DC and AC Circuits. Objective. Theory. 1. Direct Current (DC) R-C Circuit

Transcription

1 [International Campus Lab] Objective Determine the behavior of resistors, capacitors, and inductors in DC and AC circuits. Theory Reference Young & Freedman, University Physics (14 th ed.), Pearson, R-C Circuit (p.886~890) 30.4 The R-L Circuit (p.1024~1028) 31.3 The L-R-C Series Circuit (p.1052~1056) 31.5 Resonance in AC Circuits (p.1060~1062) Symbols: : Resistance (or resistor) : Capacitance (or capacitor) : Inductance (or inductor) : Emf of the power source : Instantaneous terminal voltage or potential difference : Maximum terminal voltage (or voltage amplitude) : Instantaneous current : Maximum current (or current amplitude) : Instantaneous charge on the capacitor : Maximum charge on the capacitor : Frequency : Angular frequency, 2 : Reactance : Impedance 1. Direct Current (DC) R-C Circuit A circuit that has a resistor and a capacitor in series, as shown in Fig.1, is called an R-C circuit. We idealize the battery to have a constant emf and zero internal resistance. We begin with the capacitor initially uncharged (Fig.1a). At some initial time 0 we close the switch, permitting current around the loop to begin charging the capacitor (Fig.1b). The potential differences across the capacitor and the resistor are 0 and at 0 respectively. The initial current through the circuit is given by Ohm s law:. Fig. 1 Charging a capacitor. (a) Just before the switch is closed, the charge is zero. (b) When the switch closed (at 0), the current jumps from zero to /. As time passes, approaches and the current approaches zero. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 1 / 22

2 As the capacitor charges, increases and decreases, corresponding to a decrease in current. The sum of these two voltage is constant and equal to. After a long time the capacitor becomes fully charged, the current decreases to zero. Then the entire battery emf appears across the capacitor and. Let represent the charge on the capacitor and the current in the circuit at some time after the switch has been closed. The instantaneous potential differences and are Using these in Kirchhoff s loop rule, we find (1) 0 (2) (3) Integrating both sides of equation (5) and solving for and, we find 1 / 1 / (6) / / (7) Fig.2a and 2b are the graphs of Eqs. (7) and (6) respectively. After a time equal to, the current in the circuit has decreased to 1/ of its initial value. At this time, the capacitor charge has reached 1 1/ of its final value. The product is therefore a measure of how quickly the capacitor charges. We call the time constant, or the relaxation time, of the circuit, denoted by : (8) As the charge increases, the capacitor charge approaches its final value. The current decreases and eventually becomes zero. When 0, equation (3) gives When is small, the capacitor charges quickly. If is small, it s easier for current to flow, and the capacitor charges more quickly. (4) Note that the final charge does not depend on. We can derive general expressions for as functions of time. Substituting / into equation (3) yields Now after the capacitor has acquired a charge, we remove the battery from the circuit and connect point a and c (Fig.3). The capacitor then discharges through the resistor, and its charge eventually decreases to zero. 1 (5) Fig. 2 Current and capacitor charge as function of time for the circuit of Fig.1. Fig. 3 Discharging a capacitor. (a) Before the switch is closed at the time 0, the capacitor charge is and the current is zero. (b) At time after the switch is closed, the capacitor charge is and the current is. The actual current direction is opposite to the direction shown; is negative. After a long time, and both approach zero. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 2 / 22

3 In Fig.3 we make the same choice of the positive direction for current as in Fig.1. Then Kirchhoff s loop rule gives equation (3) but 0; that is, The current is now negative; this is because positive charge is leaving the left-hand capacitor plate in Fig.3b, so the current is in the opposite to that shown in the figure. (9) 2. Direct Current (DC) R-L Circuit A circuit that includes both a resistor and an inductor, and possibly a source of emf, is called an R-L circuit, as shown in Fig.5. The inductor helps to prevent rapid changes in current. Let be the current at some time after switch S is closed, and let / be its rate of change at that time. The potential differences and across the resistor and the inductor at that time are At 0, when, the initial current is /. To find, we rearrange and integrate equation (9), then we find Using these in Kirchhoff s loop rule, we find (16) / (10) / (11) 0 (17) 1 (18) We graph the current and the charge in Fig.4. Substituting equations (6), (7), (10), (11) and into equation (1) yields the terminal voltages across the capacitor and resistor as functions of time: 1 charging capacitor (12) charging capacitor (13) discharging capacitor (14) discharging capacitor (15) Fig. 5 An R-L circuit. Fig. 4 Current and capacitor charge as function of time for the circuit of Fig.3. Fig. 6 (a) Graph of versus for growth of current in an R-L circuit with and emf in series. The final current is ; after one time constant, the current is 11/ of this value. (b) Graph of versus for decay of current in an R-L circuit. After one time constant, the current is 1/ of its initial value. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 3 / 22

4 As Fig.6a shows, the instantaneous current first rises rapidly, then increases more slowly and approaches the final value / asymptotically. At a time equals to /, the current has risen to 1 1/ of its final value. The quantity / is therefore a measure of how quickly the current builds toward its final value; this quantity is called the time constant for the circuit, denoted by : (19) Now suppose switch S in Fig.5 has been closed for a while and the current has reached the value. We open S and close S at the same time. Then the current through and decays smoothly, as shown in Fig.6b. The Kirchhoff s-rule loop equation is obtained from equation (17) with 0, then the current varies with time according to (20) The terminal voltages across the capacitor and resistor are Note 1 0 (21) 0 (22) 0 (23) 0 (24) 3. Resistor in an AC Circuit When a sinusoidal current cos flows through, as shown in Fig.7a, the instantaneous voltage across is cos (31) The maximum voltage, the voltage amplitude, is the coefficient of the cosine function: Hence we can also write (32) cos (33) The current and voltage are both proportional to cos, so the current is in phase with the voltage. Equation (32) shows that the current and voltage amplitudes are related in the same way as in a DC circuit. Fig.7b shows graphs of and as functions of time. The vertical scales for current and voltage are different, so the relative heights of the two curves are not significant. The corresponding phasor diagram is given in Fig.7c. Because and are in phase and have the same frequency, the current and voltage phasors rotate together; they are parallel at each instant. Their projections on the horizontal axis represent the instantaneous current and voltage, respectively. Equations (16)-(24) is valid only for an idealized inductor which has negligible resistance. A real inductor generally has resistance due to the resistance of a long fine wire. Thus an inductor can be considered as L-R L combination in series. Then equations (21)~(24) become 1 0 (25) 0 (26) 0 (27) 0 (28) Fig. 7 Resistance connected across an ac source. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 4 / 22

5 4. Capacitor in an AC Circuit Next, we connect a capacitor with capacitance to the source, as in Fig.7a, producing cos through the capacitor. Integrating and using, we get or sin (34) sin (35) cos 90 (36) The current is proportional to the rate of change of voltage, so the voltage and current are out of phase by a quarter-cycle, as in Fig.8b. The peaks of voltage occur a quarter-cycle after the corresponding current peaks, and we say that the voltage lags the current by 90. Equation (35) shows that the maximum voltage is Then (39) The SI unit of is the ohm, the same as for resistance, because is the ratio of a voltage and a current. is inversely proportional both to and to. Capacitors tend to pass high-frequency current and to block low-frequency current and DC. A device that preferentially passes signals of high frequency is called a high-pass filter. 5. Inductor in an AC Circuit Finally, we connect a pure inductor with inductance and zero resistance, as in Fig.9a. Although there is no resistance, there is a potential difference between the inductor terminals because the current varies with time, giving rise to a selfinduced emf /. Thus the voltage is given by /, the negative of the induced emf. (37) cos sin (40) To put this expression in a form similar to equation (32), we define the capacitive reactance of a capacitor as 1 (38) or cos 90 (41) The voltage and current are out of phase by a quarter-cycle. Since the voltage peaks occur a quarter-cycle earlier than the current peaks, we say the voltage leads the current by 90. Fig. 8 Capacitor connected across an ac source. Fig. 9 Inductance connected across an ac source. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 5 / 22

6 From equation (40), the amplitude is 5. Alternating Current (AC) L-R-C Series Circuit (42) We define the inductive reactance of an inductor as (43) Using, we can write equation (42) in a form similar to (32) (44) To analyze a series circuit containing,,, and an AC source, we use a phasor diagram, as shown in Fig.10. The instantaneous total voltage across all three components is equal to the source voltage at that instant. The phasor representing this total voltage is the vector sum of the phasors for the individual voltage. We use the symbols,, and for the instantaneous voltage across,, and, and the symbols,, and for the maximum voltages. We denote the instantaneous and maximum source voltages by and. Then,,,, and. The inductive reactance is a description of the selfinduced emf that opposes any change in the current through the inductor. Thus, according to equation (43), the inductive reactance and self-induced emf increase with more rapid variation in current (that is, increasing angular frequency ) and increasing inductance. Since is proportional to frequency, a high-frequency voltage applied to the inductor gives only a small current. Inductors are used in some circuit applications to block high frequencies while permitting lower frequencies or DC to pass through. A circuit device that uses an inductor for this purpose is called a low-pass filter. The voltage across a resistor is in phase with the current and its maximum value is. The phasor in Fig.10b, in phase with the current phasor, represents the voltage across the resistor. The voltage across an inductor leads the current by 90 and its voltage amplitude is. The phasor in Fig.10b represents the voltage across the inductor. The voltage across a capacitor lags the current by 90 and its voltage amplitude is. The phasor in Fig.10b represents the voltage across the capacitor. So the magnitude of the phasor, which represents the total voltage across the series of elements, is (47) We define the impedance of an AC circuit as the ratio of the voltage amplitude across the circuit to the current amplitude in the circuit. From equation (47), the impedance of the L-R-C series circuit is (48) So we can rewrite equation (47) as (49) Fig. 10 An L-R-C series circuit with an ac source. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 6 / 22

7 Equation (49) has a form similar to, with impedance in an AC circuit playing the role of resistance in a DC circuit. Just as direct current tends to flow the path of least resistance, so alternating current tends to follow the path of lowest impedance. This is the angular frequency at which the inductive and capacitive reactances are equal, so at resonance, 1 1 (53) We can see the meaning of impedance for a series circuit by substituting and 1 into equation (48). The resonance frequency is 2. This is the frequency at which the greatest current appears in the circuit for a given source voltage amplitude. 1 (50) Impedance is actually a function of,,, as well as of the angular frequency. Hence for a given amplitude of the source voltage applied to the circuit, the amplitude of the resulting current will be different at different frequencies. In the phasor diagram shown in Fig.10b, the angle between the voltage and current phasors is the phase angle of the source voltage with respects to the current. tan 1 If the current is cos, then the source voltage is (51) The current at any instant is the same in and. However, the instantaneous voltages across and always differ in phase by 180. At the resonance frequency,, and and are equal. Then, the total voltage across the L-C combination in Fig.10a is exactly zero. The voltage across the resistor is then equal to the source voltage. So at the resonance frequency the circuit behaves as if the inductor and capacitor were not there at all. The phase of the voltage relative to the current is given by equation (51). Fig.11b shows the variation of with angular frequency. If we can vary or of a circuit, we can also vary the resonance frequency. This is exactly how a radio or television receiving set is tuned to receive a particular station. cos (52) 6. Resonance in AC Circuits The impedance of the L-R-C series circuit depends on the frequency, as equation (50) shows. Fig.11 shows graphs of,,,, and as functions of. As the frequency increases, increases and decreases, hence there is always one frequency at which 0. At this frequency, has its smallest value, equal simply to the resistance and the maximum value of occurs. This peaking of the current amplitude at a certain frequency is called resonance. The angular frequency at which the resonance peak occurs is called the resonance angular frequency. Fig. 11 How variations in the angular frequency of an ac circuit affect (a) reactance, resistance and impedance, and (b) impedance, current amplitude and phase angle. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 7 / 22

8 Equipment 1. List Item(s) Qty. Description PC / Capstone software 1 Records, displays and analyzes the data measured by various sensors. Interface 1 Data acquisition interface designed for use with various sensors, including power supplies which provide up to 15 watts of power. Voltage Sensor 2 Measures terminal voltages. L-R-C Circuit Board 1 Contains various electronic components including resistors, capacitors, and an inductor. Patch Cords (with banana plugs) 3 Carry an electric signal. Multimeter 1 Measures voltage, current, and resistance. 2. Details (1) L-R-C Circuit Board (2) Multimeter 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 8 / 22

9 Procedure Experiment 1. DC R-C Circuit (1) Build an R-C circuit and connect the circuit to the interface. (2) Set up the PASCO Capstone program. 1 Connect the red banana plug patch cord from the red [OUTPUT1] port of the interface to the bottom jack of the 330 μf capacitor on the L-R-C circuit board. 2 Connect the black patch cord from the black [OUTPUT1] port of the interface to the bottom jack of 100 Ω resistor. (2-1) Add sensors. Click the ports which you plugged the sensors or patch cords into and select [Voltage Sensor] or [Output Voltage Current Sensor] from the list. 3 Connect the yellow patch cord across the inductor. 4 Connect one voltage sensor to [Analog input A] on the interface and attach the leads across the capacitor, making sure the red cable of the voltage sensor is connected to the bottom jack of the capacitor. 5 Connect the other voltage sensor to [Analog input B] on the interface and attach the leads across the resistor, making sure the black cable of the voltage sensor is connected to the bottom jack of the resistor. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 9 / 22

10 (2-2) Configure the signal generator. We will supply a positive 4V square wave signal through the circuit. Set the parameters of [Signal Generator Output 1] as below. [Waveform] : Square [Frequency] : 1 Hz 1s [Amplitude] : 2V 2V~2V [Voltage Offset] : 2V 2V 0V~4V [Auto] (starts or stops the generator automatically.) Measurement will be started manually and continued only for 3 seconds. Thus, set the parameters of [Stop Condition] only, as below. [Condition Type] : Time Based [Record Time] : s (2-5) Define an equation. We will add a graph of total voltage across the capacitor and the resister to verify Kirchhoff s loop rule. Click [Calculator] in the [Tools] palette and enter any name (for example, Vtotal) and equal sign = in a new line of [Calculations]. Now you will enter the sum of the measurements of two voltage sensors. If you type [, a pop-up list opens. Complete the equation by selecting [Voltage, Ch A (V)] and [Voltage, Ch B (A)] from the list. Enter the [Units]. (2-3) Adjust the sample rates of the measurements. Select [5.00 khz] for all sensors in the [Controls] palette. (2-4) Set the auto recording conditions. Click [Recording Conditions] in the [Controls] palette. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 10 / 22

11 (2-6) Create graph displays. Select following measurement for the -axis of each graph. Click and drag the [Graph] icon from the [Displays] palette into the workbook page. You need 4 graph displays. Output Voltage, Ch 01 : input signal through the circuit Voltage, Ch A : terminal voltage across the capacitor Voltage, Ch B : terminal voltage across the resistor Vtotal : sum of two terminal voltages. See the step (2-5). Select [Times] for the -axis of all graphs. (3) Begin recording measurements. Click the [Record] button at the left end of the [Controls] palette to begin collecting data. The position of each graph depends on your configuration. Change the unit of time to [ms] to read more precise values of time. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 11 / 22

12 (4) Analyze the graphs. Adjust the scale of the graph. You can read off the coordinates of the data using [Show coordinate ] icon. The graph of [Voltage, Ch A (A)], which is the terminal voltage / across the capacitor, represents on the capacitor, since is proportional to. (You can also add the graph of as explained above.) Attach the graphs to the table below and compare them with Fig.2 and Fig.4 of the [Theory]. Charging Discharging or or (5) Analyze your results. (5-1) Current through the R-C circuit The graph of [Voltage, Ch B (A)], which is the terminal voltage across the resistor, represents through the circuit, since is proportional to. If you want to plot the graph of, you can use [Calculator]. Enter the equation 100Ω in a new line of [Calculations] and add one more graph of. (5-2) Kirchhoff s loop rule The graph of [Output Voltage, Ch 01] shows the emf of the power source. The graphs of [Voltage, Ch A (V)] and [Voltage, Ch B (A)] show / and respectively. Measure the voltages from the graphs for any 5 points of time and verify Kirchhoff s loop rule. 0 (2) / The graph of [Vtotal] shows the sum of and. Compare it with the graph of and verify the rule is always valid. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 12 / 22

13 (5-3) Capacitance of the capacitor Equations (12) and (14) represent the terminal voltage of the charging or discharging capacitor. 1 / charging capacitor (12) / discharging capacitor (14) Note Most resistors and capacitors have a tolerance rating expressed as a percentage. The tolerance value is the extent to which the actual value is allowed to vary from its nominal value. The tolerance is usually indicated by colored stripes or codes on the surface. The most common tolerance variation is 1~10% for resistors and 5~20% for capacitors. If you know, you can find using your result and the equations above. After a time equal to, the capacitor charge has risen to 1 1/ of its final value, i.e. the terminal voltage across the capacitor reaches 63.2% of the input voltage. Find the time from the graph, and then calculate the capacitance of the capacitor using following equations. 11/ 1 / or Similarly, find the time at which the charge of discharging capacitor reaches 1/ 36.8% of the initial value. Calculate using following equations. (6) Repeat the experiment with 100 μf capacitor. Using 100 μf and 100 Ω, repeat the steps (3)-(5). (7) Compare the results. Compare the graphs of the terminal voltages across 330 μf and 100 μf capacitors. [ viewing of multiple runs Select ] or [ ] activates viewing multiple runs together. 1/ / or. result average marked charging discharging Attach the graphs and answer the following questions. Describe the difference of the graphs. Q What is the physical meaning of the difference? Explain your result using the time constant A 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 13 / 22

14 Experiment 2. DC R-L Circuits (2-1) Configure the signal generator. (1) Build an R-L circuit and connect the circuit to the interface. 1 Connect the red banana plug patch cord from the red [OUTPUT1] port of the interface to the right jack of the 8.2 mh inductor on the L-R-C circuit board. 2 Connect the black patch cord from the black [OUTPUT1] port of the interface to the bottom jack of 100 Ω resistor. 3 Connect one voltage sensor to [Analog input A] on the interface and attach the leads across the inductor, making sure the red cable of the voltage sensor is connected to the right jack of the inductor. 4 Connect the other voltage sensor to [Analog input B] on the interface and attach the leads across the resistor, making sure the black cable of the voltage sensor is connected to the bottom jack of the resistor. 5 Insert the iron core inside the inductor. [Waveform] : Square [Frequency] : 0.01s [Amplitude] : 2V 2V~2V [Voltage Offset] : 2V 2V 0V~4V [Auto] (starts or stops the generator automatically.) If the input voltage alternates before the terminal voltage approaches a certain value, set [Frequency] at 50Hz. (2-2) Adjust the sample rates of the measurements. Select [. ] for all sensors in the [Controls] palette. If you set the sample rates too high, the computer system may stop with an error message. If you have any problem proceeding the experiment with 50 khz, decrease the sample rates. However, remember that too low rate precludes the accurate analysis of the result. (2) Set up the PASCO Capstone program. Follow the steps of the previous experiment, EXCEPT NEXT. (Before you modify configuration, you should save your data of the R-C experiment with a different file name!) 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 14 / 22

15 (2-3) Set the auto recording conditions. (4) Analyze the graphs. Click [Recording Conditions] in the [Controls] palette. Note The measurements will be started manually and continued for only 0.03 seconds. Set the parameters of [Stop Condition] as below. (The computer system may stop if the measuring time is too long.) The terminal voltage across the inductor theoretically decreases from the input voltage, 4 V, however, your graph will probably not, since the interface cannot detect the sudden change of voltage. [Condition Type] : Time Based [Record Time] :. (5) Analyze your results. (3) Begin recording measurements. (5-1) [Optional] Internal resistance of an inductor The graphs show that the terminal voltages across the inductor and the resistor do not approach the values as we predicted. The position of each graph depends on your configuration. (If the input voltage alternates before the terminal voltage approaches a certain value, set [Frequency] at 50Hz. See step (2-1)) These result from the internal resistance of the inductor. For more accurate observations, we should consider. Measure the asymptotic values of and and calculate using. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 15 / 22

16 (5-2) Current through the R-L circuit The graph of [Voltage, Ch B (A)], which is the terminal voltage across the resistor, represents through the circuit, since is proportional to. Similarly, find the time at which the current across the resistor is decreased to 1/ 36.8% of the initial value. Calculate using following equations. 1/ or If you want to plot the graph of, use [Calculator] as explained in previous experiment. Attach the graphs and compare them with the Fig or result average marked Note 0 0 (5-3) Kirchhoff s loop rule Follow the instructions of the previous experiment. (5-4) Inductance of the inductor 1 0 (21) 0 (22) 0 (23) 0 (24) If you know, you can find using your result and the equations above. At a time equal to, the current has risen to 1 1/, or 63.2%, of its final value, i.e. the terminal voltage across the resistor reaches 63.2% of the input voltage. Find the time from the graph, and calculate the inductance of the inductor using following equations. (If you cannot find the exact value of time, calculate the time by supposing two adjacent points are in a linear relationship.) For more accurate calculation, we should consider the internal resistance of the inductor. 1 0 (25) 0 (26) 0 (27) 0 (28) In this case, the equation in step (5-4) becomes 11/ 1 or You should also consider that the experimental value of the terminal voltage across the resistor is lower than the predicted value of an idealized circuit. (6) Repeat measurements with different conditions. Repeat the steps (3)-(5) with the combinations of 33 Ω and 8.2 mh, and 10 Ω and 8.2 mh. 11/ 1 or Repeat the experiment without iron core inside the inductor. Compare the results and explain why. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 16 / 22

17 Experiment 3. Resistor in an AC Circuit (2-3) Create a scope display. (1) Connect 10 Ω to the interface signal output. Click and drag the [Scope] icon from the [Displays] palette into the workbook page. (2-4) Add an independent -axis on the right of the graph. (2) Set up the PASCO Capstone program. (2-1) Add an [Output Voltage Current Sensor]. (2-2) Configure [Signal generator]. [Waveform] : Sine [Frequency] : 100 Hz 0.01s [Amplitude] : 5V 5V~5V [Auto] (2-5) Define the axes. -axis: Time(s) -axis(left): Output Voltage (V) -axis(right): Output Current (A) 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 17 / 22

18 (2-6) Select [Fast Monitor Mode] in the controls palette. (4) Analyze your results. [Fast Monitor Mode] displays data without recording. Check that [Recode] button is changed to [Monitor] button. Measure the phase difference between cos and cos. Compare your result with Fig.7 in [Theory]. (3) Start monitoring data. (3-1) Click [Monitor] in the controls palette. Experiment 4. Capacitor in an AC Circuit Connect 100 μf to the interface signal output. (3-2) Click [Activate trigger] in the toolbar to horizontally align repetitions of the signal. [Waveform] : Sine [Frequency] : 100 Hz 0.01s [Amplitude] : 5V 5V~5V Color in legend matches plot color. Measure the phase difference between cos and cos. Compare your result with Fig.8 in [Theory]. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 18 / 22

19 Experiment 5. Inductor in an AC Circuit Experiment 6. AC L-R-C Series Circuit Connect 8.2 mh to the interface signal output. (1) Build an L-R-C series circuit. [Waveform] : Sine [Frequency] : 0.002s [Amplitude] : 5V 5V~5V Connect 8.2 mh, 10 Ω, and 100 μf in series. (2) Set up the PASCO Capstone program. Follow the setups of the experiment 3~5. Also, the following additional setup is required. (2-1) Create a table display. Click and drag the [Table] icon from the [Displays] palette into the workbook page. Select [User-Entered Data] and enter Frequency Hz and Curret A on each column. Measure the phase difference between cos and cos. Compare your result with Fig.9 in [Theory]. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 19 / 22

20 (2-2) Create a graph. (3) Start monitoring data. -axis: Frequency (Hz) (User-Entered Data) -axis: Current (A) (User-Entered Data) Set [Frequency] of [Signal Generator] at. Click [Monitor] and then click [Stop]. Measure the amplitude of the alternating current using [Show data coordinates ]. You now have [Scope], [Table] and [Graph] displays on the workbook page. Enter the frequency and the current amplitude on the table. (2-3) Change the unit of [Current] to [ma] in the [Scope] display for more accurate measurements (5) Repeat measurement, increasing the frequency. Increase the frequency in 10 Hz increments until 200 Hz and continue to enter values of frequency and current, In the range of slight variation in current, Increase the frequency in 5 Hz or 2 Hz increments 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 20 / 22

21 (6) Analyze your result. The graph shows the current through the circuit as a function of frequency. Find the resonance frequency of the circuit from the graph and compare your measured value with the theoretical value. (It is recommended to calculate the theoretical value using the measured and in the experiment 1 and 2. The actual value of the components may differ from the nominal value.) result theory 1 2 Attach the results on your report and answer the following questions. (7) Observe the variation of the phase difference. Compare the phase differences between and in the condition of,, and. Q At the resonance frequency; 1. Which graph among the figure 7-9 shows the phase relationship between and? 2. Find the impedance of the circuit using equation (50), How is the behavior of the combination of the capacitor and the inductor in the circuit? Explain using a phase diagram as in Fig.10 of theory. A 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 21 / 22

22 Result & Discussion Your TA will inform you of the guidelines for writing the laboratory report during the lecture. End of Lab Checklist Please put your equipment in order as shown below. Delete your data files and empty the trash can from the lab computer. Turn off the Computer and the Interface. Return the Iron Core of the inductor to its storage location on the L-R-C circuit board. Handle the L-R-C Circuit Board with care. The plastic case is fragile. Turn off the Multimeter. 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA ( ) Page 22 / 22