Question
Download Solution PDFWhat is the average value of a sinusoidal wave having maximum voltage value of 520V?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFRepresentation of a sinusoidal wave
In the figure, A represents the maximum value and T represents the time period.
The average value of a sine wave is given by:
\(V_{o(avg)}={2V_m\over \pi}\)
The RMS value of a sine wave is given by:
\(V_{o(RMS)}={V_m\over \sqrt{2}}\)
Calculation
Given, Vm = 520 V
\(V_{o(avg)}={2\times 520\over \pi}\)
Vo(avg) = 330.72 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.