Voltmeter in simple battery + resistor circuit

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Discussion Overview

The discussion revolves around understanding the function of a voltmeter in a simple battery and resistor circuit. Participants explore how to interpret the voltage readings from the voltmeter, particularly when placed between components in a series circuit.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • One participant presents a circuit with a battery and resistors, calculating total resistance and current but expresses confusion about what the voltmeter measures when placed in the middle of the circuit.
  • Another participant clarifies that the voltmeter measures the voltage difference between its two connections, specifically referencing the top connection between resistors and the ground at the bottom.
  • A participant reflects on their understanding, suggesting that the voltage drops as it passes through the resistors, calculating a specific voltage difference across the voltmeter's terminals.
  • Another participant confirms the understanding of the voltage difference calculated by the previous participant.

Areas of Agreement / Disagreement

Participants generally agree on the function of the voltmeter and the concept of voltage drop across resistors, but there is an initial uncertainty regarding the interpretation of measurements when the voltmeter is placed in different positions within the circuit.

Contextual Notes

The discussion does not resolve all uncertainties about the implications of adding additional components, such as another battery, on the measurements taken by the voltmeter.

Who May Find This Useful

Students learning about basic electrical circuits, particularly those studying series circuits and the role of measuring instruments like voltmeters.

DrOnline
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Homework Statement


23uziqc.jpg


Find the current in the circuit and the voltage on the volt meter

Homework Equations


E=22.5 V
Rtot= 1kΩ + 2.4kΩ + 5.6kΩ = 9kΩ
I=U/R=22.5V/9kΩ=2.5mA

A voltmeter has super high resistance, to prevent any current from passing through it. I know that.

So I have the resistance, I have the voltage at the source, and the current in the circuit. Tons of information to work with.

The Attempt at a Solution



Here comes the problem. I don't understand the reasoning that explains what the volt meter is measuring. How do I know if I am measuring the left or the right side? What if there was a battery on the right side too, what then?


But what am I measuring? if you put a volt meter over E, or over R1, I get it. You get the voltage over those specific components, but when the volt meter is in the middle, am I measuring the left side, or the right?

Can somebody help me make sense of this? It's... a sad state of affairs that I don't get this, so please! :)
 
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DrOnline said:

Homework Statement


23uziqc.jpg


Find the current in the circuit and the voltage on the volt meter

Homework Equations


E=22.5 V
Rtot= 1kΩ + 2.4kΩ + 5.6kΩ = 9kΩ
I=U/R=22.5V/9kΩ=2.5mA

A voltmeter has super high resistance, to prevent any current from passing through it. I know that.

So I have the resistance, I have the voltage at the source, and the current in the circuit. Tons of information to work with.

The Attempt at a Solution



Here comes the problem. I don't understand the reasoning that explains what the volt meter is measuring. How do I know if I am measuring the left or the right side? What if there was a battery on the right side too, what then?


But what am I measuring? if you put a volt meter over E, or over R1, I get it. You get the voltage over those specific components, but when the volt meter is in the middle, am I measuring the left side, or the right?

Can somebody help me make sense of this? It's... a sad state of affairs that I don't get this, so please! :)

Good work so far. When the voltmeter is shown like that, it is measuring the voltage difference between its top connection between the resistors with respect to the ground at the bottom.

You've figured out the current flowing through the resistors. Now calculate the voltage drop across each resistor (which should still add up to 22.5V). You will then see how much voltage gets dropped across the righthand 2 resistors...
 
Hmm.. I think I just got it.. sometimes writing out the challenge clears some of the fog.

How about i see it like this... the voltage is 22.5V as it leaves the battery, then it continues to drop as it passes over the series. As it passes by R3, the voltage drops to zero.

This means at the bottom of the voltmeter there is 0 V, and at the Top, you have E-R1*I = 22.5V - 2.5V = 20V..

The difference between the two connectors of the voltmeter is 20V..

Amirite?
 
Perfect!
 
Thanks, Berkeman!
 

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