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d.c. shunt motor

Hello Guys,
im stuck in the question below about electricity and cant find a solution for it. I've also attached my workings. Im not sure if this is a right thread to post this but here i go :
The output of a d.c. shunt motor when supplied at 200V and running at 960rev/min is 2.5kW. The supply current for this load condition is 14A. The armature and field resistances are 0.4Ω and 160Ω respectively. Calculate for this condition
(i)the armature current
(ii) back e.m.f
Thanks Guys.
Original post by Alen.m
Hello Guys,
im stuck in the question below about electricity and cant find a solution for it. I've also attached my workings. Im not sure if this is a right thread to post this but here i go :
The output of a d.c. shunt motor when supplied at 200V and running at 960rev/min is 2.5kW. The supply current for this load condition is 14A. The armature and field resistances are 0.4Ω and 160Ω respectively. Calculate for this condition
(i)the armature current
(ii) back e.m.f
Thanks Guys.


Yes, you can post engineering / electrical / electronic problems in here.

The first thing to note is the total current:

Itot=Iarm+Ifield=14AmpsI_{tot} = I_{arm} + I_{field} = 14 \mathrm {Amps}

and the field current

Ifield=ERfield=200160=1.25AmpsI_{field} = \frac{E}{R_{field}} = \frac{200}{160} = 1.25 \mathrm {Amps}

rearranging the first eq

Iarm=ItotIfieldI_{arm} = I_{tot} - I_{field}

Iarm=141.25=12.75AmpsI_{arm} = 14 - 1.25 = 12.75 \mathrm {Amps}


For the back e.m.f.

Eb=EEarmE_{b} = E - E_{arm}

Earm=IarmRarmE_{arm} = I_{arm}R_{arm}

Earm=12.75x0.4=5.1VE_{arm} = 12.75 \mathrm {x}0.4 = 5.1 \mathrm {V}

substituting

Eb=2005.1=194.9VE_{b} = 200 - 5.1 = 194.9 \mathrm {V}
(edited 6 years ago)
Reply 2
Original post by uberteknik
Yes, you can post engineering / electrical / electronic problems in here.

The first thing to note is the total current:

Itot=Iarm+Ifield=14AmpsI_{tot} = I_{arm} + I_{field} = 14 \mathrm {Amps}

and the field current

Ifield=ERfield=200160=1.25AmpsI_{field} = \frac{E}{R_{field}} = \frac{200}{160} = 1.25 \mathrm {Amps}

rearranging the first eq

Iarm=ItotIfieldI_{arm} = I_{tot} - I_{field}

Iarm=141.25=12.75AmpsI_{arm} = 14 - 1.25 = 12.75 \mathrm {Amps}


For the back e.m.f.

Eb=EEarmE_{b} = E - E_{arm}

Earm=IarmRarmE_{arm} = I_{arm}R_{arm}

Earm=12.75x0.4=5.1VE_{arm} = 12.75 \mathrm {x}0.4 = 5.1 \mathrm {V}

substituting

Eb=2005.1=194.9VE_{b} = 200 - 5.1 = 194.9 \mathrm {V}

Thanks very much for your reply.

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