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

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter7: Energy Of A System
Section: Chapter Questions
Problem 7.9P
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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 1/21/4, 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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