The voltage across a voltmeter Olymp 09

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Homework Help Overview

The discussion revolves around a problem related to the voltage reading of a voltmeter connected across a battery, specifically in the context of internal resistances and circuit analysis. The subject area includes concepts from electrical circuits and voltage measurement.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore different methods to calculate the voltmeter reading, including considering internal resistances and the relationship between voltage, current, and resistance. Questions arise regarding the interpretation of circuit connections and the application of Ohm's law.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the validity of their approaches. Some guidance has been provided regarding the interpretation of the voltmeter reading and the significance of internal resistance, while others express confusion about the methods used.

Contextual Notes

There is mention of specific values and resistances, but the overall setup and certain assumptions about the circuit configuration remain unclear. Participants are also navigating the implications of measuring voltages in the presence of internal resistances.

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The question I'm asking about is Olymp 09

My attempt : 2.36 V

Vb1 / (R + r1 + r2) = 2 / 2.5 = 0.8 A
Vb2 / (R + r1 + r2) = 5 / 2.5 = 2 A

The reading of the voltmeter = the voltage across r1 + the voltage across r2 and R = (0.8)(2.2) + (2)(0.3)
= 2.36 V

I'm not sure of my answer, and even if it's correct I'm not completely convinced how is my solution is right.
 
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The voltmeter sits across a 2V battery. What do you suppose it will read?
 
rude man said:
The voltmeter sits across a 2V battery. What do you suppose it will read?
Not 2 V, as we have to consider the internal resistances.

@ElmorshedyDr: What is meant with "in the opposite circuit [...]"?
I don't understand your approach to convert those currents to voltages across the 0.3 Ohm resistor.
 
mfb said:
Not 2 V, as we have to consider the internal resistances.

Right. Didn't see that.
 
So the answer is voltmeter reading = VB + V across the internal resistance = 2 + (1.2)(0.3) = 2.36 V ??
 
ElmorshedyDr said:
So the answer is voltmeter reading = VB + V across the internal resistance = 2 + (1.2)(0.3) = 2.36 V ??

Correct.
 
ElmorshedyDr said:
View attachment 70418

The question I'm asking about is Olymp 09

My attempt : 2.36 V

Vb1 / (R + r1 + r2) = 2 / 2.5 = 0.8 A
Vb2 / (R + r1 + r2) = 5 / 2.5 = 2 A

The reading of the voltmeter = the voltage across r1 + the voltage across r2 and R = (0.8)(2.2) + (2)(0.3)
= 2.36 V

I'm not sure of my answer, and even if it's correct I'm not completely convinced how is my solution is right.
Your working is basically correct, but your technique is not what I was expecting. Can you explain the method you have used?
 
rude man said:
Correct.
But I'm still confused, why can't we just calculate the total voltage and the total resistance and then find the total current ad multiply it by the internal resistance of B1 to find the reading, since it is connected in parallel with internal resistance of the battery which would be 0.36V
 
Sure, and the result is the same.
 
  • #10
ElmorshedyDr said:
But I'm still confused, why can't we just calculate the total voltage and the total resistance and then find the total current ad multiply it by the internal resistance of B1 to find the reading, since it is connected in parallel with internal resistance of the battery which would be 0.36V
That can't be correct because the voltmeter is not connected directly across the internal resistance of the battery. The voltmeter reading is the sum of two voltages---you can see how it also captures the battery's EMF.

The best we can do is use the voltmeter to measure the terminal voltage of the battery, which means it's the scalar addition of two potentials that we measure here.

I'm stll interested in your explanation of the first method you showed. It is correct, BTW.
 

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