Finding the Current Through a Resistor (Working With Parallel and Series)

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The discussion revolves around calculating current through resistors in series and parallel configurations. The initial attempt found the equivalent resistance to be 5.9 ohms with a total current of 1.53 Amperes, but the calculated current through a 12-ohm resistor was incorrect. After reviewing the calculations and considering significant figures, the correct current was determined to be approximately 0.17 A. The importance of significant figures in the final answer was emphasized, as some values were provided with only one significant figure. The participants confirmed the accuracy of the revised calculation.
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Homework Statement
What is the Current Through A Resistor in Parallel?
Relevant Equations
I = V/R
Resistors in Parallel => 1/R = 1/R1 + 1/R2
Resistors in Series => R = R1 + R2
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Here is the actual question.

And here is my attempt at a solution
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In Summary I did the following
  1. Found the Equivalence Resistance to Be 5.9 ohms and the Current throughout the entire resistor to be 1.53 Amperes
  2. Worked backwards from my resistor simplifications. When the resistors were in series I solved for V because they should have the same Current. When they were in parallel I solved for Current because they had the same voltage.
  3. Eventually I worked my way back to the parallel resistors of 4.3 and 12 ohms and got .166 Amperes of current running through the 12 ohms resistor.
  4. This answer, however, was wrong. I found this method from this video on Khan Academy https://www.khanacademy.org/science...-example-finding-current-voltage-in-a-circuit.
 

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Your work looks correct to me. I carried an extra significant figure through the calculation and got an answer of 0.1679 A. But some of the resistors and the battery voltage are given with only 1 significant figure! So, I think two significant figures in the answer should suffice: 0.17 A.
 
TSny said:
Your work looks correct to me. I carried an extra significant figure through the calculation and got an answer of 0.1679 A. But some of the resistors and the battery voltage are given with only 1 significant figure! So, I think two significant figures in the answer should suffice: 0.17 A.

You are exactly right thanks a bunch. I was going mad trying to figure out where i was going wrong.
 
Yes, 0.168A
 
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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