How to include resistance of wire and voltmeter in Kirchhoff equations

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SUMMARY

This discussion addresses the impact of wire resistance (0.01 Ω) and voltmeter internal resistance (10 MΩ) on Kirchhoff's equations in circuit analysis. It concludes that the wire resistance can be treated as negligible when placed in series with larger resistors (50, 100, or 200 Ω), as its effect on current and voltage measurements is minimal. Similarly, the voltmeter's internal resistance, when modeled in parallel with the resistors, does not significantly alter the measurements due to its high value compared to the resistors. Overall, the modifications result in no significant changes to the measured voltages and currents in the circuit.

PREREQUISITES
  • Understanding of Kirchhoff's laws
  • Basic knowledge of series and parallel resistor configurations
  • Familiarity with Ohm's Law
  • Experience with circuit analysis techniques
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Students in electrical engineering, circuit designers, and anyone involved in laboratory experiments requiring precise measurements of current and voltage in circuits.

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Homework Statement


I'm doing a lab report and the last question is, how would your measurements (currents and voltages across each resistor) be affected if the resistance of the wires is 0.01 Ω and the internal resistance of the voltmeter is 10 MΩ?

EM_Lab_3_Kirchhoff1.jpg

Homework Equations


The Attempt at a Solution


I'm going to take a guess here. For the wire, do I just pretend there is a resistor of 0.01 Ω right next to one of the power supplies? Or do I have to add it in each of the three branches? For the voltmeter, do I add a 10 MΩ resistor in parallel to each of the three resistors, find their equivalent resistance and then use that for the Kirchhoff current equations? Thanks.
 
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In short you would need to replace all wires with 0.01Ohm resistors and place a 10MOhm resistor between the leads of your otherwise ideal voltmeter.

So in this circuit the end result is to put 0.01 Ohm resistors in series with all of the resistors. Putting a 0.01 Ohm in series with a 50, 100 or 200 Ohm resistor will have virtually no effect because 50 >> 0.01.

When you use the modified voltmeter to measure the voltage across any of the resistors the 10MOhm resistor will be in parallel with the resistor. This will have no significant effect because 10MOhm >> 50, 100 or 200 Ohm.

So in this circuit there will be no significant change to the measured voltages an currents. In other circuits with different component values there could be a significant change.
 
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