# A 440-V shunt motor has armature resistance of 0.8 Ω and field resistant 200 Ω. What will be the back e.m.f when giving an output of 7.46 kW at 85 percent efficiency.

A 440-V shunt motor has armature resistance of 0.8 Ω and field resistant 200 Ω. What will be the back e.m.f when giving an output of 7.46 kW at 85 percent efficiency.

### Right Answer is: 425.8 V

#### SOLUTION

Given data

Supply Voltage V = 440 V
Armature Resistance Ra = 0.8Ω
Field Resistance Rsh = 200Ω
Motor Output ηout = 7.46 kW = 7460 Watt
Motor Efficiency η = 85% = 0.85

The efficiency of the motor is given as

η = Motor Output power ⁄ Motor Input power

Motor Input Power = Motor Output Power ⁄ Efficiency

= 7460 ⁄ 0.85 = 8776.4 Watt

Now the Motor Line current or the Input current IL = P ⁄ V

IL = 8776.4 ⁄ 440 = 19.95 A

The Line current of DC shunt motor is the sum of Armature current and Shunt field current

IL = Ia + Ish

And Shunt Field Resistance is given as

Ish = V/Rsh = 440/200 = 2.2 A

∴ Ia = IL − Ish

Ia = 19.95 − 2.2 = 17.75 A

The Back EMF of DC Motor is given by

Eb = V − IaRa

= 440 − 17.75 × 0.8
= 440 − 14.2

= 425.8 V

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