A 4-pole DC generator having wave wound armature winding has 51 slots, each containing 20 conductors. If the machine is running at 1500 rpm and flux per pole is 7 mWb, the voltage generated in the machine is:

This question was previously asked in
JSSC JE Re-Exam Official Paper-I (Held On: 04 Nov, 2022)
View all JSSC JE Papers >
  1. 363 V
  2. 360 V
  3. 255 V
  4. 357 V

Answer (Detailed Solution Below)

Option 4 : 357 V
Free
JSSC JE Full Test 1 (Paper 1)
5.7 K Users
120 Questions 360 Marks 120 Mins

Detailed Solution

Download Solution PDF

Calculation:

No. of slots = 51

Each slots contains 20 conductors.

Total number of conductors, Z = 51× 20 = 1020

Flux per pole, \(\phi = 7\ mWb\)

For wave winding, A = 2

Emf generated, \(E = \frac{{\phi ZNP}}{{60A}}\)

\(= \frac{{4\times7 \times {{10}^{ - 3}} \times 1020 \times 1500 }}{{60 \times 2}}\)

= 357 V

The correct answer is option "4"

Latest JSSC JE Updates

Last updated on Sep 23, 2024

-> JSSC JE Additional Result has been released for the Jharkhand Diploma Level Combined Competitive Examination-2023 (Regular and Backlog Recruitment) This is for the Advertisement No. 04/2023 and 05/2023. 

-> The JSSC JE notification was released for 1562 (Regular+ Backlog) Junior Engineer vacancies.

-> Candidates applying for the said post will have to appear for the Jharkhand Diploma Level Combined Competitive Examination (JDLCCE).

-> Candidates with a diploma in the concerned engineering branch are eligible for this post. Prepare for the exam with JSSC JE Previous Year Papers.

More Generator EMF Equation Questions

Get Free Access Now
Hot Links: teen patti master plus teen patti master 2024 mpl teen patti teen patti tiger teen patti wealth