Instantaneous Values of Capacitor

On occasion it will be necessary to determine the voltage or current at a particular instant of time that is not an integral multiple of $\tau$, as in the previous sections. For example, if
$$v_C = 20(1 - e ^{-t/(2 \times 10^{-3}})$$
the voltage $v_C$ may be required at t = 5 ms, which does not correspond to a particular value of $\tau$. [Fig. 1] reveals that $(1 - e^{-t/\tau})$ is approximately $0.93$ at $t = 5 ms = 2.5\tau$, resulting in $v_C = 20(0.93) =18.6 V$.
Additional accuracy can be obtained simply by substituting $t = 5 ms$ into the equation and solving for $v_C$. Thus,
$$ \begin{split} v_C &= 20(1 - e^{-5ms/2ms}) \\ &= 20(1 - e^{-2.5})\\ &= 20(1 - 0.082)\\ &= 20(0.918)\\ &= 18.36 V \end{split}$$
Fig. 1: Universal time constant chart.
The results are close, but accuracy beyond the tenths place is suspect using [Fig. 1]. The above procedure can also be applied to any other equation introduced in this chapter for currents or voltages.

Time to reach a particular voltage or current

There are also occasions when the time to reach a particular voltage or current is required. The procedure is complicated somewhat by the use of natural logs ($log_e$, or $ln$), but today's calculators are equipped to handle the operation with ease.
There are two forms that require some development. First, consider the following sequence:
$$\begin{split} v_C &= E(1 - e^{-t/\tau}) \\ {v_C \over E} &= (1 - e^{-t/\tau}) \\ 1- {v_C \over E} &= e^{-t/\tau} \\ \log_e(1- {v_C \over E}) &= \log_e(e^{-t/\tau}) \\ \log_e(1- {v_C \over E}) &= {-t \over \tau} \\ {t \over \tau} &= - \log_e(1- {v_C \over E}) \\ t &= -\tau \, \log_e(1- {v_C \over E}) \end{split}$$
$$- \log_e({x \over y}) = + \log_e({y \over x}) $$
$$ \bbox[10px,border:1px solid grey]{t = \tau \, \log_e({E \over E - v_C })} $$

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