A capacitor, charged to 10 V, is being discharged through a resistance R.  At the end of 1 s the voltage across the capacitor is 5 V.  What will be the voltage after 2 s?

$q=q_0{e}^{-t/CR}$

$\frac{q_0}{q}=e^{t/CR}$

${\log }_e{q}_0/q=\frac{t}{CR}$

$t=CR{\log }_e\left(\frac{q_0}{q}\right)$

$Att=1$

$1=CR{\log }_e\left(\frac{10}{5}\right)\text{-----(1)}$

$At \text{\hspace{0.33em}}t=2$

$2=CR{\log }_e\left(\frac{10}{v}\right)\text{----(2)}$

dividing 2 by 1

$\begin{array}{rl} & 2=\frac{{\log }_{e}10-{\log }_{e}V}{{\log }_{e}2}\\ & 2{\log }_{e}2={\log }_{e}10-{\log }_{e}V\\ & {\log }_{e}V={\log }_{e}10-{\log }_e{2}^2={\log }_e\left(\frac{10}{4}\right)\\ & V=\frac{10}{4}=2.5V\end{array}$

$v=2.5volt$