You set up the circuit shown in Fig. 26.20, where $\mathit{C}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{F}$. At time t= 0, you close the switch and then measure the charge qon the capacitor as a function of the current iin the resistor. Your results are given in the table:

(a) Graph qas a function of i. Explain why the data points, when plotted this way, fall close to a straight line. Find the slope and y-intercept of the straight line that gives the best fit to the data. (b) Use your results from part (a) to calculate the resistance Rof the resistor and the emf E of the battery. (c) At what time tafter the switch is closed is the voltage across the capacitor equal to 10.0 V? (d) When the voltage across the capacitor is 4.00 V, what is the voltage across the resistor?

The Kirchhoff’s rule will help to determine the equation across the loop,

$$\begin{gathered} \epsilon -Ri-q/C=0 \\ q=\epsilon C-RCi \end{gathered}$$

The graph will be a straight line as can be seen in equation above with slope is equal to –RC and y-intercept is equal to ɛC. The graph is depicted below,

The best-fit slope of this graph is $-1.233\times {10}^{-3}\hspace{0.33em}C/A$, and the y-intercept is $7.054\times {10}^{-5}\hspace{0.33em}C$

From part (a),

$\begin{array}{r}\mathrm{RC}=\text{negetive of slope}\\ =\left(-1.233\times {10}^{-3}\mathrm{C}/\mathrm{A}\right)\end{array}$

So,

$\begin{array}{r}R=\frac{1.233\times {10}^{-3}\mathrm{C}/\mathrm{A}}{\mathrm{C}}\\ =\frac{1.233\times {10}^{-3}\mathrm{C}/\mathrm{A}}{5.00\times {10}^{-6}\mathrm{F}}\\ =246.6\mathrm{\Omega }\\ \approx 247.0\mathrm{\Omega }\end{array}$

The y-intercept in the obtained straight line is,

$\begin{array}{r}\epsilon C=7.054\times {10}^{-5}\mathrm{C}\\ \epsilon =\frac{7.054\times {10}^{-5}\mathrm{C}}{5.00\times {10}^{-6}\mathrm{F}}\\ =15.9\mathrm{V}\end{array}$

Thus, the resistance is approximately $247.0\Omega$ and the emf produced is 15.9 V .

The transient voltage expression is,

$\begin{array}{r}{\mathrm{V}}_{\mathrm{c}}=\epsilon \left(1-{\mathrm{e}}^{-\mathrm{t}/R\mathrm{c}}\right)\\ \frac{V_C}{\epsilon }=1-{\mathrm{e}}^{-\mathrm{t}/RC}\\ =\left(10.0\mathrm{V}\right)\left(15.9\mathrm{V}\right)\end{array}$

Solve for time,

$\begin{array}{r}\mathrm{t}=\left(247\mathrm{\Omega }\right)\left(5.00\mu \mathrm{F}\right)\ln \left(0.3714\right)\\ =1223\mu \mathrm{s}\\ \approx 1.22\mathrm{ms}\end{array}$

Thus, the time lapse is 1.22 ms .

The voltage across the resistor can be found as,

$\begin{array}{r}{V}_R=\epsilon -V_C\\ =15.9\mathrm{V}-4.00\mathrm{V}\\ =11.9\mathrm{V}\end{array}$

Thus the voltage drop across the resistor is 11.9 V.