Solve Combination Circuit Homework: Req, IT, Current & Voltage

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The discussion focuses on solving a combination circuit homework problem involving the calculation of equivalent resistance (Req), total current (IT), and the voltage and power across each resistor. The initial attempt incorrectly combined resistors in parallel without accounting for intervening resistors, leading to an erroneous total current calculation. A correction was suggested to combine resistors E, F, and G in series first, then address the parallel combination with BC. This clarification helped the student understand the correct approach to solving the circuit. Ultimately, the discussion emphasizes the importance of accurately identifying series and parallel relationships in circuit analysis.
dropgigawatt
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Homework Statement



Determine the Req, IT, and the current and voltage across each resistor. Collapse each drawing into one resistor that equals Req. Find the total power used by the circuit. Also determine the voltage, current, and power for each individual resistor.

The one I'm having trouble with:
pAfgQsr.jpg


Homework Equations



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

The Attempt at a Solution



I combined B and C into BC (4 Ω) and then combined that with F to make BCF (2 Ω). This would give me the following resistances: RA = 2 Ω, RE = 6 Ω, RBCF = 2 Ω, RG = 2 Ω, and RD = 5 Ω. That would make Req = 17 Ω since all these are now in series.

When trying to calculate the current of the whole circuit... I = V/Req = 40/17 = 2.35 A

My teacher posted that the total current should be 4 A. I'm not sure where I messed up.
 
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dropgigawatt said:

Homework Statement



Determine the Req, IT, and the current and voltage across each resistor. Collapse each drawing into one resistor that equals Req. Find the total power used by the circuit. Also determine the voltage, current, and power for each individual resistor.

The one I'm having trouble with:
pAfgQsr.jpg


Homework Equations



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

The Attempt at a Solution



I combined B and C into BC (4 Ω)
That's a fair start. :approve:

and then combined that with F to make BCF (2 Ω).
There is where the problem is!

You can't treat BC and F as being parallel with each other. There are resistors E and G in the way which prohibit that!

However, what you can do is combine E, F and G, all in series.

After that, you have parallel resistors that you can combine: BC and EFG. (And then deal with a couple more series resistors A and D even after that.)
 
AH! That makes so much sense! Thanks!
 
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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