Copy of Lab 4 - RC Circuits

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Lab 4 - RC Circuits Overview In this lab you will construct electrical circuits that contain capacitors . A capacitor is an electrical element that can store electrical charge. The amount of charge a capacitor can store depends upon its capacitance. If a voltage is applied across a capacitor, and amount of charge is stored on the capacitor, then its capacitance is: (1) Now you are going to construct circuits using capacitors. In Lab 2 you constructed circuits with resistors. When you apply a voltage across a resistor, a current flows through the resistor according to Ohm’s Law . The more voltage applied, the more current will flow. Likewise, when you apply a voltage across a capacitor, charge will be stored in the capacitor. What do you get when you combine a resistor and a capacitor together in a circuit? You get an RC circuit ! An important property of an RC circuit is its charging/discharging time constant : (2) In this lab you will construct different RC circuits and measure the time constant for each one. Part 1 - Short Time Constant From Eq. 2 up above, it is easy to see the duration of the time constant depends on the size of the capacitance of the capacitor in the circuit. A large capacitor can hold more charge, so it takes longer to discharge the capacitor. A smaller capacitor holds less charge so the time to discharge the capacitor is shorter. Construct the following circuit using a 10k resistor in series with a 56 Farad capacitor. Remember, 1 Farad = 1 Farad.

You may notice that the terminals of the capacitor are different lengths. When you connect the capacitor to the circuit, the longer terminal should be connected to the higher potential and the short terminal should connect to the lower potential. Once your circuit is connected, it should look like this:

● Red alligator clip connected to D6. ● Blue alligator clip connected to A7. ● White alligator clip connected to GND. The time constant is measured in seconds. In Eq. 2, you can see is the product of measured in Farads and measured in Ohms . 1. Using the units of Farads and Ohms , and show is measured in units seconds. 2. Calculate the value of the time constant for your circuit with = 10k and = 56 F. = .56 Setup the iOLab and application: ● Launch the iOlab application on your computer. ● Make sure the dongle is connected to the USB port of your computer and the iOLab is turned on. ● Check the A7 sensor. ● On the toolbar click the settings button (cog) > Expert Mode > Output configuration. ● At the bottom of the iOLab app window the D6 output should appear. ● For the D6 Output select 3.3V and then click On. Allow the capacitor to charge for about 10 seconds. Now it’s time to take some data: ● Click the Record button and record the A7 voltage for about 5 seconds. ● While still recording, change the D6 output to Off. ● Continue recording for about 10 - 15 seconds, then click Stop. Your data should look something like this:

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3. Take a screenshot of your iOLab application with the data and paste here: When the voltage supplied by the iOLab, the voltage across the capacitor does NOT instantaneously drop to zero. The voltage does decrease over time because the charge stored decreases, BUT not instantaneously. If you look closely at your data, you may notice the decrease in voltage follows a curve. This is an exponential curve. The voltage across the capacitor decreases with time: (3)

Now let’s record some data. In the application when you move the mouse arrow, a vertical line will track with the arrow. In the upper left corner of the graph, you should see the value of time and voltage for that position of the vertical line. 4. Record two sets of time and voltage across the discharging capacitor: One set at a high point along the curve, just after turning off D6, and the other set at a low point on the curve. Do not measure the voltage along the flat part of the curve. Record your value in the table below. Time (sec) Voltage (V) first point ( , ) 5.78 2.897 second point ( , ) 7.5 .179 In Eq. (3) above we see how the voltage across the discharging capacitor is related to the time. At the first point the voltage is: (4) At the second point the voltage is: (5) You can combine this two equations and show the time constant is: (6) 5. Combine Eqs. (4) and (5) and show is equal to Eq. (6). Here is the procedure: a. Divide Eq. (4) by Eq.(5) V1/V2 = (e^-t1/T) / (e^-t2/T) b. Take the natural log of both sides of the equation. ln(V1/V2) = ln( (e^-t1/T) / (e^-t2/T) ) c. Use the logarithmic identity ln(V1) - ln(V2) = -t1/T + t2/T d. Show your work here: T = (t2 - t1) / (ln(V1) - ln(V2)) 6. Use your values of ( , ) and ( , ), and calculate the value of you measured in your DC circuit. Record your value of in table below: for 56 F Capacitor (sec)

.618 7. How does your measure value compare the value of calculated in question 2? |0.618-0.56|= 0.058s Part 2 - Long Time Constant Now replace the 56 F capacitor with a 220 F capacitor. Remember, the capacitor has a long and a short terminal. The long terminal should be on the higher potential and the short terminal should be on the lower potential. 8. Calculate the value of the time constant for your circuit with = 10k and = 220 F. T = 2.2 Repeat the same procedure of recording data: ● For the D6 Output select 3.3V and then click On. Allow the capacitor to charge for about 10 seconds. ● Click the Record button and record the A7 voltage for about 5 seconds. ● While still recording, change the D6 output to Off. ● Continue recording for about 20 - 30 seconds, then click Stop. 9. Take a screenshot of your iOLab application with the data and paste here: 10. Record two sets of time and voltage across the discharging capacitor: One set at a high point along the curve, just after turning off D6, and the other set at a low point on the curve. Do not measure the voltage along the flat part of the curve. Record your value in the table below. Time (sec) Voltage (V)

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first point ( , ) 6.57 2.991 second point ( , ) 11.11 0.44 11. Use your values of ( , ), ( , ), and Eq. (6), and calculate the value of you measured in your RC circuit. Record your value of in table below: for 220 F Capacitor (sec) 2.369 12. How does your measure value compare the value of calculated in question 8? 2.369-2.2 = .169 Part 3 - Parallel Capacitors Just like you did in Lab 2 with resistors, you can combine capacitors in different ways to make a new capacitor. First combine the 56 F capacitor with the 220 F capacitors in parallel with each other. Your circuit should look like this: ● Red alligator clip connected to D6. ● Blue alligator clip connected to A7. ● White alligator clip connected to GND. When capacitors are combined in parallel, the equivalent capacitance is the sum of the individual capacitors:

, (7) 13. If two capacitors are connected in parallel, what must be true about the voltage across each capacitor? Explain your answer. The voltage across each capacitor is the same because the potential difference is 0 and the potential between them is the same. 14. Using your answer to question 13, conservation of electric charge (Hint: Kirchhoff’s Junction Rule), and the equation , derive the equation for equivalent capacitance for capacitors in parallel. Show your work. Q eq = C eq V Q eq = Q1 + Q2 <- conservation of electric charge Q1 = C1V Q2 = C2V C eq V = C1V + C2V C eq = C1 + C2 15. Will the time constant be longer or shorter when capacitors are combined in parallel? Explain your answer. It becomes longer because when capacitors are combined in parallel the capacitance of the circuit increases. 16. Calculate the equivalent capacitance of the 56 F and 220 F capacitors combined in parallel. 56 + 220 = 276 F 17. Calculate the value of the time constant for your circuit with = 10k and the equivalent parallel capacitor. = 2.76 Repeat the same procedure of recording data: ● For the D6 Output select 3.3V and then click On. Allow the capacitor to charge for about 10 seconds. ● Click the Record button and record the A7 voltage for about 5 seconds. ● While still recording, change the D6 output to Off. ● Continue recording for about 40 - 50 seconds, then click Stop. 18. Take a screenshot of your iOLab application with the data and paste here:

19. Record two sets of time and voltage across the discharging capacitor: One set at a high point along the curve, just after turning off D6, and the other set at a low point on the curve. Do not measure the voltage along the flat part of the curve. Record your value in the table below. Time (sec) Voltage (V) first point ( , ) 6.22 3.036 second point ( , ) 11.87 .899 20. Use your values of ( , ), ( , ), and Eq. (6), and calculate the value of you measured in your RC circuit. Record your value of in table below: for Parallel Capacitor (sec) 4.64 21. How does your measure value compare the value of calculated in question 15? 4.64 - 2.76 = 1.88 Part 4 - Series Capacitors Finally you can combine the capacitors in series . combine the 56 F and 220 F capacitors in series with each other. Your circuit should look like this:

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● Red alligator clip connected to D6. ● Blue alligator clip connected to A7. ● White alligator clip connected to GND. When capacitors are connected in series, the equivalent capacitance is: (8) 22. If two capacitors are connected in series , what must be true about the charge stored in each capacitor? Explain your answer. Since the current is equivalent because they are in series, each capacitor must store the same amount of charge, so capacitors in series have equivalent charges. 23. Using your answer to question 22, conservation of energy (Hint: Kirchhoff Loop Rule), and the equation , derive the equation for equivalent capacitance for capacitors in series. Show your work. V eq = Q / C eq V1 = Q/C1 V2 = Q/C2 Q/C eq = Q/C1 + Q/C2 1/C eq = 1/C1 + 1/C2

24. Will the time constant be longer or shorter when capacitors are combined in series? Explain your answer. Shorter because the total capacitance, C, will be less than that of when there is a single capacitor. 25. Calculate the equivalent capacitance of the 56 F and 220 F capacitors combined in series. 1/(1/56 + 1/220) = 44.64 F 26. Calculate the value of the time constant for your circuit with = 10k and the equivalent series capacitor. = .4464 Repeat the same procedure of recording data: ● For the D6 Output select 3.3V and then click On. Allow the capacitor to charge for about 10 seconds. Mine wasn’t charging properly and only reached a peak of 1.3V for some reason, but I did what I could. ● Click the Record button and record the A7 voltage for about 5 seconds. ● While still recording, change the D6 output to Off. ● Continue recording for about 10 - 20 seconds, then click Stop. 27. Take a screenshot of your iOLab application with the data and paste here: 28. Record two sets of time and voltage across the discharging capacitor: One set at a high point along the curve, just after turning off D6, and the other set at a low point on the curve. Do not measure the voltage along the flat part of the curve. Record your value in the table below.

Time (sec) Voltage (V) first point ( , ) 6.9 1.255 second point ( , ) 8.38 .114 29. Use your values of ( , ), ( , ), and Eq. (6), and calculate the value of you measured in your RC circuit. Record your value of in table below: for Series Capacitors (sec) .617 30. How does your measure value compare the value of calculated in question 22? .617 - .4464 = .171

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