The voltage drop across the capacitor rises from 0 to ℰ. Note that ℰ is never actually known in the measurement. In fact, the oscilloscope voltage is decalibrated, so that, whatever ℰ is, ℰ is at the top line while zero is at the bottom line. We don't measure voltage levels, but rather 1/2, 1/4, and 1/8 the maximum.Kirchhoff's voltage law give: ℰ = IR + Q/C or the following:dQdt=−1RC(Q−EC)dQdt=−1RC(Q−ℰC) The solution for the capacitor voltage isVC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC) Calculate the theoretical half-life (in μs) of the circuit. The quantities are:R = 0.7 kΩC = 6.3 μF R,3.

Answered: The voltage drop across the capacitor…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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The voltage drop across the capacitor rises from 0 to ℰ. Note that ℰ is never actually known in the measurement. In fact, the oscilloscope voltage is decalibrated, so that, whatever ℰ is, ℰ is at the top line while zero is at the bottom line. We don't measure voltage levels, but rather ¹/₂, ¹/₄, and ¹/₈ the maximum.

Kirchhoff's voltage law give: ℰ = IR + ^Q/_C or the following:

dQdt=−1RC(Q−EC)dQdt=−1RC(Q−ℰC)

The solution for the capacitor voltage is

VC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC)

Calculate the theoretical half-life (in μs) of the circuit. The quantities are:

R = 0.7 kΩ
C = 6.3 μF

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