Question

A capacitor with an initial potential difference of$\mathbf{100}\mathit{V}$is discharged through a resistor when a switch between them is closed at$\mathbf{t}\mathbf{=}\mathbf{0}$. At$\mathit{t}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathit{s}$, the potential difference across the capacitor is$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{V}$. (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at$\mathbf{t}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{0}\mathbf{s}$?

Short answer

a) The time constant of the circuit is$2.17s$.

b) The potential difference across the capacitor at $t=17s$is$3.96\times {10}^{-2}\text{V}$.

Step-by-step solution

The given data

a) Initial potential difference of the capacitor, $V_0\left(0\right)=100\text{V}$

b) Potential difference at across the capacitor, $V_t\left(10\right)=1.00\text{V}$

c) Time value,$t=17\text{s}$

Understanding the concept of time constant

Using the charging and discharging concept of the capacitor, we can get the required time constant of the RC circuit. Now, for a different time, the discharging as continued, the potential difference will differ and thus is given using the conditions substituted in the voltage formula.

Formulae:

The charge equation of a RC circuit,$q=q_0{e}^{-t/\tau }$ (i)

Where,$\tau$is the time constant of the RC circuit.

The charge stored within the plates of a capacitor, $q=CV$ (ii)

a) Calculation of the time constant

$$\begin{gathered} V=V_0{e}^{-t/\tau }......................\left(a\right) \\ 1=100e^{-10/\tau } \\ \tau =10\ln \left(100/1\right) \\ =2.17\text{s} \end{gathered}$$

Hence, the value of the time constant is$2.17s$.

 b) Calculation of the potential difference at t=17s

At$\tau =2.17\text{s}$

$$\begin{gathered} V=100e^{-17/2.17} \\ =100e^{-7.83} \\ =3.96\times {10}^{-2}\text{V} \end{gathered}$$

Hence, the value of the potential difference is$3.96\times {10}^{-2}\text{V}$.