# Series and Parallel Resistance

Discussion in 'Homework Help' started by xBlz1n, May 5, 2011.

1. ### xBlz1n Thread Starter New Member

Apr 30, 2011
6
0
Hi everyone, I am just beginning to learn Electronics within my AP Physics class and am having some trouble with series and parallel circuit calculations as well as bread-board layouts. Appreciate any help I receive.

I created a few photo shop images to help organize the problems I am having.

First, I have a basic circuit with a 12v battery and 5 resistors. For each resistor I need to find the currents through and voltages across. Below is the circuit diagram and table.

I was able to calculate R equivalent by adding R3, R4, and R5 in parallel to obtain 1ohms. I then added R1 and R2 with the 1ohms to obtain a equivalent resistance of 12ohms. With the information I now know that the current out of the battery is 1amp. The voltage across R1 is 2v considering V = iR = (1A)(2ohms) = 2v. Therefore, the voltage drop is 2v. Where i get stuck is determining the voltage and current for the other resistors. When i go to calculate the current through R3 I know (v/R) = i. I plug in the numbers, (10v/2ohms) = 5A!?! No way, If anything it is .5A because the current splits. Can anyone help direct me through completion of this problem? I assume I am on the correct path.

Also, if anyone has any additional time could you help me determine the breadboard layouts for these circuits?

Thanks in advance to anyone who provides any suggestions or help, I greatly appreciate it.

2. ### beenthere Retired Moderator

Apr 20, 2004
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3. ### xBlz1n Thread Starter New Member

Apr 30, 2011
6
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I have used that page for reference but it poses no use to me at the moment. Thanks anyway.

4. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
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From your description it is actually very difficult to determine which one of the two questions you are showing your work for.

Neither circuit shows a parallel combination of R3, R4 and R5. Please use the reference pages that beenthere has referred to in order to correctly identify the parallel and series resistor combinations. Which resistors are in series and parallel? Knowing that is essential to the problem!!

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5. ### xBlz1n Thread Starter New Member

Apr 30, 2011
6
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I looked at the page again and am still having trouble, I can easily determine which is in series and parallel on the first circuit I posted. R3, R4, and R5 are parallel with each other? Can anyone solve the problem? If I saw the answers I could trial and error. I haven't received any useful suggestions yet so I am still stuck.

I found the equivalent resistance to be 12ohms, if that is correct then I justified which was series and parallel correctly. Then, I labeled the voltages throughout the circuit. When I go to calculate the current of R3, I get 5A by doing the math. I can already concur that 5A is incorrect considering the amps coming out of the battery is 1A, anyone know what I'm doing wrong, I did (10V/2ohms) = i = 5A.

EDIT: I gave the first circuit my best effort, could anyone possibly view my answers to see if I am correct?

Last edited: May 5, 2011
6. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
OK, now I can see the image of the circuit in question... When you determine the voltage for the v=ir calculation, you need to determine the voltage drop across the resistor itself. R3 has 10v on one side and 9v on the other. So, the voltage drop across R3 is only 1v and not 10v, or 9v.

7. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
How is the bread board drawing setup? Are some of the circles connected like a physical bread board?

8. ### xBlz1n Thread Starter New Member

Apr 30, 2011
6
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Yes it would be setup like a regular breadboard without the battery. I also posted the answers I got for the first circuit, care to take a look?

9. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
I guess it depends on your breadboard orientation. If each column of dots are connected together then you are correct in your design.

10. ### xBlz1n Thread Starter New Member

Apr 30, 2011
6
0
Awesome, thanks a lot for the help, it really helped! By far the most useful website I have came across. Thanks again!