91Ó°ÊÓ

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:

$\mathrm{q}=\mathrm{Î\mu °ä}(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}})..............................\left(1\right)$

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

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

It is also given that the capacitance of $\mathrm{C}=12.4\mathrm{μ¹ó}$the resistance connected through it is$\mathrm{R}=0.895×{10}^6\mathrm{Î\copyright }$ 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:$\mathrm{Î\mu }=\mathrm{V}$

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

$$\begin{gathered} \mathrm{Ï„}=\mathrm{RC} \\ =(0.895×{10}^6\mathrm{Î\copyright })(12.4×{10}^{-6}\mathrm{F}) \\ =11.1\mathrm{S} \end{gathered}$$

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

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

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

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

Substituting the values of$\mathrm{Î\mu }$ ,T and$\mathrm{Ï„}$ to get the instantaneous current in the circuit at any time t:

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

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

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

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