Alen.m
Badges: 11
Rep:
?
#1
Report Thread starter 2 years ago
#1
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.
Attached files
0
reply
uberteknik
  • Study Helper
Badges: 21
Rep:
?
#2
Report 2 years ago
#2
(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:

I_{tot} = I_{arm} + I_{field} = 14 \mathrm {Amps}

and the field current

I_{field} = \frac{E}{R_{field}} = \frac{200}{160} = 1.25 \mathrm {Amps}

rearranging the first eq

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

I_{arm} = 14 - 1.25 = 12.75 \mathrm {Amps}


For the back e.m.f.

E_{b} = E - E_{arm}

E_{arm} = I_{arm}R_{arm}

E_{arm} = 12.75 \mathrm {x}0.4 = 5.1 \mathrm {V}

substituting

E_{b} = 200 - 5.1 = 194.9 \mathrm {V}
0
reply
Alen.m
Badges: 11
Rep:
?
#3
Report Thread starter 2 years ago
#3
(Original post by uberteknik)
Yes, you can post engineering / electrical / electronic problems in here.

The first thing to note is the total current:

I_{tot} = I_{arm} + I_{field} = 14 \mathrm {Amps}

and the field current

I_{field} = \frac{E}{R_{field}} = \frac{200}{160} = 1.25 \mathrm {Amps}

rearranging the first eq

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

I_{arm} = 14 - 1.25 = 12.75 \mathrm {Amps}


For the back e.m.f.

E_{b} = E - E_{arm}

E_{arm} = I_{arm}R_{arm}

E_{arm} = 12.75 \mathrm {x}0.4 = 5.1 \mathrm {V}

substituting

E_{b} = 200 - 5.1 = 194.9 \mathrm {V}
Thanks very much for your reply.
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Bournemouth University
    Undergraduate Open Day Undergraduate
    Wed, 19 Feb '20
  • Buckinghamshire New University
    Postgraduate and professional courses Postgraduate
    Wed, 19 Feb '20
  • University of Warwick
    Warwick Business School Postgraduate
    Thu, 20 Feb '20

People at uni: do initiations (like heavy drinking) put you off joining sports societies?

Yes (166)
66.4%
No (84)
33.6%

Watched Threads

View All