Question
Download Solution PDFA 230 V motor has an armature circuit resistance of 0.8 ohms. If the full load armature current is 25 A, find the amount of back EMF induced in the armature.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 1):(210 V )
Concept:
In a DC generator, generated emf is given by
Eg = Vt + IaRa
In a DC motor, back emf is given by
Eb = V – IaRa
Where,
Ia is the armature current
Ra is armature resistance
Calculation:
Armature resistance (Ra) = 0.8 ohm
Voltage (V) = 230 V
Armature current (Ia) = 25 A
Eb = V – IaRa = 230 – 25(0.8) = 230 - 20 = 210 V
Last updated on May 9, 2025
-> PGCIL Diploma Trainee result 2025 will be released in the third week of May.
-> The PGCIL Diploma Trainee Answer key 2025 has been released on 12th April. Candidates can raise objection from 12 April to 14 April 2025.
-> The PGCIL DT Exam was conducted on 11 April 2025.
-> Candidates had applied online from 21st October 2024 to 19th November 2024.
-> A total of 666 vacancies have been released.
-> Candidates between 18 -27 years of age, with a diploma in the concerned stream are eligible.
-> Attempt PGCIL Diploma Trainee Previous Year Papers for good preparation.