# Why is a logarithmic best-fit trendline utilized in plot of capacitor voltage versus time?

### 1 Answer

- Anonymous5 years agoBest Answer
As an example, consider the "exponential discharge" of a capacitor (C) from an initial (maximum) voltage of (V) through a resistor (R):

As the discharge current flows, the charge on the capacitor reduces thus the voltage across the capacitor (v) reduces, thus the resistor voltage and hence the discharge current also reduce, so that the magnitude of the rate of change of capacitor voltage reduces with time

i.e. the gradient ( dv/dt ) of the graph is continually changing.

the capacitor voltage ( v ) is given by the non-linear exponential function:

v = V e ^ ( - t / C R )

[ NB note lower case v on LHS of eqn. and upper case V on RHS ]

see image in link below.

If we take natural logs of both sides of the exponential function, we get:

ℓn v = ℓn V - t / CR

which has form: y = a x + b

where

y = ℓn v

a = - 1/CR

x = t

b = ℓn V

thus plotting ℓn v against t can be expected (in principle) to give a straight line graph with (negative) gradient determined by the reciprocal of the time constant (CR) of the circuit.

If we obtain corresponding values of v and t experimentally, then plot ℓn v against time .....

(alternatively, plot v against t on "log-lin" graph paper - see link #2)

..... we can expect to get a set of plotted points closely displaced about an easily recognizable straight line graph (with points displaced some distance determined by the magnitude of the experimental error).

Source(s): http://cnx.org/resources/910358bd1d72420ba4fcdcf2f... http://www.mathnstuff.com/math/spoken/here/2class/...