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Chapter 28: Problem 69

A triangle wave swings linearly between voltages \(-V_{\mathrm{p}}\) and \(+V_{\mathrm{p}}\) Show that the rms voltage of a triangle wave is \(V_{\mathrm{p}} / \sqrt{3}\)

Short Answer

Expert verified
The rms voltage of a triangle wave is \(V_{p} / \sqrt{3}\)

Step by step solution

01

Defining the RMS Voltage

Root Mean Square (RMS) voltage is defined as the square root of the mean of the square of the function. The expression for RMS voltage, \( V_{rms} \), is given by \( V_{rms} = \sqrt {\frac{1}{T} \int_{0}^{T} [f(t)]^2 dt} \) where \( f(t) \) represents any periodic function and T is the time period.
02

Defining Triangle Wave Function

A triangle wave function varies linearly with time and swings between \( -V_p \) and \( +V_p \). It can be defined as \( f(t) = \frac{4V_p}{T} |t - \frac{T}{2}| - V_p \), where \( T \) is the time period i.e. the time taken by waveform to complete one cycle and \( V_p \) is the peak voltage.
03

Substituting in RMS Equation

Substitute the triangle wave function \( f(t) \) into the RMS voltage expression. Thus, we have: \( V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} [(\frac{4V_p}{T} |t - \frac{T}{2}| - V_p) ^2] dt} \)
04

Evaluating the Integral

The square of the absolute value of \( t-\frac{T}{2} \) can be written as \( (\frac{4V_p}{T}t - 2V_p)^2 \). Thus, the integral becomes: \( V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} [ (\frac{4V_p}{T}t - 2V_p) ^2 ] dt} \) . Using integral calculus, this evaluates to \( \frac{V_p^2}{3} \)
05

Taking the Square Root

Taking the square root of both sides gives the RMS voltage of the triangle wave as \( V_{rms} = \frac{V_p}{\sqrt{3}} \)

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Most popular questions from this chapter

A 15 - \(\mu\) F capacitor carries \(1.4 \mathrm{A}\) rms. What's its minimum safe voltage rating if the frequency is (a) \(60 \mathrm{Hz}\) and (b) \(1.0 \mathrm{kHz} ?\)

What's meant by the statement, "A capacitor acts like a DC open circuit"?

An AC voltage of fixed amplitude is applied across a series \(R L C\) circuit. The components are such that the current at half the resonant frequency is half the current at resonance. Show that the current at twice the resonant frequency is also half the current at resonance.

Find the rms current in a 1.0 - \(\mu \mathrm{F}\) capacitor connected across \(120-\mathrm{V} \mathrm{rms}, 60-\mathrm{Hz}\) AC power.

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