217 Views

In a PMDC motor, when armature is rotated at 100rpm, an emf of 2V is developed across its terminals. For this motor, calculate the armature current flow required to generate a torque of 0.4Nm.


nivedithapdvg 20th Oct, 2020
Answer (1)
Priya 26th Oct, 2020

Required torque (T) = 0.4 Nm

Eb = 2V

Mechanical Power delivered = P = T*w = Eb* Ia

Eb=back emf

Ia = Amateur current

w = Angular frequency

w = 2pi * n /60 = 2pi*100/ 60 = 10.47 rad/seconds

from the previous equation

Ia = T*w/Eb = 0.4 * 10.472 / 2 = 2.0944 A

Related Questions

UPES Integrated LLB Admission...
Apply
Ranked #28 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS University Rankings | 16.6 LPA Highest CTC
Jindal Global Law School Admi...
Apply
Ranked #1 Law School in India & South Asia by QS- World University Rankings | Merit cum means scholarships | Application Deadline: 31st Jan'25
Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities In QS Asia Rankings 2025 | Scholarships worth 210 CR
Great Lakes PGPM & PGDM 2025
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 1st Dec 2024
ISBR Business School PGDM Adm...
Apply
180+ Companies | Highest CTC 15 LPA | Average CTC 8 LPA | Ranked as Platinum Institute by AICTE for 6 years in a row | Awarded Best Business School...
Nirma University Law Admissio...
Apply
Grade 'A+' accredited by NAAC
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books