A 12.4-µF capacitor is connected through a 0.895-MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 20 s

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

Charge on a charging capacitor as a function of time is given by:

$\mathrm{q}=\mathrm{\epsilon C}(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}})..............................\left(1\right)$

And the function between time and current in a circuit is given by:

$\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{R}}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}$

It is also given that the capacitance of $\mathrm{C}=12.4\mathrm{\mu F}$the resistance connected through it is$\mathrm{R}=0.895\times {10}^6\mathrm{\Omega }$ and the potential difference connected across the capacitor and the resistor is V = 60.0 V

The constant potential difference connected to the capacitor and the resistance be our emf of the R-C circuit:$\mathrm{\epsilon }=\mathrm{V}$

Let’s first calculate the time constant of the circuit:

$$\begin{gathered} \mathrm{\tau }=\mathrm{RC} \\ =(0.895\times {10}^6\mathrm{\Omega })(12.4\times {10}^{-6}\mathrm{F}) \\ =11.1\mathrm{S} \end{gathered}$$

Now, putting the values of into the equation (1) we get:

$$\begin{gathered} \mathrm{q}\left(\mathrm{t}\right)=(60.0\mathrm{V})(12.4\times {10}^{-6}\mathrm{F})\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{11.1\mathrm{S}}}\right) \\ \mathrm{q}\left(\mathrm{t}\right)=(7.44\times {10}^{-4}\mathrm{C})\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{11.1\mathrm{S}}}\right) \end{gathered}$$

Now, substituting the values of given times t to get the values of charges at each instant:

$$\begin{gathered} \mathrm{q}(\mathrm{t}=0)=(7.44\times {10}^{-4}\mathrm{C})(1-1)=0 \\ \mathrm{q}(\mathrm{t}=5.00)=(7.44\times {10}^{-4}\mathrm{C})\left(1-{\mathrm{e}}^{\frac{-5\mathrm{S}}{11.1\mathrm{S}}}\right) \\ =22.70\times {10}^{-4}\mathrm{C} \\ \mathrm{q}(\mathrm{t}=10)=(7.44\times {10}^{-4}\mathrm{C})\left(1-{\mathrm{e}}^{\frac{-10\mathrm{S}}{11.1\mathrm{S}}}\right) \\ =4.42\times {10}^{-4}\mathrm{C} \\ \mathrm{q}(\mathrm{t}=20.0)=(7.44\times {10}^{-4}\mathrm{C})\left(1-{\mathrm{e}}^{\frac{-20\mathrm{S}}{11.1\mathrm{S}}}\right) \\ =6.21\times {10}^{-4}\mathrm{C} \\ \mathrm{q}(\mathrm{t}=100)=(7.44\times {10}^{-4}\mathrm{C})\left(1-{\mathrm{e}}^{\frac{-10\mathrm{S}}{11.1\mathrm{S}}}\right) \\ =7.44\times {10}^{-4}\mathrm{C} \end{gathered}$$

Substituting the values of$\mathrm{\epsilon }$ ,T and$\mathrm{\tau }$ to get the instantaneous current in the circuit at any time t:

$$\begin{gathered} \mathrm{i}\left(\mathrm{t}\right)=\frac{60.0\mathrm{V}}{0.895\times {10}^6\mathrm{\Omega }}{\mathrm{e}}^{\frac{-\mathrm{t}}{11.1\mathrm{S}}} \\ \mathrm{i}\left(\mathrm{t}\right)=(6.74\times {10}^{-5}\mathrm{A}){\mathrm{e}}^{\frac{-\mathrm{t}}{11.1\mathrm{S}}} \end{gathered}$$

Now, we enter each value of the given time t and get the current at each of the instants:

$$\begin{gathered} \mathrm{i}(\mathrm{t}=0)=(6.74\times {10}^{-5}\mathrm{A})\left(1\right) \\ =6.74\times {10}^{-5}\mathrm{A} \\ \mathrm{i}(\mathrm{t}=5)=(6.74\times {10}^{-5}\mathrm{A}){\mathrm{e}}^{\frac{-5\mathrm{S}}{11.1\mathrm{S}}} \\ =4.30\times {10}^{-5}\mathrm{A} \\ \mathrm{i}(\mathrm{t}=10)=(6.74\times {10}^{-5}\mathrm{A}){\mathrm{e}}^{\frac{-10\mathrm{S}}{11.1\mathrm{S}}} \\ =2.74\times {10}^{-5}\mathrm{A} \\ \mathrm{i}(\mathrm{t}=20)=(6.74\times {10}^{-5}\mathrm{A}){\mathrm{e}}^{\frac{-20\mathrm{S}}{11.1\mathrm{S}}} \\ =1.11\times {10}^{-5}\mathrm{A} \\ \mathrm{i}(\mathrm{t}=100)=(6.74\times {10}^{-5}\mathrm{A}){\mathrm{e}}^{\frac{-100\mathrm{S}}{11.1\mathrm{S}}} \\ =8.25\times {10}^{-9}\mathrm{A} \end{gathered}$$

The graphs of the resultant part of the above instances for between and is given below: