Instantaneous Values

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 $$vC = 20(1 - e ^{-t/(2 \times 10^{-3}})$$ the voltage $v_C$ may be required at t = 5 ms, which does not correspond to a particular value of $\tau$. Figure 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, $$ \begin{split} v_C &= 20(1 - e^{-5ms/2ms}) \\ &= 20(1 - e^{-2.5})\\ &= 20(1 - 0.082)\\ &= 20(0.918)\\ &= 18.36 V \end{split}$$ 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 other voltages. 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 (loge, 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: $$\begin{split} v_C &= E(1 - e^{-t/\tau}) \\ {v_C \over E} &= (1 - e^{-t/\tau}) \\ 1- {v_C \over E} &= e^{-t/\tau} \\ \log_e(1- {v_C \over E}) &= \log_e(e^{-t/\tau}) \\ \log_e(1- {v_C \over E}) &= {-t \over \tau} \\ {t \over \tau} &= - \log_e(1- {v_C \over E}) \\ t &= -\tau \, \log_e(1- {v_C \over E}) \end{split}$$ but $$- \log_e({x \over y}) = + \log_e({y \over x}) $$ therefore; $$ \bbox[10px,border:1px solid grey]{t = \tau \, \log_e({E \over E - v_C })} $$