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Chapter 28: Problem 74

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 (Vrms) of a triangle wave is \( V_{p} / \sqrt{3} \)

Step by step solution

01

Understand the Root Mean Square (RMS)

The root mean square of a waveform (in our case a triangle wave) is a mathematical tool that provides a measure of the magnitude of the waveform. The RMS voltage of a waveform is calculated as the square root of the mean value of the squares of the waveform voltage variations over a period. The formula for RMS value is: \( V_{rms}= \sqrt{ \frac{1}{T} \int_ {0}^ {T} [f(t)]^ {2} dt}\) where \(f(t)\) is the function describing the waveform over one period \(T\).
02

Understand the Triangle Wave Function

A triangle wave is a non-sinusoidal waveform where the voltage swings linearly between high (+Vp) and low (-Vp) extreme values. We can divide a triangle wave into two components based on its swing, allowing us to compute two integrals: \[V_{rms}^ {2} = \frac{1}{T} \int_ {0}^ {T/2} [Vp t/ (T/2)]^ {2} dt + \frac{1}{T} \int_ {T/2}^ {T} [- Vp t/ (T/2) + Vp]^ {2} dt\]
03

Perform the Integration

The integral can now be separated and be performed in order to make implementation better: \[V_{rms}^ {2} = \frac{1}{T} [\int_ {0}^ {T/2} (Vp^ {2} t^ {2} / T) dt + \int_ {T/2}^ {T} [Vp^ {2} t^ {2} / T - 2 Vp^ {2} t + Vp^ {2}] dt]\]
04

Calculate the RMS Value

Solving these equations, we will get the rms voltage of a triangle wave: \( V_{rms} = V_{p} / \sqrt{3}\)

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Most popular questions from this chapter

You're concerned about a circuit that will be used in a remote communications installation. The series RLC circuit with \(R=5.5 \Omega, L=180 \mathrm{mH},\) and \(C=0.12 \mu \mathrm{F}\) is connected across a sine-wave generator. The inductor can safely handle 1.5 A of current. The peak generator output when it's tuned to resonance will be \(8.0 \mathrm{V} .\) Will the inductor current stay within a safe limit?

For \(R L C\) circuits in which the resistance isn't too high, the \(Q\) factor may be defined as the ratio of the resonant frequency to the difference between the two frequencies where the power dissipated in the circuit is half the power dissipated at resonance. Using suitable approximations, show that this definition leads to \(Q=\omega_{0} L / R,\) with \(\omega_{0}\) the resonant frequency.

A series \(R L C\) circuit with \(R=47 \Omega, L=250 \mathrm{mH},\) and \(C=\) \(4.0 \mu \mathrm{F}\) is connected across a sine-wave generator whose peak output voltage is independent of frequency. Find the frequency range over which the peak current will exceed half its value at resonance.

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.

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