Solve Network of Resistors: R1, R2, R3, R4, V=6V

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Homework Help Overview

The discussion revolves around a circuit problem involving a network of resistors (R1, R2, R3, R4) with a voltage source of 6V. Participants are tasked with determining the current through specific resistors and finding an equivalent resistance for the entire circuit.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to combine resistors in series and parallel to find equivalent resistance and current through R1, R2, and the overall circuit. Some participants question the accuracy of the potential drop across R2 and suggest redrawing the circuit for clarity.

Discussion Status

Participants have made progress on several parts of the problem but are encountering difficulties with the calculations for current through R2. Guidance has been offered regarding the need to consider the correct voltage drop across R2, indicating a productive direction in the discussion.

Contextual Notes

There is an indication that the voltage across R2 is not simply the source voltage of 6V, which raises questions about the assumptions made in the calculations. Participants are exploring the implications of this on their current calculations.

sarah895
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Homework Statement



This is the diagram:
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/summer/homework/Ch-21-DC-Circuits/resistor_network/ex1s95p3.gif

R1 = 3W, R2 = 6W, R3 = 11W, R4 =8W, V=6V

A) What is the current through the resistor, R1 in the above circuit? -2A

B) What single, equivalent resistor could replace all of the resistors in this circuit?

C) What is the current supplied by the battery?

D) What is the current through the resistor, R2 ?


Homework Equations



1/R (parallel) = 1/R1 + 1/R2 + 1/R3 + 1/R4
R (series) = R1 + R2 + R3 + R4
E = IR

The Attempt at a Solution



I solved part a. For part be, it seemed to me that R1 and R4 were connected in series, so I combined those two resistors by adding the resistances. Then I added that resistance to R2 and R3 using the equation for parallel resistors.

So, I had: 1/R = (1/11W) + (1/6W) + (1/11W) = 2.87W. This answer is wrong, and I am not sure what to do. Any help would be greatly appreciated :)
 
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Okay, so I solved parts b and c also. Now I am working on part d.

I thought that I2 would = V/R2 = 6/6 = 1, but this is wrong. Any ideas?

Thanks,
Sarah
 
You need the correct potential drop across the resistor. I'll redraw the circuit for you, convince yourself that the one I have drawn is equivalent. I hope it will then be clearer to you.

Redrawing the circuit into this form is often a good idea when solving such problems. It's now very clear which parts are in parallel, and which are in series.
 

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sarah895 said:
Okay, so I solved parts b and c also. Now I am working on part d.

I thought that I2 would = V/R2 = 6/6 = 1, but this is wrong. Any ideas?

Thanks,
Sarah

The voltage across R2 is necessarily different from the 6V. Because the voltage at that node between R2||R4 and R3 is determined by 6v * R3 / (R3 + (R2||R4)). Subtract that from 6v to give the drop across R2. With that voltage drop divided by R2 you should have I through R2.
 

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