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The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, $V_{\text { rms. }}$ However, the number of interest for analyzing currents driven by the cell is the peak-to-peak voltage difference $\left(V_{\mathrm{pp}}\right),$ that is, the voltage difference between the two extremes of the electrodes excursion. What is the value of $V_{\mathrm{pp}}$ in terms of $V_{\mathrm{rms}} ?$A. $V_{\mathrm{rms}} / \sqrt{2} \quad \mathrm{B} . V_{\mathrm{rms}} / 2 \sqrt{2}$ C. $\sqrt{2} V_{\mathrm{rms}} \quad$ D. 2$\sqrt{2} V_{\mathrm{rms}}$

$V_{\mathrm{PP}}=2 \sqrt{2} V_{\mathrm{ms}}$the option $(\mathrm{D})$ is correct.

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

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in this problem where you want the peak to peak voltage in terms of the armistice voltage. Now, in a typical figure, we have the amplitude of the voltage, which is that today, by this capital V here. And we know it's related to the voltage are missed by this morning. So the whole to jar mess is thie amplitude of the voltage over there to hear. And then, based on a figure, the voltage peak to peak is equal to two times B and we want the voltage peak to peak in terms of the voltage, our mess. So let me take this formula and sell for V. I get that just by multiplying by her too. And now I know obvious We're going to take this and some student there. And so I get bolted. Peak to peak is equal to two times two. That was a voltage armas. And this ends up being answered Choice D, and that completes it

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