What will be the peak value of a sinusoidal voltage whose RMS value is 115√2 volts?

This question was previously asked in
MPPGCL JE Electrical 28 April 2023 Shift 2 Official Paper
View all MPPGCL Junior Engineer Papers >
  1. 345 V
  2. 230 V
  3. 460 V
  4. 115 V

Answer (Detailed Solution Below)

Option 2 : 230 V
Free
MPPGCL JE Electrical Full Test 1
3.9 K Users
100 Questions 100 Marks 120 Mins

Detailed Solution

Download Solution PDF

Concept

The RMS value of an alternating current is the steady (D.C) current that produces the same amount of heat as the A.C. 

The RMS value of any signal is given by:

\(RMS=\sqrt{{1\over T}\int_{-\infty}^{\infty}x^2(t)\space dt}\)

where, T = Time period of the signal

The relationship between RMS and peak value for an AC supply is given by:

\(V_{RMS}={V_m\over \sqrt{2}}\)

where, Vm = Peak value (maximum value)

Calculation

Given, VRMS = 115√2 V

\(115\sqrt{2}={V_m\over \sqrt{2}} \space \)

Vm = 230 V

Latest MPPGCL Junior Engineer Updates

Last updated on May 29, 2025

-> MPPGCL Junior Engineer result PDF has been released at the offiical website.

-> The MPPGCL Junior Engineer Exam Date has been announced.

-> The MPPGCL Junior Engineer Notification was released for 284 vacancies.

-> Candidates can apply online from 23rd December 2024 to 24th January 2025.

-> The selection process includes a Computer Based Test and Document Verification.

-> Candidates can check the MPPGCL JE Previous Year Papers which helps to understand the difficulty level of the exam.

Get Free Access Now
Hot Links: teen patti star apk teen patti king lotus teen patti teen patti vungo teen patti 500 bonus