Question

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

Short answer

(a) The charge on the capacitor at the following times after the connections are made is

At 0s=0, at $5.0\mathrm{s}=2.70\mathrm{x}{10}^{-4}\mathrm{C}\mathrm{at}10.0\mathrm{s}=4.42\mathrm{x}{10}^{-4}\mathrm{C}\mathrm{at}20.0\mathrm{s}=6.21\mathrm{x}{10}^{-4}\mathrm{C}\mathrm{and}\mathrm{at}100\mathrm{s}=7.44\mathrm{x}{10}^{-4}\mathrm{C}$

(b) The charging current at the same instants are as follows: at localid="1668329306161" $0\mathrm{s}=6.67\mathrm{x}{10}^{-5}\mathrm{A},\mathrm{at}5.0\mathrm{s}=4.30\mathrm{x}{10}^{-5}\mathrm{A},\mathrm{at}10.0\mathrm{s}=2.74\mathrm{x}{10}^{-5}\mathrm{A}\mathrm{at}20\mathrm{s}=1.11\mathrm{x}{10}^{-5}\mathrm{A}\mathrm{and}\mathrm{at}100\mathrm{s}=8.25\mathrm{x}{10}^{-9}\mathrm{A}$

Step-by-step solution

What is given to us

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

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

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

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

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

Calculating the charge for the given instance

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

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

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

Now, putting the values of$\epsilon ,C,\tau$ into the equation (i) we get:

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

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

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

Calculating the current at the given instances

Putting the values of $\epsilon ,T,\tau$, to get the instantaneous current in the circuit at anu time t:

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

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

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

The graphs of the given data

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