Ammeter from parallel to series, current change

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When an ammeter is connected in parallel with two identical resistors, it reads 0.40A due to the division of current. If connected in series, the total resistance increases, resulting in a lower current reading. The textbook states that the correct reading in series would be 0.20A, which reflects the halving of current due to the equal resistance of the two resistors. Understanding the behavior of resistors in circuits is crucial for grasping these concepts. This highlights the importance of proper ammeter placement in circuit analysis.
Faris A
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Homework Statement


In the circuit shown, a student mistakenly connects a perfect ammeter(i.e. One of negligible resistance), in parallel with one of two identical resistors. The ammeter reads 0.40A.

What would the reading have been if he had correctly connected the ammeter in series with the two resistors? The cell also has negligible resistance.

Homework Equations


Relevant circuit (http://imgur.com/Uo6w6Um)

The Attempt at a Solution


I have no idea, the textbook says the answer is 0.20A but I have no idea why.
 
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Faris A said:
I have no idea,
Faris A said:
two identical resistors
Think about it.
 
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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