91Ó°ÊÓ

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

$q=\mathrm{Î\mu }C\left(1-e^{\frac{-t}{RC}}\right)$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(¾±)

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

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

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

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

Let’s first calculate the time constant of the R-C circuit:

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

Now, putting the values of$\mathrm{Î\mu },C,\mathrm{Ï„}$ into the equation (i) we get:

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

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

$$\begin{gathered} q(t=0)=\left(7.44×{10}^{−4}\mathrm{C}\right)(1−1)=0 \\ \begin{array}{rl}q(t& =5.00)=\left(7.44×{10}^{−4}\mathrm{C}\right)\left(1−e^{\frac{−5}{11.1}}\right)\end{array} \\ =22.70×{10}^{−4}\mathrm{C}\begin{array}{}\end{array} \\ \begin{array}{rl}q(t& =10)=\left(7.44×{10}^{−4}\mathrm{C}\right)\left(1−e^{\frac{−10}{11.1}}\right)\end{array}\begin{array}{}\end{array} \\ =4.42×{10}^{−4}\mathrm{C} \\ \begin{array}{rl}q(t& =20.0)=\left(7.44×{10}^{−4}\mathrm{C}\right)\left(1−e^{\frac{−20}{111.}}\right)\end{array} \\ =6.21×{10}^{−4}\mathrm{C} \\ \begin{array}{rl}\mathrm{q}(\mathrm{t}& =100)=\left(7.44×{10}^{−4}\mathrm{C}\right)\left(1−{\mathrm{e}}^{\frac{−10}{11.1}}\right)\end{array} \\ =7.44×{10}^{−4}\mathrm{C} \end{gathered}$$

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

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

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

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

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