Chapter 28, Problem 28.78AP

The circuit shown in Figure P28.78 is set up in the laboratory to measure an unknown capacitance C in series with a resistance R = 10.0 MΩ powered by a battery whose emf is 6.19 V. The data given in the table are the measured voltages across the capacitor as a function of lime, where t = 0 represents the instant at which the switch is thrown to position b. (a) Construct a graph of In (ε/Δv) versus I and perform a linear least-squares fit to the data, (b) From the slope of your graph, obtain a value for the time constant of the circuit and a value for the capacitance.

Δv(V)	t(s)	In (ε/Δv)
6.19	0
5.56	4.87
4.93	11.1
4.34	19.4
3.72	30.8
3.09	46.6
2.47	67.3
1.83	102.2

To determine

To draw: The graph of $\ln \left(\epsilon /\Delta v\right)$ versus time $t$ and perform a linear least squares fit to the data.

Answer to Problem 28.78AP

The graph of $\ln \left(\epsilon /\Delta v\right)$ versus time $t$, where $\ln \left(\epsilon /\Delta v\right)$ on $y$ axis and time $t$ on $x$ axis is,

And the equation of best fit is $\ln \left(\epsilon /\Delta v\right)=\left(0.01178\right)t+0.08798$.

Explanation of Solution

Introduction:

The linear least square is a technique of fitting a statistical model to a set of variables. This shows the data points linearly in terms of unknown parameters.

Given: The capacitance is $C$ and the resistance $R$ is $10.0\text{M}\Omega$, the emf $\epsilon$ of a battery is $6.19\text{V}$, the time $t$ at which the switch is thrown to position $b$ is $0$ and the data given in the table is,

$\Delta v\left(\text{V}\right)$	$t\left(\text{s}\right)$
$6.19$	$0$
$5.55$	$4.87$
$4.93$	$11.1$
$4.34$	$19.4$
$3.72$	$30.8$
$3.09$	$46.6$
$2.47$	$67.3$
$1.83$	$102.2$

Write the equation of line which gives a linear relationship.

$y=mx+c$

Here,

$y$ is the ordinate.

$x$ is the abscissa.

$m$ is the slope of the line.

$c$ is the intercept on $y$ axis.

Take $\ln \left(\epsilon /\Delta v\right)$ on $y$ axis and time on $x$ axis due to which $y=\ln \left(\epsilon /\Delta v\right)$ and $x=t$

Substitute $\ln \left(\epsilon /\Delta v\right)$ for $y$ and $t$ for $x$ in above equation.

$\ln \left(\epsilon /\Delta v\right)=mt+c$ (1)

Calculate the values of $\ln \left(\epsilon /\Delta v\right)$ for each values of $\Delta v$.

$\Delta v\left(\text{V}\right)$	$x=t\left(\text{s}\right)$	$y=\ln \left(\epsilon /\Delta v\right)$
$6.19$	$0$	$0$
$5.55$	$4.87$	$0.109$
$4.93$	$11.1$	$0.227$
$4.34$	$19.4$	$0.355$
$3.72$	$30.8$	$0.509$
$3.09$	$46.6$	$0.694$
$2.47$	$67.3$	$0.918$
$1.83$	$102.2$	$1.218$

Write the formula of slope.

$m=\frac{N\left({\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}\right)-\left({\displaystyle \sum x_{\text{i}}}\right)\left({\displaystyle \sum y_{\text{i}}}\right)}{N\left({\displaystyle \sum x_{\text{i}}^2}\right)-{\left({\displaystyle \sum x_{\text{i}}}\right)}^2}$

Here,

$N$ is the number of points.

The number of data points $N$ is $8$.

Calculate $\left({\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}\right)$, $\left({\displaystyle \sum x_{\text{i}}}\right)$, $\left({\displaystyle \sum y_{\text{i}}}\right)$ and $\left({\displaystyle \sum x_{\text{i}}^2}\right)$ from the table.

$x=t\left(\text{s}\right)$	$y=\ln \left(\epsilon /\Delta v\right)$	$x^2$	$xy$
$0$	$0$	$0$	$0$
$4.87$	$0.109$	$23.7169$	$0.53083$
$11.1$	$0.227$	$123.21$	$2.5197$
$19.4$	$0.355$	$376.36$	$6.887$
$30.8$	$0.509$	$948.64$	$15.6772$
$46.6$	$0.694$	$2171.56$	$32.3404$
$67.3$	$0.918$	$4529.29$	$61.7814$
$102.2$	$1.218$	$10444.84$	$124.4796$
${\displaystyle \sum x_{\text{i}}}=282.27$	${\displaystyle \sum y_{\text{i}}}=4.03$	${\displaystyle \sum x^2}=18617.617$	${\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}=244.216$

Substitute $8$ for $N$, $244.216$ for $\left({\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}\right)$, $282.27$ for $\left({\displaystyle \sum x_{\text{i}}}\right)$, $4.03$ for $\left({\displaystyle \sum y_{\text{i}}}\right)$ and $18617.617$ for $\left({\displaystyle \sum x_{\text{i}}^2}\right)$ in above equation to find $m$.

$\begin{array}{c}m=\frac{\left(8\right)\left(244.216\right)-\left(282.27\right)\left(4.03\right)}{\left(8\right)\left(18617.617\right)-{\left(282.27\right)}^2}\\ =0.01178\end{array}$

Thus, the slope of line is $0.01178$.

Write the formula of intercept.

$c=\frac{\left({\displaystyle \sum x_{\text{i}}^2}\right)\left({\displaystyle \sum y_{\text{i}}}\right)-\left({\displaystyle \sum x_{\text{i}}}\right)\left({\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}\right)}{N\left({\displaystyle \sum x_{\text{i}}^2}\right)-{\left({\displaystyle \sum x_{\text{i}}}\right)}^2}$

Substitute $18617.617$ for $\left({\displaystyle \sum x_{\text{i}}^2}\right)$, $4.03$ for $\left({\displaystyle \sum y_{\text{i}}}\right)$, $282.27$ for $\left({\displaystyle \sum x_{\text{i}}}\right)$, $244.216$ for $\left({\displaystyle \sum x_{\text{i}}{y}_{\text{i}}}\right)$ and $8$ for $N$ in above equation to find $c$.

$\begin{array}{c}c=\frac{\left(18617.617\right)\left(4.03\right)-\left(282.27\right)\left(244.216\right)}{\left(8\right)\left(18617.617\right)-{\left(282.27\right)}^2}\\ =0.08798\end{array}$

Thus, the value of intercept on $y$ axis is $0.08798$.

Substitute $0.01178$ for $m$ and $0.08798$ for $c$ in equation (1).

$\ln \left(\epsilon /\Delta v\right)=\left(0.01178\right)t+0.08798$

Thus, the equation of best fit is $\ln \left(\epsilon /\Delta v\right)=\left(0.01178\right)t+0.08798$.

The graph of $\ln \left(\epsilon /\Delta v\right)$ versus time $t$, where $\ln \left(\epsilon /\Delta v\right)$ on $y$ axis and time $t$ on $x$ axis is,

Figure 1

To determine

The value of time constant of the circuit and the value of the capacitance.

Answer to Problem 28.78AP

The value of time constant of the circuit is $84.8\text{s}$ and the value of the capacitance is $8.48\text{μF}$.

Explanation of Solution

Given: The capacitance is $C$ and the resistance $R$ is $10.0\text{M}\Omega$, the emf $\epsilon$ of a battery is $6.19\text{V}$ and the time $t$ at which the switch is thrown to position $b$ is $0$.

Write the equation of time constant.

$\tau =RC$ (2)

Here,

$\tau$ is the time constant.

$R$ is the resistance.

$C$ is the capacitance.

The time constant from the slope is,

$\tau =\frac{1}{m}$

Substitute $0.01178$ for $m$ in above equation to find $\tau$.

$\begin{array}{c}\tau =\frac{1}{0.01178}\\ =84.8\text{s}\end{array}$

Thus, the time constant is $84.8\text{s}$.

Substitute $84.8\text{s}$ for $\tau$ and $10.0\text{M}\Omega$ for $R$ in equation (2) to find $C$.

$\begin{array}{l}\left(84.8\text{s}\right)=\left\{10.0\text{M}\Omega \left(\frac{{10}^6\Omega }{1\text{M}\Omega }\right)\right\}C\\ C=8.48\times {10}^{-6}\text{F}\left(\frac{1\text{μF}}{{10}^{-6}\text{F}}\right)\\ C=8.48\text{μF}\end{array}$

The capacitance is $8.48\text{μF}$.

Conclusion:

Therefore, the value of time constant of the circuit is $84.8\text{s}$ and the value of the capacitance is $8.48\text{μF}$.