Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A simple electric motor has an armature resistance of 1Ω and runs from a dc source of 12 volt. When running unloaded it draws a current of 2A. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn (in ampere)?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 7.00.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Let initial e.m.f. induced =e∴ Initial current i=E−eR ⇒ 2=12−e1⇒e=12−2=10 V As e ∝ω when speed is halved, the value of induced e.m.f. becomes ⇒e2=102=5 V∴ New value of current =i'=E−eR=12−51=7 A
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
personalised 1:1 online tutoring