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**1. Homework Statement**

when running at its rated load, a 40kW series motor takes 74A and 550V. The rated speed is 750 rev/min. the armature and field resistances are 0.35 Ω and 0.15 Ω respectivly. when the load torque is doubled the rated value, the current is 110A. dertermin the moror speed and the power output at 100% torque overload.

Answers given, worked solution not

ans = 538 Rev/min and 57.4kW

**2. Homework Equations**

Work= Torque x (2pi x speed over 60)

torque = K(t) I(a) ψ (k(t) =constant and I(a)= armature current)

EMF =KNψ (k= constant)

EMF= v-IR

**3. The Attempt at a Solution**

Work= Torque x (2pi x speed over 60)

40000= T x ((2pi x 750) / 60)

T=509.29

EMF =KNψ (condition 1)

EMF =KNψ (condition 2)

divide both and cancel

k is constant as its the same motor

flux is constant (as flux is not given a value, i assume that its the same)

EMF(1) over EMF(2) = N(1) over N(2)

put both to the power -1 so as to find N(2) easier

EMF(2) over EMF(1) = N(2) over N(1)

((550 - (74 x 0.35+0.15)) over (550 - (74 x 0.35+0.15))) x 750 = N(2)

N(2) = 723.6 rev/min which is not the value of the answer

seeing as i can't get the speed right i can't go onto finding the power developed, however using the speed provided in the answer i have done it and correctly

Work= 2Torque x (2pi x speed over 60)

Work= 2Torque x (2pi x 538 over 60)

=57386W

=57.4kW