Using the exact exponential treatment, find how much time is required to discharge a$\mathbf{250}\mathbf{-}\mathbf{\mu F}$capacitor through a$\mathbf{500}\mathbf{-}\mathbf{\Omega }$resistor down tolocalid="1656395883292" $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}$of its original voltage.

A resistor is an electrical component that prevents current from flowing through a circuit. It's more like friction that prevents energy from flowing.

A capacitor is an electronic component that stores electrical charges. In electrical and electronic circuits, it generally opposes current changes.

Assume a charged capacitor with capacitance$\mathrm{C}=250\mathrm{\mu F}\left(\frac{{10}^{-5}\mathrm{F}}{1\mathrm{\mu F}}\right)=2.50\times {10}^{-4}\mathrm{F}$linked to a resistor$\mathrm{R}=500\mathrm{\Omega }$in this problem.

We work out how long it will take for the capacitor to drop to 1% of its initial voltage. We begin by determining the time constant:

$$\begin{gathered} \mathrm{\tau }=\mathrm{RC} \\ =\left(500\mathrm{\Omega }\right)\times \left(2.50\times {10}^{-4}\mathrm{F}\right) \\ =0.{125}_{\mathrm{s}} \end{gathered}$$

The formula for calculating the voltage of the discharging capacitor is:

$V\left(t\right)=V_0{e}^{-t/\tau }$

The time T required for the voltage to drop to $\mathrm{V}\left(\mathrm{t}=\mathrm{T}\right)={\mathrm{V}}_{\mathrm{f}}=0.01{\mathrm{V}}_0$is calculated as follows,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}={\mathrm{V}}_0{\mathrm{e}}^{-\mathrm{T}/\mathrm{\tau }} \\ 0.01{\mathrm{V}}_0={\mathrm{V}}_0{\mathrm{e}}^{-\mathrm{T}/\mathrm{\tau }} \\ {\mathrm{e}}^{-\mathrm{T}/\mathrm{\tau }}=0.01 \\ -\frac{\mathrm{T}}{\mathrm{\tau }}=\ln \left(0.01\right) \\ \mathrm{T}=-\mathrm{\tau ln}\left(0.01\right) \\ \mathrm{T}=0.576\mathrm{s} \end{gathered}$$

Therefore, the time required to discharge is$0.576s$.