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Capacitor in the AC circuit LEP 4.4.05 -01 Related topics Inductance, Kirchhoffβs 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 Ξ© PEK carbon resistor 1 W 5% 10 Ξ© PEK carbon resistor 1 W 5% 100 Ξ© 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. www.phywe.com P244001 PHYWE Systeme GmbH & Co. KG © All rights reserved 1 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 ππβπ measured on the oscilloscope are divided by 2β2. In accordance with relation (1), the total current πΌ in the circuit can be deduced by measuring the voltage π across the resistor with resistance π . πΌ = πβπ (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 generatorβs signal should not be higher than 5 V and should have no offset. The digital frequency generatorβs 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 PHYWE Systeme GmbH & Co. KG © All rights reserved P2440501 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 π Ξ© = ππΆ = ππΆ (2) From this relation the capacitance πΆ of a capacitor can be deduced, if the impedance ππΆ 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) β possibly with variable sweep rate β 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 π = ππΆ = 100 Ξ©. The capacitance can then be calculated analogously to task 2. Theory If a capacitor of capacitance πΆ and the charge π on the plates as well as an ohmic resistor of resistance π are connected in a circuit, the sum of the potential differences across the capacitor and the resistor is equal to the terminal voltage ππ‘ which gives relation (3). π ππ‘ = πΌ β π + πΆ (3) This corresponds to Kirchhoffβs 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 Kirchhoffβs voltage law for the studied circuits, where UΞ© = πΌ β π is π the voltage across the resistor and ππΆ = πΆ the voltage across the capacitor. ππΆ + πΞ© β ππ‘ = 0 (4) Noting the fact that πΌ= ππ , ππ‘ from differentiation of (3) follows www.phywe.com P244001 PHYWE Systeme GmbH & Co. KG © All rights reserved 3 TEP 4.4.05 -01 Capacitor in the AC circuit πππ‘ππ‘ ππ‘ ππΌ 1 = π ππ‘ + πΆ . If the alternating voltage ππ‘ has the frequency π = 2ππ and the waveform ππ‘ = π0 cos ππ‘, (5) inserting (5) into equation (3) gives the following solution for the current πΌ: πΌ = πΌ0 cos(ππ‘ β π) (6) There π is the phase displacement which is given by 1 ππΆπ tan π = β (7) and πΌ0 = π0 βπ 2 +( (8) 1 2 ) ππΆ 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. πΆπ‘ππ‘ = 1β 1 β (9) πΆπ‘ππ‘ = β Ci (10) Ci 4 PHYWE Systeme GmbH & Co. KG © All rights reserved P2440501 LEP 4.4.05 -01 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 πΆ1 gives relation (11). 152 ππΆ (π)βΞ© = πβkHz β 4 Fig. 4: (11) Impedance values for various frequencies for the two capacitors πͺπ and πͺπ . Equation (11) gives the relation of the rational fit for πͺπ . 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 πΆ1 = 1.05 µF. 1 ππΆ βΞ© = 2π πβHzβC 1 βF 152 = πβkHz βΊ πΆ1 βµF = 1β152 β 2π β 103 (12) www.phywe.com P244001 PHYWE Systeme GmbH & Co. KG © All rights reserved 5 TEP 4.4.05 -01 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 π = 9.5 πβkHz β 1.37 Fig. 5: (13) Frequency-dependent phase shift between terminal voltage and total current for πͺπ . Relation (13) gives the fitted function. Task 4: Determine the total capacitance of several capacitors connected in parallel and in series. For series (π ) and parallel (π) connection with π = 100 Ξ© 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 πβkHz 2.00 2.25 1.00 0.30 0.50 0.25 PHYWE Systeme GmbH & Co. KG © All rights reserved πΆ βµF 0.8 0.7 1.6 5.3 3.2 6.4 P2440501