Съдържание

Условие Background Materials and apparatus Instructions Отговори Решение

APhO 2003, Експеримент — Задача 1. Determination of Capacitance

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4th Asian Physics Olympiad — Experimental Competition — 25 April 2003. I. Determination of Capacitance · 10 т.

Условие

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. 1.1 (a), 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 $\varepsilon_0$ , $C$ , $R$ and frequency of the mains $f$ . Graphical analysis of this relationship can be used to determine $C$ .

Series AC circuit: a source labelled εo sin ωt (circle with a sine symbol) on the left, a capacitor labelled C above it, and a variable resistor labelled R (resistor with an arrow) on the right side of the loop. — Fig. 1.1 (a): AC Circuit for determination of capacitance C

Schematic of the equipment: a transformer fed from 220 VAC, 50 Hz; its secondary is connected through a capacitor C at the top to terminal D, and the bottom wire ends at terminal E; three resistors labelled RC, RB, RA stand side by side on the right, each with terminals at both ends. — Fig. 1.1 (b): A schematic diagram of the equipment used

Materials and apparatus

capacitor
three resistors of known values with ±5% errors ( $R_A = 680\,\Omega$ , $R_B = 1500\,\Omega$ and $R_C = 3300\,\Omega$ ) as shown in figure 1.1 (b).
step-down isolation transformer for alternating voltage source of $f = 50\ \mathrm{Hz}$
digital voltmeter
electrical connectors
linear graph papers

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

Instructions

a) Derive the expression for the average power dissipation $\overline{P}$ in resistor $R$ in terms of $\varepsilon_o$ , $R$ , $C$ and $\omega$ . [1 т.]

b) Deduce the condition for which $\overline{P}$ is a maximum. [1 т.]

c) Convert the dependence found in a) into a linear dependence of certain quantities $\alpha$ and $\beta$ . [1 т.]

d) Measure the root mean square (effective) voltage $V$ across $R$ for each of all possible combinations of $R_A$ , $R_B$ and $R_C$ . [2,5 т.]

e) Plot $\overline{P}$ versus $R$ and from this graph compute the value of capacitance $C$ . [2 т.]

f) From c), draw the graph of $\alpha$ versus $\beta$ and determine capacitance $C$ . [2 т.]

g) Estimate the uncertainties in the values of $C$ obtained in e) and f). [0,5 т.]

Отговори

Решение

\overline{P} = I^2 R = \frac{\tfrac{1}{2}\varepsilon_0^{\,2}}{R^2 + \left(\dfrac{1}{\omega C}\right)^2}\,R \qquad \text{(1.0 point)}

\frac{d}{dR}\overline{P} = 0 \qquad \text{(0.3 point)}

\frac{d}{dR}\overline{P} = \frac{d}{dR}\left(\frac{\tfrac{1}{2}\varepsilon_0^{\,2}}{R^2 + \left(\dfrac{1}{\omega C}\right)^2}\,R\right) \qquad \text{(0.4 point)}

= \tfrac{1}{2}\varepsilon_0^{\,2}\,\frac{R^2 + \left(\dfrac{1}{\omega C}\right)^2 - R(2R)}{\left(R^2 + \left(\dfrac{1}{\omega C}\right)^2\right)^2}

condition for $\overline{P}_{\max}$ :

R = \frac{1}{\omega C} \qquad \text{(0.3 point)}

c. ( 1 point)

\overline{P} = \frac{\dfrac{1}{2}\varepsilon_0^{\,2}}{R^2 + \left(\dfrac{1}{\omega C}\right)^2}\,R = \frac{\dfrac{1}{2}\varepsilon_0^{\,2} R}{R^2\left[1 + \left(\dfrac{1}{R\omega C}\right)^2\right]} = \frac{\dfrac{1}{2}\varepsilon_0^{\,2}}{R\left[1 + \left(\dfrac{1}{R\omega C}\right)^2\right]}

\Rightarrow \frac{1}{R\overline{P}} = \frac{2}{\varepsilon_0^{\,2}}\left[1 + \frac{1}{R^2}\,\frac{1}{\omega^2 C^2}\right]

\frac{1}{R\overline{P}} = \frac{1}{V^2} = \frac{2}{\varepsilon_0^{\,2}} + \frac{2}{\varepsilon_0^{\,2}}\left(\frac{1}{\omega C}\right)^2 \frac{1}{R^2}

Note: The linear graph will be $\dfrac{1}{R\overline{P}}$ or $\dfrac{1}{V^2}$ versus $\dfrac{1}{R^2}$ . If $a$ is the slope and $b$ is the intercept with the Y axis, then

\frac{1}{\omega^2 C^2} = \frac{a}{b} \quad\Rightarrow\quad C = \frac{1}{\omega}\sqrt{\frac{b}{a}}

An alternative method:

\frac{V^2}{R^2} = \frac{\tfrac{1}{2}\varepsilon_0^{\,2}}{R^2 + \left(\dfrac{1}{\omega C}\right)^2}

\frac{R^2}{V^2} = \left(R^2 + \left(\frac{1}{\omega C}\right)^2\right)\frac{2}{\varepsilon_0^{\,2}}

\frac{1}{V^2} = \left[1 + \left(\frac{1}{\omega C}\right)^2 \frac{1}{R^2}\right]\frac{2}{\varepsilon_0^{\,2}} = \frac{2}{\varepsilon_0^{\,2}} + \frac{2}{\varepsilon_0^{\,2}}\left(\frac{1}{\omega C}\right)^2 \frac{1}{R^2}

R^2 = \left(\frac{1}{2}\varepsilon_0^{\,2}\right)\left(\frac{R}{V}\right)^2 - \left(\frac{1}{\omega C}\right)^2

Note: The graph will be $R^2$ versus $\left(\dfrac{R}{V}\right)^2$ and $C$ is determined from the Y-intercept.

No.	Resistor(s)	$R$ (Ω)	$V$ (V)	$\overline{P} = \frac{V^2}{R}$ (W)
1	$R_A$	680	9.86	0.144
2	$R_B$	1500	17.36	0.202
3	$R_C$	3300	22.81	0.159
4	$R_A + R_B$	2180	20.49	0.193
5	$R_A$ // $R_B$	468	7.28	0.111
6	$R_B + R_C$	4800	23.98	0.122
7	$R_B$ // $R_C$	1032	13.78	0.186
8	$R_C + R_A$	3980	23.66	0.141
9	$R_C$ // $R_A$	564	8.42	0.126
10	$R_A + R_B + R_C$	5480	24.40	0.109
11	( $R_A$ // $R_B$ )+ $R_C$	3768	23.43	0.147
12	( $R_B$ // $R_C$ )+ $R_A$	1712	18.63	0.202
13	( $R_C$ // $R_A$ )+ $R_B$	2064	20.15	0.195
14	( $R_A$ // $R_B$ )// $R_C$	410	6.22	0.094
15	( $R_A$ + $R_B$ )// $R_C$	1313	16.18	0.200
16	( $R_B$ + $R_C$ )// $R_A$	596	8.82	0.131
17	( $R_C$ + $R_A$ )// $R_B$	1089	14.36	0.190

(2.5 points):(data points = 17; 2.5, $\gt 13$ ; 2.0, $\gt 9$ ; 1.5; $\gt 3$ ; 1.0, $\le 3$ ; 0.5)

e) (1.5 points):good graph (0.5 point): correct value

Graph of P (W) versus R (Ω): 17 measured points joined by a curve rising from about 0.094 W at 410 Ω to a maximum of about 0.20 W near 1600 Ω and falling to about 0.11 W at 5480 Ω; dotted lines mark the maximum at 0.20 W and R ≈ 1600 Ω. Axes: P from 0.00 to 0.25, R from 0 to 6000.

R \text{ at } \overline{P}_{\max} = 1600\,\Omega \Rightarrow C = \frac{1}{\omega R} = \frac{1}{2\pi \times 50 \times 1600} = 1.9 \times 10^{-6}\ \mathrm{F} = 1.9\,\mu\mathrm{F}

f) Linear graph.

$R$ (Ω)	$V$ (V)	$\overline{P} = \frac{V^2}{R}$ (W)	$\frac{1}{R\overline{P}}$ (ΩW) $^{-1}$	$\frac{1}{R^2}$ (×10 $^{-6}$ Ω $^{-2}$ )
410	6.22	0.094	0.0259	5.948
468	7.28	0.111	0.0193	4.565
564	8.42	0.126	0.0141	3.143
596	8.82	0.131	0.0128	2.815
680	9.86	0.144	0.0102	2.162
1032	13.78	0.186	0.0052	0.938
1089	14.36	0.190	0.0048	0.843
1313	16.18	0.200	0.0038	0.580
1500	17.36	0.202	0.0033	0.444
1712	18.63	0.202	0.0029	0.341
2064	20.15	0.195	0.0025	0.234
2180	20.49	0.193	0.0024	0.210
3300	22.81	0.159	0.0019	0.091
3768	23.43	0.147	0.0018	0.070
3980	23.66	0.141	0.0018	0.0631
4800	23.98	0.122	0.0017	0.0434
5480	24.40	0.109	0.0017	0.0333

Linear graph of 1/(RP̄) in (ΩW)^-1 versus 1/R² in (×10^-6 Ω^-2): points on a straight fitted line labelled y = 0.004x + 0.0015; vertical axis 0.000 to 0.030, horizontal axis 0.00 to 7.00.

slope = $a$ = 0.004 × 10 $^6$ Ω/W , Y-intercept = $b$ = 0.0015 (ΩW) $^{-1}$ :

\frac{1}{\omega^2 C^2} = \frac{a}{b} \quad\Rightarrow\quad C = \frac{1}{\omega}\sqrt{\frac{b}{a}} = 1.95 \times 10^{-6}\ \mathrm{F} = 1.95\,\mu\mathrm{F}

An alternative method of linear graph

$R$ (Ω)	$V$ (V)	$\overline{P} = \frac{V^2}{R}$ (W)	$\left(\frac{R}{V}\right)^2$ (Ω /V) $^2$	$R^2$ (×10 $^6$ Ω $^2$ )
410	6.22	0.094	4345	0.17
468	7.28	0.111	4133	0.22
564	8.42	0.126	4487	0.32
596	8.82	0.131	4566	0.36
680	9.86	0.144	4756	0.46
1032	13.78	0.186	5609	1.07
1089	14.36	0.190	5751	1.19
1313	16.18	0.200	6585	1.72
1500	17.36	0.202	7466	2.25
1712	18.63	0.202	8445	2.93
2064	20.15	0.195	10492	4.26
2180	20.49	0.193	11320	4.75
3300	22.81	0.159	20930	10.89
3768	23.43	0.147	25863	14.20
3980	23.66	0.141	28297	15.84
4800	23.98	0.122	40067	23.04
5480	24.40	0.109	50441	30.03

Graphical analysis:

\text{Y-intercept} = \left(\frac{1}{\omega C}\right)^2 = 2.5428 \times 10^6\ \Omega^2

\frac{1}{\omega C} = 1.595 \times 10^3\ \Omega \quad\Rightarrow\quad C = 1.99 \times 10^{-6}\ \mathrm{F} = 1.99\,\mu\mathrm{F}

(1.5 points): good graph

(0.5 point): correct value

a) Estimation of the uncertainty in the values of $C$ obtained in e) (0.25 point)

Estimation of the uncertainty in the values of $C$ obtained in f) (0.25 point)

Linear graph of R² (×10^6 Ω²) versus (R/V)² (Ω/V)²: points on a straight fitted line labelled y = 6.4428 x - 2.5428, crossing the vertical axis below zero; vertical axis -5.00 to 35.00, horizontal axis -1.000 to 6.000.

Оригинал в Архива: exp1.pdf · официални решения: exp1_sol.pdf