Unit 3 MLR Example

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University Of Arizona *

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Electrical Engineering

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Dec 6, 2023

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Kolob Farnsworth Unit 3 Lab Week 5 Section 3 Lab Report: Abstract: This section aimed to demonstrate the relationship between voltage, current, and resistance while it also showed how the DMM functioned and how to determine the internal resistance of the device. The section started off with a step-by-step walkthrough of how to set up an equation that calculated the resistance of the multimeter based on the total voltage, the voltage through that multimeter, and the resistance of a secondary component. We then set up the circuit and read the voltage displayed by the multimeter and calculated its resistance. We concluded that the DMM has an internal resistance of 10.5M W . The large resistance ensured that for circuits with low resistance outside of the DMM, the voltage reading would be accurate. We learned that large resistors throw off the DMM’s readings. Procedure: This section aimed to calculate the internal resistance of the DMM, which it used to calculate the voltage potential of a part of a circuit it was measuring. We used a circuit that featured the DMM in series with a second resistor. The first thing we did was set up the equation needed to calculate the resistance of the DMM. We found an equation for the current of the circuit in terms of the circuit voltage, the resistance of the DMM, and the resistance of a second resistor. A second equation was then written for the current through the DMM in terms of the voltage through the DMM and the resistance of the DMM. These two equations were set equal. We then solved for the resistance of the DMM in terms of the DMM’s voltage, the voltage of the system, and the resistance of the second resistor. We then set up the circuit with a second resistor of 10 M W and a circuit voltage of 10V. We recorded the voltage displayed on the DMM and calculated the device’s internal resistance. Results: For the total current of the circuit, we got the equation: ࠵? = ࠵? ࠵? ! + ࠵? "## And the equation for current through the DMM: ࠵? = ࠵? "## ࠵? "## Setting these equal, we then solved for R DMM : ࠵? ࠵? ! + ࠵? "## = ࠵? "## ࠵? "## ࠵? = ࠵? "## ࠵? "## ࠵? ! + ࠵? "##
࠵? ࠵?࠵?࠵? = ࠵? ࠵? ࠵? ࠵?࠵?࠵? ࠵? − ࠵? ࠵?࠵?࠵? For the circuit with a 10M W resistor and 10V circuit voltage, the DMM displayed a voltage of 5.12V. This gave us a resistance for the DMM of 10.5M W . Discussion: This lab focused primarily on two concepts: The first was the relationship between voltage, current, and resistance; the second was how resistors acted in series versus how they acted in parallel. This section focused on a circuit with two resistors, one being the DMM. The lab showed that with components in series, the resistance for the circuit was the sum of the resistance for each component and that the current supplied by the power source was the voltage supplied divided by the total resistance. Additionally, we knew that for resistors in series, each resistor had the same current flowing through it as the circuit. We used these facts to write the two equations: I = V/(R 1 +R DMM ) and I = V DMM /R DMM . We then set those two equations equal knowing that current of each component was equal to the total current. We learned that the DMM works not by measuring the voltage potential directly, but rather by measuring the current passing through it and calculating the voltage with its internal resistor. This section showed that if there was a resistor in the circuit the DMM was measuring, the larger of a resistance it had, the less accurate the voltage reading on the DMM was. This explained why the DMM uses such a large internal resistance, to increase the percentage of the total voltage potential through the device. Extrapolating from this, we can conclude that the most effective version of a DMM would have infinite resistance.
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