Calculating Correct Voltmeter Reading in a Circuit

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Discussion Overview

The discussion revolves around calculating the correct voltmeter reading in a circuit with given resistor values and a supply voltage. Participants are attempting to verify the voltmeter reading based on their calculations and the configuration of resistors in the circuit.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant states that the voltmeter reading should equal the voltage across the 4.7 kΩ resistor, calculating it as 8.798 V based on their formula.
  • Another participant questions the understanding of series and parallel resistor reduction, implying a potential error in the first calculation.
  • A subsequent reply suggests an alternative calculation method, reducing R1 and R2 in series and R3 and R4 in parallel, resulting in a new voltage calculation of 4.389 V.
  • The last post indicates that the latest calculation appears correct, but does not confirm the accuracy of the voltmeter reading.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct voltmeter reading, with multiple calculations presented and some confusion regarding the methodology for combining resistors.

Contextual Notes

There are unresolved assumptions regarding the circuit configuration and the method of combining resistances, which may affect the calculations presented.

orangeincup
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Homework Statement


The voltmeter in the figure below reads 6.2 V. If the vs = 12 V and R1 = 560 Ω, R2 = 470 Ω,
R3 = 680 Ω, and R4 = 4.7 kΩ, is the voltmeter reading correct?

Homework Equations


V*Rx/(Ry+Rx)

The Attempt at a Solution


So I thought the voltage in the voltmeter would equal the voltage across the 4.7kohm resistor

So 12V*4700/(4700+560+470+680)= 8.798 V
So my answer would be no.

Is this correct or am I doing it wrong?
 

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So 12V*4700/(4700+560+470+680)= 8.798 V

HUH ?

Do you understand how to reduce series and parallel resistors?
 
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Opps, should it look more like this?

R1+R2=1030 ohms
R3||R4=594 ohms

12 V * 594/(1030+594)= 4.389 V

Which is exactly half of my old answer for some reason.
 
Now it looks right
 
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