A $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{k\Omega }$resistor and a capacitor are connected in series, and then a12.0 Vpotential difference is suddenly applied across them. The potential difference across the capacitor rises to5.00 Vin $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\mu s}$. (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

1. The resistance of the resistor, $\mathrm{R}=15.0\mathrm{k\Omega }$
2. The potential difference across the combination of resistor and capacitor is series, ${\mathrm{V}}_{\mathrm{o}}=12.0\mathrm{V}$
3. The rise in potential difference in $\mathrm{t}=1.30\mathrm{\mu s}$,${\mathrm{V}}^{\text{'}}=5\mathrm{V}$

Using the concept of charging and discharging a given combination of the resistor and capacitor, we can see that an amount of charge is stored within the capacitor. Thus, the potential difference is created due to this charge stored within the capacitor plates and changes with the time constant of the circuit. Using the given relation, we can get the time constant value that further is used to calculate the unknown capacitance.

Formulae:

The potential difference of a RC circuit,$\mathbf{V}\mathbf{=}{\mathbf{V}}_{\mathbf{o}}\left(1-e^{-t/\tau }\right)$ (i)

The time constant of the RC circuit, $\mathbf{\tau }\mathbf{=}\mathbf{RC}$ (ii)

Using the given values in equation (i), the time constant of the RC circuit can be given as:

$\begin{array}{rcl}5\mathrm{V}& =& 12\mathrm{V}\left(1-{\mathrm{e}}^{-1.30\mathrm{\mu s}/\mathrm{\zeta }}\right)\\ \frac{5}{12}& =& \left(1-{\mathrm{e}}^{-1.30\mathrm{\mu s}/\mathrm{\zeta }}\right)\\ {\mathrm{e}}^{-1.30\mathrm{\mu s}/\mathrm{\zeta }}& =& 1-\frac{5}{12}\\ & =& \frac{7}{12}\\ \mathrm{\zeta }& =& \left(1.30\mathrm{\mu s}\right)\ln \left(\frac{12}{7}\right)\\ & =& 2.41\mathrm{\mu s}\end{array}$

Hence, the value of time constant is $2.41\mathrm{\mu s}$.

Now, using the above value in equation (ii), the capacitance of the capacitor in the RC circuit can be given as follows:

$\begin{array}{rcl}\left(2.41\times {10}^{-6}\mathrm{s}\right)& =& \left(15\times {10}^3\mathrm{\Omega }\right)\mathrm{C}\\ \mathrm{C}& =& \frac{2.41\times {10}^{-6}\mathrm{s}}{15\times {10}^3\mathrm{\Omega }}\\ & =& 0.161\times {10}^{-9}\mathrm{F}\\ & =& 161\mathrm{pF}\end{array}$

Hence, the capacitance value is 161 pF.