Potential Difference between two points of a circuit

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The discussion focuses on calculating the potential difference (Vab) between two points in a circuit using Kirchhoff's Law and Ohm's Law. The solution outlines the steps taken to find the current through the circuit, resulting in approximately 0.29 Amps. The potential difference is calculated by considering voltage drops across resistors and contributions from power supplies, ultimately arriving at a value of 11.95 V between points a and b. The calculation process includes accounting for voltage losses and gains as current flows through various components. The overall approach and reasoning are confirmed as sound, although the math is not explicitly verified.
lylos
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Homework Statement


Given the following circuit, what is Vab? Explain...

circuit.jpg



Homework Equations


Kirchoff's Law and Ohm's Law


The Attempt at a Solution



Here is my solution... Is this describing it correctly?

Kirchoff’s Loop Law to find current:
-10 Ω (I) + 30 Ω V – 100 Ω (I) – 5 Ω (I) + 15 V – 40 Ω (I) = 0
45 V = 155 Ω (I)
I = 45 V / 155 Ω or .29 Amps

Potential Difference between points a and b:
1. At point a, we will assume there to be zero potential.
2. By going over the 5 Ω resistor, we lose 1.45 V (5 Ω * .29 A), thus leaving us at - 1.45 V.
3. The 15 V power supply adds 15 V, leaving us with 13.55 V.
4. The 40 Ω resistor takes away another 11.6 V (40 Ω * .29 A), leaving us with 1.95 V.
5. The 5 Ω resistor that we come in contact with next doesn’t take anything away as it doesn’t have current running through it, so we’re left with 1.95 V.
6. Now, we add another 10 V to the potential difference between the point a and point b due to the power supply.
7. We now have a combined Potential Difference between point a and point b to be 11.95 V.
 
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I didnt check your math but the rest seems good to me.
 
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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