Lab 3 physics 222 PDF

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Brianna Jett and Caitlin Priester [email protected], [email protected] PHYS 221 - 001 8 September 2020 Laboratory 3 Report Goal: To investigate the behavior of simple circuits that contain resistors and capacitors. Experiment 1 Record the values for each member of your group with dry and with wet thumbs. Dry: 0.95MOhms 1.02m Ohms Wet: 1.76MOhms, 1.84MOhms Predict the current that would flow through your body if you held one terminal of a 1.5 V battery in one of your hands and the other terminal in the other hand. The current would be the voltage, 1.5 V, divided by the resistance of our body which is what we measured above. So using our numbers from above, you would do 1.5 V/ (0.95x10^6) = 1.58 x10^-6 A. Predict the current that would flow through your body if you held one terminal of a 110 V power source in one of your hands and the other terminal in the other hand. The current would be the voltage, 110 V, divided by the resistance of our wet body which is what we measured aboved. The resistance would be lower, so the current would be higher. Using the same numbers from above, we get 110V / (1.76 x10^6) = 6.25 x10^-5 A. Experiment 2

Are the 33 Ω and 100 Ω resistor ohmic resistors? Why or why not? Base your conclusions on your measurements. Yes, they are ohmic resistors. This is because both of the graphs show a linear relationship between current and voltage (the large graph). Do the resistances found from the slope of the voltage versus current graphs agree with the nominal resistances written on the RLC board? What are the percentage differences? Yes, the percentage difference for the first graph is 0%. The percentage difference for the second graph is approximately 1.4%-1.6%.

Is the bulb an ohmic resistor? Why or why not? Base your conclusions on your measurements. Describe your observations. 7.5 V bulb - just a regular light bulb, the light was flashing off and on and it never looked like it was completely on. It was more dim, and then off, and then dim, and then off again. This bulb is a non ohmic because to be ohmic it needs to follow ohm's law, which is V = IR. A normal filament bulb has its own resistance. To be ohmic, the object needs to have the same resistance when voltage and current are divided for any values of them. Additionally, as more current flows

through the lightbulb filament, it heats up and makes it harder for the electrons to flow through. So a bulb is non-ohmic. Can you determine the resistance of the bulb? The slope is the resistance. The slope is m= 16.5, which is a non ohmic light bulb. Describe your observations. Of the 150 Ohms bulb. This light bulb flashed green, then red, then green, then red. Change the signal generator amplitude to 5 V and repeat.

Describe your observations. This light bulb flashed green, then red, then green, then red. Is the LED an ohmic resistor? Why or why not? Base your conclusions on your measurements. Yes, the LED is an ohmic resistor. To be ohmic, the object needs to have the same resistance when voltage and current are divided for any values of them. Additionally, as more current flows through the lightbulb filament, it heats up and makes it harder for the electrons to flow through.

Online Activity Use figure 1. Find t½, the time it takes to charge the capacitor so that it holds half of the maximum charge at half its maximum voltage. From the graph, we see that the maximum voltage on the capacitor is 10 V because it is the voltage when the curve becomes asymptotic. Then, half this voltage is 5 V, and we can get the time when the voltage is 5 V from the graph, the time is around 0.14s Find t½, the time it takes for the charging current to drop to half of its maximum value. Do you get the same value for t½? From the graph, the maximum current through the resistor (current at t=0) is 5 mA. Half of this is 2.5 mA, and we can get the time to reach this current from the graph as we did earlier: There we see it is also about 0.14 s, so we can say they are equal. Use figure 1. Do you obtain the same values of t½, from these plots? Yes, for both Q and i, the t½ will be the same which is explained in part one. Use t½ = τ*ln2 = 0.693*τ to find the time constant τ of this RC circuit. 0.693 * 0.14 sec = 0.20 sec. If R = 2 kΩ , what is the value of C in this circuit? τ = RC = 20 second so, C= 100 uF What is the battery voltage V? The battery voltage v= v0 = 10 V....