Potential Difference with Series and Parallel Resistors

AI Thread Summary
The discussion centers on calculating the potential difference across resistor R5 in a circuit with an ideal battery of 14.0 V and multiple resistors. The user correctly identifies that R1 and R2 are in parallel, with their equivalent resistance in series with R3, while R4 and R5 are in series. A key point made is that the voltage across any leg of a parallel circuit is the same, meaning R4 and R5 share the total voltage drop of 14 V. To find the potential difference across R5, the user is advised to first determine the current through that leg, which will then allow for calculating the voltage across each resistor. This approach is crucial for solving the problem effectively.
thebert010
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Homework Statement


In Figure 27-72, an ideal battery with an emf of = 14.0 V is connected to a network of resistances with R1 = 7.00 , R2 = 12.0 , R3 = 4.00 , R4 = 1.00 and R5 = 6.00 . What is the potential difference across R5?

FIGURE:

|----**R1**---|
| |---**R3**---|
|---**R2**----| |
| |--|
| | |
|---**R4**-------**R5**----| |
| | |
|-----------+\emf\-----------|--|

Homework Equations



V=iR
Parallel R = R12 = (1/R1 + 1/R2)^-1
Series R = R12 = R1 + R2

The Attempt at a Solution


I just typed a realllllly long attempt at my solution and the page crashed and I lost everything, and I'm at work...so briefly:
R1 and R2 are parallel so R12 = (1/7 + 1/12)^-1 = 4.42ohm
R12 and R3 are in Series so R123 = 4.42ohm + 4ohm = 8.42 ohm
R4 and R5 are in Series so the same = R45 = 7ohm
R123 and R45 are parallel so R12345 = (1/8.42 + 1/7)^-1 = 3.822 ohm

after attempting various things involving V = iR (where R's in Series have same i and R's parallel have same V) I cannot seem to get the potential difference across R5...PLEASE HELPPPPPP

thanks so much
 
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thebert010 said:

Homework Statement


In Figure 27-72, an ideal battery with an emf of = 14.0 V is connected to a network of resistances with R1 = 7.00 , R2 = 12.0 , R3 = 4.00 , R4 = 1.00 and R5 = 6.00 . What is the potential difference across R5?

FIGURE:

|----**R1**---|
| |---**R3**---|
|---**R2**----| |
| |--|
| | |
|---**R4**-------**R5**----| |
| | |
|-----------+\emf\-----------|--|

Homework Equations



V=iR
Parallel R = R12 = (1/R1 + 1/R2)^-1
Series R = R12 = R1 + R2

The Attempt at a Solution


I just typed a realllllly long attempt at my solution and the page crashed and I lost everything, and I'm at work...so briefly:
R1 and R2 are parallel so R12 = (1/7 + 1/12)^-1 = 4.42ohm
R12 and R3 are in Series so R123 = 4.42ohm + 4ohm = 8.42 ohm
R4 and R5 are in Series so the same = R45 = 7ohm
R123 and R45 are parallel so R12345 = (1/8.42 + 1/7)^-1 = 3.822 ohm

after attempting various things involving V = iR (where R's in Series have same i and R's parallel have same V) I cannot seem to get the potential difference across R5...PLEASE HELPPPPPP

thanks so much




awesome so the circuit isn't showing up...BUT...take my word for it that the R's I said were parallel/series actually are...I'm pretty confident that part is absolutely correct
 
Hi thebert010,

Just as a point of clarification for me, what is the relationship between R1 and R3 and R2 and R3? Or in other words, are R1 and R3 in parallel or series and what about R2 and R3?

Thanks,

KEØM
 
Hi!
so R1 and R2 are parallel...their equivalent is then in series with R3.

At the same time, R4 and R5 are in series.

===R1=|
|...|==R3==|
===R2=|....|
|......|===|
|===R4===R5==|...|
|===+\EMF\-=======|

sorry that's the best i can do :-(
 
I think I can help. First thing you need to realize is that for each leg of a parallel circuit, the emf (or difference of potential) across any leg is going to be the same (14V in this case). This means you can "isolate" resistors R4 and R5 and just look at that one leg. So you know that 14 volts are being dropped across this leg. Since you have only one variable you don't know, (i) that's where you want to start. Once you find the current through that leg, you can then find out what the voltage across each resistor is.
 
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