Voltmeter Reading for Ohm's Law Problem

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The discussion revolves around solving an Ohm's Law problem where the provided answer is ɛ/2, but the original poster struggles to reach this conclusion. They mention using KVL and note that an ideal voltmeter has infinite resistance, preventing current flow into it. A suggestion is made to improve the clarity of shared images or to type out the work for better understanding. Additionally, the importance of using LaTeX for mathematical expressions is highlighted. The thread is ultimately locked, indicating no further responses will be accepted.
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Homework Statement


2iW6Doc


Homework Equations


Ohm's Law (V=IR)

The Attempt at a Solution


FMFlQ

These are my two attempts at getting to the solution. I'm really finding it difficult to tackle this question (our teacher gave it as an optional challenge). The provided answer on the worksheet is ɛ/2, but I have no clue how to get there. Any help is much appreciated.
 

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An ideal voltmeter is one that has infinite resistance. Use KVL, the resistor to the rightmost (nearest to terminal A) is short circuited anyway, so you are left with two resistors of equal resistance.

All measuring devices conncected to a circuit adveresly effect it, ideal ones dont. Since a voltage measurement is always made parallel to elements there must not be any current flowing into the voltmeter I believe, and this is achieved by having an infinite resistance. :-)
 
Hi Gex,

Unfortunately your images are essentially unreadable. Please make a new thread and type out your work or provide better pictures (preferably the former). I highly recommend learning a little bit of LaTeX. It greatly helps with inserting math into your posts. We have a primer for it here: https://www.physicsforums.com/help/latexhelp/

Thread locked.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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