- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 1 (40.52 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 2 (40.62 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 3 (48.25 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 4 (57.06 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 5 (37.5 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 6 (35.7 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 7 (71.15 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 8 (50.98 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 9 (53.57 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 10 (67.03 KiB) Viewed 13 times
- Equipment 1 1 1 1 1 Resistor Capacitor Inductor Network Ui 5210 Voltage Sensor Ui 5100 Short Patch Cord Set Se 7123 Not 11 (72.51 KiB) Viewed 13 times
78 RC Circuit To determine how the charge on a capacitor decays in time, use Kirchhoff's Loop Rule for Figure 1: Vo=Vc + VR (2) Solving Equation (1) for the voltage across the capacitor gives Vc = Q/C (3) The voltage across the resistor is given by Ohm's Law VR = IR Therefore,, Vo = Q/C+IR (5) Since the applied voltage V, is zero when the capacitor is discharging. Equation (5) reduces to Q/C+IR=0 Since the current is dQ 1= dt Equation (6) becomes the differential equation dQ +0=0 dt Solving Equation (8) for Qgives 78-Page 2 of 5 (6) Q=Qmaxe Plugging Qinto Equation (2) gives the voltage across the capacitor as a function of time cod Written by Ann Hanks V(t) = (10) where V.QC. The rate that voltage across a capacitor (and the charge stored in the capacitor) decreases depends on the resistance and capacitance that are in the circuit. If a capacitor is charged to an initial voltage, Ve, and is allowed to discharge through a resistor, R. the voltage, V, across the capacitor will decrease exponentially The half-life, tio is defined to be the time that it takes for the voltage to decrease by half V(t) = V₁/2 = V₂e-¹/c Solving for the half-life gives tx-RC In 2. half life (11) (12) PASCO
78 RC Circuit V(L) The product RC is called the capacitive time constant and has the units of seconds. Pre-lab Questions Ya Show that the capacitive time constant RC has units of seconds. 2. If the capacitance in the circuit is doubled, how is the half-life affected? half-life 0.256 3. If the resistance in the circuit is doubled, how is the half-life affected? If the charging voltage in the circuit is doubled, how is the half-life affected? To plot the equation V(t)= Ve so the graph results in a straight line, what quantity do you have to plot vs. time? What is the expression for the slope of this straight line? (2) half lif = 0.112 ms Setup 1. Construct the circuit shown in Figure 2. The voltage source is Signal Generator #1 on the 850 Universal Interface. C-3900 pF and R-47 k C R Written by Ann Hanks 78- Page 3 of 5 VA (Data for Analysis) 3.990 V 2U8 t= 1.4 88 1.996 V 161.744. 1.014 6-1.856 @half-life: 20.13489 t=1.976 Figure 2: RC Circuit Diagram 2. Click on Signal Generator #1 to connect the internal Output Voltage-Current Sensor. Set the signal generator to a 350 Hz positive square wave with 4 V amplitude. Set the signal generator on Auto. 3. Plug the Voltage Sensor into Channel A. Connect the Voltage Sensor across the capacitor. Procedure 1. Set up an oscilloscope display with the Voltage Ch.A and the Output Voltage on the same axis. Select the Fast Monitor Mode. Click Monitor and adjust the scale on the oscilloscope so there is a complete cycle, so the capacitor fully charges and discharges. PASCO
78 RC Circuit 2. Increase the number of points (using the tool on the scope toolbar) to the maximum allowed. Then take a snapshot of both voltages shown. Rename the snapshots "3900pF". Analysis 1. Create a graph with Voltage Ch.A and the Output Voltage vs. time. Select the voltages for the 3900 pF run on the graph. 2. Using the Coordinates Tool, measure the time it takes for the voltage to decay to half of its maximum. This time is the half-life. It may be necessary to reduce the snap-to-pixel distance to 1 in the properties of the Coordinates Tool (right click on the tool to access the properties). 78- Page 4 of 5 3. Measure the time it takes for the voltage to decay to one-quarter of its maximum. This is two half-lives. Then divide this time by two to find the half-life. 4. Measure the time it takes for the voltage to decay to one-eighth of its maximum. This is three half-lives. Then divide this time by three to find the half-life. Take the average of the three measured values of the half-life. Estimate the precision of the measurement and-state-it-as- (half-life + precision). 0.256 + 0.112 + 0.12 = 0.163 V 3 Calculate the theoretical half-life given by Equation (12) and compare it to the measured value using a percent difference. Linearize the Data 1. Create a calculation x-In(V/V) and plot it versus time. Also plot the voltage across the capacitor in the same plot area. This will help you identify where the capacitor is discharging. half-life = 0.1ms Data for increased 7.967V the vollage t 2.410 sec 4.009 u to 2.550 2. In the graph, choose the 3900 pF run. Select the part of the In(V/Vo) plot where the capacitor is discharging and fit it to a straight line. Ghalf-life= 0.115 ms 1.999 3. Use the slope of the line to find the half-life. To determine how the slope is related to the t 2.670 half-life, solve Equation (10) for In(V/V). 4. Does this value agree with the value found from the previous analysis! Written by Ann Hanks helf life = 12 ms Increase the Voltage Increase the voltage amplitude to 8 V. Keep the circuit components the same. Repeat the procedure and analysis. half life alles ✔ PASCO 0.984 t= 2.790 10.514 2.900
78 RC Circuit Decrease the Capacitance Replace the 3900 pF capacitor with a 560 pF capacitor and repeat the procedure and analysis. The frequency of the signal generator should be changed to 1800 Hz. Return the output voltage to 4 V Conclusions 1. Summarize how changing the voltage and capacitance changes the half-life. 2. Include the values found for the half-lives and the % differences. Does the theoretical value lie within the range of precision of your measurements? Explain what causes the differences. Did your answers to the Pre-Lab Questions agree with the results? Written by Ann Hanks (Data for Decrease the capacitance) 3.989 V 0.6695 ms half-life = 0.02 86 ros: 78-Page 5 of 5 half life= 0.0195 half life= 0,0182 ms 1.988 u 0.6981m 0.913 V 0.7176 ms. 0.495 V to 0.7358/ns PASCO
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Voltage, Ch A (V) 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 0.5 3.980 1.0 1.996 1.014 0.489 1 1.976 Figure 4: Graph of the Voltages for Analysis 2.5 (ms) 3.0 Monitor Run 3.5 4.0 4.5
Voltage (V) 9 8 7 6 5 4 با 3 2 1 0 0.000 0.001 0:002 Time (s) 0.003 Monitor Run 0.004 < Vo H 0.005
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Voltage (V) 9 8 7 6 5 4 3 2 -1 0.0 0.5 1.0 15 Time (ms) 2.0 Monitor Run V 2.5 Vo 4
Voltage, Ch A (V) 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 3.984 1.988 0.973 0.495 {} 0.7358 L.O Monitor Run 1.5 Time (ms) 2.0 V 2.5