he 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) The voltage is zero at t = 0, t is the rising time, and you have to know when the rising begins. Now, calculate VC (in dev) when t = RC ln 2.R = 0.6 kΩC = 2.3 μF

As given in the question : VC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC) At t = RC ln 2: Vc = ℰ(1-0.5) = 0.5…

Answered: he 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

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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he 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)

The voltage is zero at t = 0, t is the rising time, and you have to know when the rising begins.

Now, calculate V_C (in dev) when t = RC ln 2.

R = 0.6 kΩ
C = 2.3 μF

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