Question 9.4.1: Estimating Capacitance from the Free Response Commercially a...

The circuit model may be derived from Kirchhoff’s voltage law, which gives
v_{s} = Ri + v_{C}       or           i = \frac{v_{s}  −  v_{C}}{R}                 (1)
For the capacitor we have
v_{C} = \frac{1}{C}  \int   i dt       or         \frac{dv_{C}}{d t } = \frac{i}{C}

Substituting from equation (1) gives the circuit model.
\frac{d v_{C}}{d t} = \frac{v_{s}  −  v_{C}}{RC}        or      RC \frac{d v_{C}}{d t} + v_{C} = v_{s}
The free response has the form
v_{C}(t) = v_{C}(0)e^{−t/RC} = v_{C}(0)e^{−t/τ}
where the time constant is \tau = RC. Taking the natural logarithm of both sides gives
\ln v_{C}(t) = \ln v_{C}(0)  −  \frac{t}{\tau}            (2)
When the logarithmic transformation is applied to the original data, we obtain the following table

These transformed data are plotted in Figure 9.4.1c. Note that the data lie close to a straight line, which is given by \ln v_{C} = −1.621t + 1.574 . This line was found with the least-squares method, but a similar line could have been obtained by using a straightedge to draw a line through the data. The least-squares method is required when there is considerable scatter in the data.
Comparing the equation for the line with equation (2), we obtain \ln v_{C}(0) = 1.574, which gives v_{C}(0) = 4.825, and 1/\tau = 1.621, which gives \tau = 0.617. Because we know that R = 10^{5}  \Omega , we obtain C = \tau/R = 0.617/10^5 = 6.17 × 10^{−6}  F.