Question #227251

A 440 volt D.C. shunt motor has an armature resistance of 0.8 ohm and field resistance of 200 ohm. If motor is taking 20 Amp current from supply then the back e.m.f. will be

Expert's answer

Given Data:-

Supply Voltage V = 440 V

Line Current ILI_L = 20A

Shunt Resistance RshR_{sh} = 200Ω

Armature Resistance RaR_a = 0.8 Ω

Armature Current IaI_a =?

Back EMF EbE_b =?

Armature current

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

IL=Ia+IshI_L=I_a+I_{sh}

And Shunt Field Resistance is given as

Ish=VRsh=440200=2.2AI_{sh}=\frac{V}{R_{sh}}=\frac{440}{200}=2.2A

∴ Ia=IL−Ish=20−2.2=17.8AI_a=I_L-I_{sh}=20-2.2=17.8A


Back EMF

The Back EMF of DC Motor is given by

Eb=V−IaRa=440−17.8×0.8=440−14.24=425.76VE_b=V-I_aR_a=440-17.8\times 0.8=440-14.24=425.76V


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