Question

Using the exact exponential treatment, find how much time is required to charge an initially uncharged $\mathbf{100}\mathbf{-}\mathbf{pF}$capacitor through alocalid="1656397542799" $\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{M\Omega }$resistor to$\mathbf{90}\mathbf{.}\mathbf{0}\mathbf{\%}$of its final voltage.

Short answer

The required time to charge an initially uncharged $100-\mathrm{pF}$capacitor through a $75.0-\mathrm{M\Omega }$resistor to $90.0\%$of its final voltage is $17.3\mathrm{ms}$.

Step-by-step solution

Definition of Concept

Voltage: A unit of measurement for electrical force (volts).

Capacitor: Device for storing electrical energy, consisting of two conductors in close proximity and insulated from each other.

Formula Used

The formula for the time constant,

$\mathrm{\tau }=\mathrm{RC}$

The formula for voltage of a charging capacitor as,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right)$

Calculate the time Constant

The values of $\mathrm{C}=100pF\left(\frac{{10}^{-12}\mathrm{F}}{1pF}\right)=1.00\times {10}^{-10}\mathrm{F}$and$\mathrm{R}=75M\Omega \left(\frac{{10}^6\Omega }{1M\Omega }\right)=7.5\times {10}^7\mathrm{\Omega }$.

Firstly, use the formula for the time constant and substitute the given values,

$$\begin{gathered} \mathrm{\tau }=\mathrm{RC} \\ =\left(7.5\times {10}^7\mathrm{\Omega }\right)\times \left(1.00\times {10}^{-10}\mathrm{F}\right) \\ =7.5\times {10}^{-3}\mathrm{s}\left(\frac{1\mathrm{ms}}{{10}^{-3}\mathrm{s}}\right) \\ =7.5\mathrm{ms} \end{gathered}$$

Therefore, the time constant is$7.5\mathrm{ms}$

Find the required time by applying the formula

Apply the formula of the voltage of a charging capacitor to calculate time,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right)$

The value of$V\left(t=T\right)=V_f$

And substitute ${\mathrm{V}}_{\mathrm{f}}=0.9{\mathrm{V}}_0$, therefore,

$$\begin{gathered} 0.9{\mathrm{V}}_0={\mathrm{V}}_0\left(1-{\mathrm{e}}^{-\mathrm{T}/\mathrm{\tau }}\right) \\ 1-{\mathrm{e}}^{-\mathrm{T}/\mathrm{\tau }}=0.9 \\ -\frac{\mathrm{T}}{\mathrm{\tau }}=\ln \left(0.1\right) \\ \mathrm{T}=-\mathrm{\tau }\left[\ln \left(0.1\right)\right] \\ \mathrm{T}=0.0173\mathrm{s}\left(\frac{1\mathrm{ms}}{{10}^{-3}\mathrm{s}}\right) \\ \mathrm{T}=17.3\mathrm{s} \end{gathered}$$

Hence, the required time is $T=17.3\mathrm{ms}$.