   Torque = 36 n-m

Hence, The torque develop for commutative compound motor is 36 n-m Answer