Wendler_K_152_lab3_

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Dec 6, 2023

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Karsyn Wendler Physics 152- Lab Section: Wednesday 2:00pm Lab #03- DC Circuits September 13, 2023 INTRODUCTION In this lab my partner Sophia and I measured current through and voltage across circuits with different arrangements of resistors. We used this data to investigate the relationships that play a role in Ohm’s Law and Kirchoff’s Laws. PART ONE: OHM’S LAW 1-A We begin part one of the experiment by setting up a circuit with a DC power supply, an ammeter, and a resistor connected in series on a breadboard. Additionally, a voltmeter is arranged on the breadboard to measure the voltage across the resistor. The set up can be seen in the photograph below: Circuit with power supply, ammeter, resistor in series. Voltmeter also attached to resistor. Now, we set the DC power supply to produce 10.0 volts across the circuit and record the readings on both the ammeter and voltmeter. We repeated this for 10 other voltages ranging from 0.0-10.0 volts from the DC power supply. The data is summarized in the table below. DC Power Supply Ammeter Reading (mA) Voltmeter Reading (V) 0.1V 0.06 ± 0.01 mA 0.12047 ± 0.0001 V 1.0V 0.59 ± 0.01 mA 1.0405 ± 0.0001 V 2.0V 1.10 ± 0.01 mA 2.0404 ± 0.0001 V 3.0V 1.65 ± 0.01 mA 3.0815 ± 0.0001 V 4.0V 2.17 ± 0.01 mA 4.0826 ± 0.0001 V 5.0V 2.71 ± 0.01 mA 5.1053 ± 0.0001 V 6.0V 3.23 ± 0.01 mA 6.0785 ± 0.0001 V 7.0V 3.80 ± 0.01 mA 7.1590 ± 0.0001 V 8.0V 4.28 ± 0.01 mA 8.1521 ± 0.0001 V 9.0V 4.62 ± 0.01 mA 9.1014 ± 0.0001 V 10.0V 5.18 ± 0.01 mA 10.1108 ± 0.0001 V

1-B Graph of 1-A data: Using this graph, the slope informs us of the average conductance (in Amperes per volt) as 5.12 × 10 ’( Siemens. Taking the reciprocal of this value gives us the resistance in ohms: ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? (Ω) = 1 5.12 × 10 ’( ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = 1953.13 Ω 1-C + 1-D Resistor Value (removed from board and measured with bench multimeter as ohmmeter): ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = 1959.74 Ω This resistance is similar to the calculation from part 1-B, but the calculation is not precise enough to fall within the uncertainty of the measured value. 1-E The bands on the resistor indicate a resistance of 2000Ω ± 5% . The R found via graph trendline in 1-B is 1953.13 which differs from the band indicated value by 2.34%. The R found via measuring in 1-C is 1959.74 which differs from the band indicated value by 2.01%. Thus, both values are within the resistance margin. 1-F Based on all of the data in part one, we can provide the following mathematical statement of Ohm’s law where V is voltage, I is current and R is resistance: ࠵? = ࠵? ࠵?

PART TWO: RESISTORS AND EQUIVALENT RESISTANCE 2-A Posed Question: If you have a group of resistors connected in series, do you think the total resistance of the group will be larger or smaller than the individual resistors? I think that the effects of multiple resistors in series will compound to create a total resistance larger than the individual resistors in the group. This seems logical to me because the single current will have to pass through all of the resistors. Posed Question : If you have a group of resistors all connected in parallel, do you think the total resistance of the group will be larger or smaller than the individual resistors? I think that when connected in parallel, multiple resistors will result in a total resistance smaller than that of the individual resistors because the current is splitting between multiple paths. 2-B, 2-C, 2-D, 2-E We began the experimental portion of part two by arranging 6 resistors of 5 different values in series on an empty portion of our breadboard. Then, we connected this to our bench multimeter, using it as an ohmmeter to measure the resistance of each resistor. Set-up and data are both summarized below:

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Order # Band Colors Theoretical Resistance Values Measured Resistance Values 1 Yellow, purple, brown, gold 470 459.964 2 Brown, black, red, gold 1,000 988.100 3 Brown, black, orange, gold 10,000 9,966.8 4 Brown, black, red, gold 1,000 986.206 5 Brown, black, brown, gold 100 98.580 6 Yellow, purple, orange, gold 47,000 46,780.9 1-6 TOTAL -- -- 59,288.6 2-D It was ok to have all other resistors remain connected in series while measuring each ones individual resistance because the current flowing into the next resistor is always the same as that flowing out of the previous one, so it would not affect any measurements. Using the individually measured resistance value from the chart above, we can sum them up to find the theoretical equivalent resistance of resistors in series: ࠵? >?@ABCD>EF (G>HA>G) = (࠵? I + ࠵? K + ࠵? L + ⋯ ) ࠵? >?@ABCD>EF (G>HA>G) = (459.964 + 988.100 + 9,966.8 + 986.206 + 98.580 + 46,780.9) ࠵? >?@ABCD>EF (G>HA>G) = 59,280.55 With only an 8 ohm difference between the measured and theoretically calculated resistance values, we can conclude that the theoretical value of the equivalent resistance does fall within the range of our experimental value. 2-F Now, we will test the second question posed in 2-A. So, we will set up 3 resistors of 3 different values in parallel on another empty region of the breadboard. Again, we will use the bench multimeter as an ohmmeter to measure the resistance of each resistor as well as the total resistance of all the resistors in parallel. This setup and data are summarized below:

Order # Band Colors Theoretical Resistance Values Measured Resistance Values 1 Yellow, purple, orange, gold 47,000 46,409.2 2 Brown, black, orange, gold 10,000 9,923.41 3 red, black, red, gold 2,000 1,987.88 TOTAL -- -- 1,599.85 Using the individual resistances from the chart above, we can calculate the equivalent resistance of resistors in parallel: ࠵? >?@ABCD>EF (QCHCDD>D) = R 1 ࠵? I + 1 ࠵? K + 1 ࠵? L + ⋯ S ’I ࠵? >?@ABCD>EF (QCHCDD>D) = R 1 46409.2 + 1 9923.41 + 1 1987.88 S ’I ࠵? >?@ABCD>EF (QCHCDD>D) = (6.25367 × 10 ’( ) ’I ࠵? >?@ABCD>EF (QCHCDD>D) = 1599.06 Because 1599.89 and 1599.06 are only a mere 0.83 ohms apart, we can assume that the theoretical value of the equivalent resistance does fall within the range of our experimental value. PART THREE: CIRCUITS AND KIRCHOFF’S LAWS 3-A Based on the photo above (a mechanical circuit analogous to an electrical one) we can propose that the currents going through R2 and R3 equate the currents coming out of R1. This is

because nothing can be lost between R1 and R2/R3. And, although they are divided, the same total amount of current that left R1 is being split amongst the R2/R3 section. ࠵? I ࠵? I ࠵?࠵?࠵?࠵?࠵? = R ࠵? K ࠵? K + ࠵? L ࠵? L S࠵?࠵?࠵?࠵?࠵?࠵?࠵? Additionally, we can propose that the total lost potential energy of a marble passing through R1 and R2 is equal to that of one passing through R1 and R3. This seems logical as gravitational potential energy is determined by height off the ground, and the heights before and after R2/R3 are the exact same/both take us back to 0 PE. Thus, using the law of gravity to support my theory, I will assume that passing through either R2 or R3 would result in the same loss of potential energy. ∆࠵? (࠵? I + ࠵? K ) = ∆࠵? (࠵? I + ࠵? L ) 3-B Considering the rules proposed in 3-A as a circuit, we can state this general rule for junctions where currents come together/split apart: ࠵? >XAGFAEY = ࠵? >EF>HAEY Understanding that the change in voltage across R1 and R2 is equal to the change in voltage across R1 and R3, and the battery counter-acts both of these current paths, we can state the following general rule for voltage across circuit elements connected in series: ࠵? Q[\>H G@QQD] = ࠵? I + ࠵? K + ࠵? E 3-C, 3-D, 3-E, 3-F Using the breadboard, we set up a circuit series connecting the power supply to a single resistor (R1) then 2 resistors parallel (R2 and R3). This arrangement can be seen in the photo:

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Then, we measured voltage, resistance and current across all series components. The data can be found in the table below: R1 R2 R3 Power Supply Voltage 4.18775 V 6.00387 V 6.00387 V 10.19192 V Resistance 459.915 Ω 660.292 Ω 660.292 Ω -- Current 8.9 mA 7.0 mA 3.0 mA 9.0 mA Discussion: 3-C – Voltage 4.18775 ࠵? + 6.00387 ࠵? = ࠵? Q[\>H G@QQD] 10.19162 ࠵? ≈ 10.19192 ࠵? Because the voltage across R2 and R3 is exactly the same, and they both sum with R1 to approximately the same voltage as the power supply, we can assume that Kirchhoff’s voltage law does work as a real, physical relation. 3-D – Resistance 459.915 Ω + 660.292 Ω = ࠵? >?@ABCD>EF 1,120.207 Ω = ࠵? >?@ABCD>EF Measured Value of Total Resistance = 1,119.990 Ω 3-E – Current (total) Using Ohm’s Law and our previous measurements we can calculate the total current of the series to be: ࠵? = ࠵? ࠵? ࠵? = 10.19192 1,119.990 ࠵? = 0.0091࠵? = 9.1࠵?࠵? Measured Value of Total Current = 9.0 ࠵?࠵? The measured and calculated currents are very similar indicating that Ohm’s Law is accurate and any small error is likely due to human error while reading the voltmeter/ammeter. 3-F – Current (resistors) ࠵? >XAGFAEY = ࠵? >EF>HAEY 8.9 ࠵?࠵? (࠵? I ) = 7.0 ࠵?࠵? (࠵? K ) + 3.0 ࠵?࠵? (࠵? L ) Because the summation of currents entering the parallel resistors R2 and R3 is approximately the same current as that exiting R1, we can confirm that Kirchhoff’s current law does work as a real, physical relation. PART FOUR: INTERNAL RESISTANCE OF A BATTERY 4-A For part four of the lab, we will measure the internal resistance of a 1.5V flashlight battery to understand what is inside of them and how they end up dead. We will begin by using

the bench voltmeter to measure the open-circuit voltage (Voc) across the battery while it is completely disconnected from anything else. ࠵?࠵?࠵? = 1.467,79 ࠵? 4-B We will now construct a circuit with the flashlight battery, a resistor and an ammeter in series. This arrangement can be seen in the picture below. The ammeter allows us to measure the current (I) flowing through the battery. ࠵? = 59.8࠵?࠵? = .0598 ࠵? 4-C While the current is flowing in the circuit, we will use the bench voltmeter to measure the voltage across the battery alone. ࠵? = 1.365,49 ࠵? The voltage is slightly lower (than the Voc) now that the current is flowing. This is because a fraction of the battery’s voltage is dropping in the closed circuit, when the battery’s internal resistance begins to effect the measurement. 4-D Using 4-A through 4-D, we can sketch the equivalent circuit of the battery, its internal and external resistance, and the ammeter.

4-E We can then derive the Kirchhoff’s Loop equation: b ࠵? = 0 ࠵? − ࠵?࠵? I + ࠵?࠵? K + ࠵?࠵? E = 0 4-F Using the data measured and the equation from 4-E, we solved for the battery’s internal resistance, ࠵? AEF>HECD : ࠵? dCFF>H] = ࠵? ef − ࠵?࠵? AEF>HECD 1.365 = 1.467 − (. 0598࠵?)࠵? AEF>HECD ࠵? AEF>HECD = −0.102 −.0598 ࠵? AEF>HECD = 1.706 I was not sure what to expect for the battery’s internal resistance value because I have no idea how old it is nor do I know how much it has already been used. Having an internal resistance like this is part of why batteries eventually die out. Unfortunately, the resistance increases over time/usage until it is no longer effective.

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