I. Determination of Capacitance Background It is known that capacitors play a significant role in the electrical circuits. There are several methods of measurements of the capacitance of a capacitor. In this experiment you are required to perform the experiment in order to determine the

Question

I. Determination of Capacitance

Background

It is known that capacitors play a significant role in the electrical circuits. There are several methods of measurements of the capacitance of a capacitor. In this experiment you are required to perform the experiment in order to determine the capacitance of an AC capacitor using a simple electrical circuitry.

In Fig. 4 - 8, a capacitor of capacitance C and a resistor of resistance R are connected in series to the alternating voltage source of mains frequency.

The electrical power which is dissipated at the resistor R depends on the values of [latex]\varepsilon_{0}[/latex], C, R and frequency of the mains f. Graphical analysis of this relationship can be used to determine C.

Graphical analysis of this relationship can be used to determine C.

Materials and apparatus

1. Capacitor.

2. Three resistors of known values with ± 5% errors ( [latex]R_{A}[/latex] = 680 Ω, [latex]R_{B}[/latex] = 1500 Ω, and [latex]R_{C}[/latex] = 3300 Ω) as shown in Fig. 4 - 9.

3. Step-down isolation transformer for alternating voltage source of f = 50 Hz.

4. Digital voltmeter.
5. Electrical connectors.
6. Linear graph papers.

Warning: The digital multimeter in this experiment will be used for measuring the rms voltage[latex]( ilde{\bf V})[/latex] across R only. Do not use it to measure in other modes.

Instructions

(a) Derive the expression for the average power dissipation [latex]\overline{{P}}[/latex] in resistor R in terms of [latex]\varepsilon_{0}[/latex] , R, C and w.

(b) Deduce the condition for which [latex]\overline{{P}}[/latex] is a maximum.

(c) Convert the dependence found in (a) into a linear dependence of certain quantities α and β.

(d) Measure the root mean square (effective) voltage V across R for each of all possible combinations of [latex]R_{A}[/latex] , [latex]R_{B}[/latex] and [latex]R_{C}[/latex].

(e) Plot [latex]\overline{{P}}[/latex] versus R and from this graph compute the value of capacitance C.

(f) From (c), draw the graph of α versus β and determine capacitance C.

(g) Estimate the uncertainties in the values of C obtained in (e) and (f).

Accepted Answer

[latex](\mathrm{a})\,\overline{{{ P}}}={I^{2} }R=\frac{\frac{1}{2}\varepsilon_{o}^{2}}{R^{2}+\left(\frac{1}{\omega C} ight)^{2}}R.[/latex]

[latex]({\bf b})\,\,\frac{\mathrm{d}}{\mathrm{d}R}\,\overline{{P}}=0\,,[/latex]

[latex]\frac{\mathrm{d}}{\mathrm{d}R}\,\overline{{P}}=\frac{\mathrm{d}}{\mathrm{d}R}\,\frac{\frac{\mathrm{l}}{2}\epsilon_{0}^{2}}{R^{2}+\left(\frac{\mathrm{l}}{\omega C} ight)^{2}}R[/latex]

[latex]={\frac{1}{2}}\epsilon_{o}^{2}{\frac{R^{2}+\left({\frac{1}{\omega C}} ight)^{2}-R(2R)}{\left[R^{2}+\left({\frac{1}{\omega C}} ight)^{2} ight]^{2}}},[/latex]

condition for [latex]{\overline P}_{\operatorname*{max}}:[/latex] [latex]R={\frac{1}{\omega C}}.[/latex]

(c) [latex]\overline{{{P}}}=\frac{\frac{1}{2}\epsilon_{0}^{2}}{R^{2}+\left(\frac{1}{\omega C} ight)^{2}}R=\frac{\frac{1}{2}\epsilon_{0}^{2}R}{R^{2}\Big[1+\left(\frac{1}{R\omega C} ight)^{2}\Big]}[/latex]

[latex]=\frac{{\frac{1}{2}}\varepsilon_{0}^{2}}{R\Big[1+\Big({\frac{1}{R_{\omega C}}}\Big)^{2}\Big]}[/latex]

[latex]\Rightarrow{\frac{1}{R \overline{P}}}={\frac{2}{\epsilon_{0}^{2}}}\left(1+{\frac{1}{R^{2}}}\,{\frac{1}{\omega^{2}C^{2}}} ight)[/latex]

[latex]\frac{1}{R\,\overline{{{P}}}}\,=\frac{1}{V^{2}}=\frac{2}{\epsilon_{0}^{2}}+\frac{2}{\epsilon_{0}^{2}}\left(\frac{1}{\omega C} ight)^{2}\frac{1}{R^{2}}.[/latex]

Note: The linear graph will be [latex]\frac{1}{R\,\overline{{{P}}}}[/latex] or [latex]\frac{1}{V^{2}}[/latex] versus [latex]\frac{1}{R^{2}}[/latex] If a is the slope and b is the intercept with the Y axis, then [latex]{\frac{1}{\omega^{2}C^{2}}}={\frac{a}{b}}\Rightarrow\!C={\frac{1}{\omega}}\,\sqrt{\frac{b}{a}}\,.[/latex]

An alternative method:

[latex]\frac{V^{2}}{R^{2}}=\frac{\frac{1}{2}\varepsilon_{o}^{2}}{R^{2}+\left(\frac{1}{\omega C} ight)^{2}},[/latex]

[latex]\frac{R^{2}}{V^{2}}=\biggl[R^{2}+\biggl(\frac{1}{\omega C}\biggr)^{2}\biggr]\frac{2}{\epsilon_{0}^{2}},[/latex]

[latex]{\frac{1}{V^{2}}}=\left[1+\left({\frac{1}{\omega C}} ight)^{2}{\frac{1}{R^{2}}} ight]{\frac{2}{\epsilon_{0}^{2}}},[/latex]

[latex]{\frac{1}{V^{2}}}={\frac{2}{\epsilon_{\mathrm{0}}^{2}}}+{\frac{2}{\epsilon_{\mathrm{0}}^{2}}}\left({\frac{1}{\omega C}} ight)^{2}{\frac{1}{R^{2}}},[/latex]

[latex]R^{2}=\frac{1}{2}\varepsilon_{0}^{2}\bigl(\frac{R}{V}\bigr)^{2}-\bigl(\frac{1}{\omega C}\bigr)^{2}.[/latex]

Note: The graph will be [latex]R^{2}[/latex] versus [latex]\left({\frac{R}{V}} ight)^{2}[/latex] and C is determined from the Y-intercept.

(d)

Table--1--

(e)

[latex]{\mathrm{R~at~}}{\overline{{P}}}_{\mathrm{max}}=1600~{\mathrm{\Omega \Rightarrow C}}={\frac{1}{\omega R}}={\frac{1}{2\pi{ imes}{\mathrm{50}}{ imes}{\mathrm{1600}}}}[/latex]

[latex]=1.9 imes10^{-6}\ \mathrm{F}=1.9\ \mu\mathrm{F}.[/latex]

(f) Linear graph.

Table--2--

Graphical analysis: slope [latex]=a\;=\;0.004 imes10^{6}\;\Omega/\mathrm{W},[/latex] Y-intercept = b = [latex]0.0015(\Omega W)^{-1}:[/latex]

[latex]{\frac{1}{\omega^{2}\,C^{2}}}={\frac{a}{b}}\Rightarrow C={\frac{1}{\omega}}{\sqrt{\frac{b}{a}}}=1.95 imes10^{-6}\,{\mathrm{F}}=1.95\,{\mathrm{~}}\mu{\mathrm{F}}.[/latex]

An alternative method of linear graph

Table--3--

Graphical analysis: Y-intercept [latex]=\left({\frac{1}{\omega C}} ight)^{2}=2.5428 imes10^{6}\,\Omega^{2}[/latex]

[latex]{\frac{1}{\omega C}}=1.595 imes10^{3}\,\Omega\Rightarrow C=1.99 imes10^{-6}\,{F}=1.99\,~{\mu{F}}.[/latex]

(g) Estimation of the uncertainty in the values of C obtained in (e). Estimation of the uncertainty in the values of C obtained in (f).

Question 4.EC.I: I. Determination of Capacitance Background It is known that......

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