- #1

Nusc

- 753

- 2

V(t)=IR(1-e^{t/RC}) \rightarrow ln|IR - V(t)| = ln|IR|-t/RC

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The first equation below has units of voltage.

the logarithm has units of voltage, but what does ln|IR| and ln|V| physically mean?

The purpose of this question is that if you plot the natural logarithm of the voltage versus time, it should yield a straight line with slope -1/RC and intercept ln|V_c|.

From which you can determine the capacitance given a known resistance.