How Do You Solve Complex Circuit Problems?

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To solve complex circuit problems, start by calculating the equivalent resistance using the formula for parallel circuits. The equivalent resistance for the top and middle paths must be determined before adding any additional resistances. Once the total current is calculated using the total voltage and equivalent resistance, apply current division to find the current through individual resistors. Understanding these steps is crucial for determining both the current and potential difference across specific resistors. Mastering these concepts will help alleviate confusion and improve problem-solving skills in circuit analysis.
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Homework Statement


Consider the circuit in the figure.
Circuits 001.jpg

Find the current in the 1.3 Ω resistor. Answer in units of A.
Find the potential difference across the 1.3 Ω resistor. Answer in units of V.
Find the current in the 10.0 Ω resistor. Answer in units of A.

Homework Equations


V = IR
Req = 1/R1 + 1/R2 + 1/R3 ...

The Attempt at a Solution


First, I calculated the equivalent resistance.
The top path - parallel circuit: 1/((1/7.7) + (1/7.7)) = 3.85 + 3.4 = 7.25​
The middle path - parallel circuit: 1/((1/4.6) + (1/10.0)) = 3.150684932 + 1.3 = 4.450684932​
Req = ?​
My teacher helped me at school and when I tried to do the problem on my own, using his steps, I forgot them all. Just yesterday I remember that I could get the equivalent resistance which is around 6.157 but I keep trying and trying but I can't seem to get that answer either. I need help on that part too.

Here is the total current: I = V / R -> I = 18.0 / 6.157 = approx. 2.92

I don't really understand how to find the current etc. It gives me headaches! Sorry the picture is kinda dark...
 
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Everything you have done is OK. Except the equation is

1/Req = 1/R1 + 1/R2 +...

Treat the top and middle paths as two parallel resistances and find their equivalent resistance before adding the final 3.4 Ohm. You will get the 6.157 Ohm value.

Then to answer questions 1 and 2 you have to work out how much of that 2.92 amperes of current will go though the middle branch.

For question 3 you conduct a similar investigation.
 
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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