Download TEP 4.4.05 -01 Capacitor in the AC circuit LEP 4.4.05

Capacitor in the AC circuit
LEP
4.4.05
-01
Related topics
Inductance, Kirchhoff’s laws, parallel connection, series connection, a. c. impedance, phase displacement, vector diagram
Principle
The impedance and phase displacement of a capacitor in an A.C. circuit are observed as functions of
frequency and capacitance. Parallel and series circuits are studied.
Material
1
1
1
1
1
1
1
1
PEK capacitor (case 2) 1 µF/400V
PEK capacitor (case 2) 2.2 µF/400V
PEK capacitor (case 2) 4.7 µF/400V
Resistor plug-in box
50 Ω
PEK carbon resistor 1 W 5% 10 Ω
PEK carbon resistor 1 W 5% 100 Ω
Connection box
Difference amplifier
Fig. 1:
39113-01
39113-02
39113-03
06056-50
39104-01
39104-63
06030-23
11444-93
1
1
2
1
1
3
4
Digital function generator
Oscilloscope, 30 MHz, 2 channels
Screened cable, BNC, l = 750 mm
Connecting cord, l = 100 mm, red
Connecting cord, l = 100 mm, blue
Connecting cord, l = 500 mm, red
Connecting cord, l = 500 mm, blue
13654-99
11459-95
07542-11
07359-01
07359-04
07361-01
07361-04
Experimental set-up with capacitor and resistor in series.
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TEP
4.4.05
-01
Capacitor in the AC circuit
Tasks
1. Measure the impedance of the capacitors as
a function of frequency and capacitance.
2. Determine the capacitances of the capacitors.
3. Measure the phase displacement between
terminal voltage and total current as a function of the frequency in the circuit.
4. Calculate the total capacitance of several
capacitors connected in parallel and in series.
Set-up
The experimental set-up is shown in Fig. 1. Since
normal voltmeters and ammeters generally measure
only rms (root mean square) values and take no ac- Fig. 2: Circuit for simultaneous display of voltage across the
coil and total current.
count of phase relationships, it is preferable to use
an oscilloscope to study the dynamics of current
and voltage in such circuits. The experiment will be
carried out with sinusoidal voltages. The rms values are obtained when the peak-to-peak values
𝑈𝑝−𝑝 measured on the oscilloscope are divided by
2√2.
In accordance with relation (1), the total current 𝐼
in the circuit can be deduced by measuring the
voltage 𝑈 across the resistor with resistance 𝑅.
𝐼 = 𝑈⁄𝑅
(1)
The circuit shown in Fig. 2 permits the simultaneous display of the total current and the voltage
across the capacitor. The phase displacement be- Fig. 3: Circuit for simultaneous display of tertween the terminal voltage and the total current minal voltage and total current.
can be measured using a similar circuit, but with
channel B measuring the terminal voltage instead
of the capacitor voltage (see Fig. 3).
Procedure
In order to achieve high reading accuracy on the oscilloscope, high gain settings should be selected. After selecting the gain setting the Y-position of the two base-lines (GND) have to be adjusted until they
coincide. The peak-to-peak amplitude of the frequency generator’s signal should not be higher than 5 V
and should have no offset.
The digital frequency generator’s antenna output has to be connected with the ground socket of the difference amplifier. For detailed descriptions of the operation of the oscilloscope and the digital function
generator please refer to the manuals.
2
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LEP
4.4.05
-01
Capacitor in the AC circuit
Task 1 and 2:
To determine the impedance of a capacitor as a function of the frequency the capacitor is connected in
series with various resistors of known value. For each resistor the frequency is varied until there is the
same potential difference across the capacitor as across the resistor. The resistance and impedance
values are then equal and equation (2) is valid.
1
𝑅Ω = 𝜔𝐶 = 𝑋𝐶
(2)
From this relation the capacitance 𝐶 of a capacitor can be deduced, if the impedance 𝑋𝐶 at one frequency is known.
Task 3:
Set up the circuit as shown in Fig. 3 to display both terminal voltage and total current of the circuit.
There are two major ways to measure the frequency-dependent phase shift between total current and
terminal voltage. If, by means of the time-base control of the oscilloscope, one half-wave of the current is
brought to the full screen width (10 cm) – possibly with variable sweep rate – the phase displacement of
the voltage can be read off directly in cm (18° / cm). Another way to determine the phase shift is to read
the interval between the terminal voltage and the voltage across the resistor (corresponding to the total
current) directly from the oscilloscope and calculate the time gap as well as the resulting phase angle.
Attention: If the second procedure is chosen, the variable sweep rate must not be used as it distorts the
timeframe up to a factor of 2.5 which makes an accurate calculation of the time gap impossible.
Task 4:
In order to determine the total capacitance of the capacitors in parallel and series, include the capacitors
into the circuit appropriately and determine the frequency for 𝑅 = 𝑋𝐶 = 100 Ω. The capacitance can then
be calculated analogously to task 2.
Theory
If a capacitor of capacitance 𝐶 and the charge 𝑄 on the plates as well as an ohmic resistor of resistance
𝑅 are connected in a circuit, the sum of the potential differences across the capacitor and the resistor is
equal to the terminal voltage 𝑈𝑡 which gives relation (3).
𝑄
𝑈𝑡 = 𝐼 ∙ 𝑅 + 𝐶
(3)
This corresponds to Kirchhoff’s second law, which states, that the directed sum of all electrical potential
differences in a closed circuit is zero. Care must be taken regarding the direction of the potential differences. The terminal voltage has the reversed direction of the voltage across the capacitor and the resistor. Relation (4) gives the formula of Kirchhoff’s voltage law for the studied circuits, where UΩ = 𝐼 ∙ 𝑅 is
𝑄
the voltage across the resistor and 𝑈𝐶 = 𝐶 the voltage across the capacitor.
𝑈𝐶 + 𝑈Ω − 𝑈𝑡 = 0
(4)
Noting the fact that
𝐼=
𝑑𝑄
,
𝑑𝑡
from differentiation of (3) follows
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TEP
4.4.05
-01
Capacitor in the AC circuit
𝑑𝑈𝑡𝑜𝑡
𝑑𝑡
𝑑𝐼
1
= 𝑅 𝑑𝑡 + 𝐶 .
If the alternating voltage 𝑈𝑡 has the frequency 𝜔 = 2𝜋𝜈 and the waveform
𝑈𝑡 = 𝑈0 cos 𝜔𝑡,
(5)
inserting (5) into equation (3) gives the following solution for the current 𝐼:
𝐼 = 𝐼0 cos(𝜔𝑡 − 𝜑)
(6)
There 𝜑 is the phase displacement which is given by
1
𝜔𝐶𝑅
tan 𝜑 = −
(7)
and
𝐼0 =
𝑈0
√𝑅2 +(
(8)
1 2
)
𝜔𝐶
is the peak current. As is seen from relation (4) the total current follows the terminal voltage.
If there are more than one capacitors in the circuit, the total capacitance can be calculated with equations (9) for series and (10) for parallel connection.
𝐶𝑡𝑜𝑡 = 1⁄ 1
∑
(9)
𝐶𝑡𝑜𝑡 = ∑ Ci
(10)
Ci
4
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Capacitor in the AC circuit
Results and Evaluation
In the following the evaluation of the obtained values is described exemplary. Your results may vary from
those presented here.
Task 1: Measure the impedance of the capacitors as a function of frequency and capacitance.
From equation (2) the rational relation between frequency of the signal and impedance of the capacitor is
obvious. Fitting a rational function to the measured values for the capacitor 𝐶1 gives relation (11).
152
𝑋𝐶 (𝜈)⁄Ω = 𝜈⁄kHz − 4
Fig. 4:
(11)
Impedance values for various frequencies for the two capacitors 𝑪𝟏 and 𝑪𝟑 .
Equation (11) gives the relation of the rational fit for 𝑪𝟏 .
Task 2: Determine the capacitance of the capacitors.
Considering relations (2) and (11) for the first capacitor leads to equation (12) which gives the capacitance of the capacitor with 𝐶1 = 1.05 µF.
1
𝑋𝐶 ⁄Ω = 2𝜋 𝜈⁄Hz∙C
1 ⁄F
152
= 𝜈⁄kHz ⟺ 𝐶1 ⁄µF = 1⁄152 ∙ 2𝜋 ∙ 103
(12)
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TEP
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Capacitor in the AC circuit
Task 3: Measure the phase displacement between the terminal voltage and total current as a function of
the frequency in the circuit.
As can be seen from equation (7) the dependence between phase shift and frequency is also rational.
The fitting procedure delivers relation (13) which approximates the measured values fairly well.
tan 𝜑 =
9.5
𝜈⁄kHz
− 1.37
Fig. 5:
(13)
Frequency-dependent phase shift between terminal voltage and total
current for 𝑪𝟏 . Relation (13) gives the fitted function.
Task 4: Determine the total capacitance of several capacitors connected in parallel and in series.
For series (𝑠) and parallel (𝑝) connection with 𝑅 = 100 Ω we find the frequencies given in table 1. Calculation is done analog to equation (12). One can easily verify that the found total capacitances follow
equations (9) and (10).
Tab. 1:
Coupling of capacitors in series or parallel connection.
series
parallel
6
Capacitors
1+3
1+2
2+3
1||3
1||2
2||3
𝜈⁄kHz
2.00
2.25
1.00
0.30
0.50
0.25
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𝐶 ⁄µF
0.8
0.7
1.6
5.3
3.2
6.4
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