# Diagonal Series/Parallel Circuit

#### mirroredeyes

I got my test back and was reviewing my question, to find out that I didn't even solve for R2,R3,R4 Current, correctly.

I only got 2.613A correct which was R1, could anyone possibly help me out. I know for parallel circuits voltage drop is equal across circuit but my R1 was in series with the combined resistors, meaning there was a voltage drop.

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#### anorlunda

Mentor
Gold Member
That image is too dim for me to see. Can you post a better one? Use brighter lighting to make the photo.

#### mirroredeyes

That image is too dim for me to see. Can you post a better one? Use brighter lighting to make the photo.

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#### DaveE

100/38 ≠ 2.613
Otherwise it looks ok to me.

#### mirroredeyes

100/38 ≠ 2.613
Otherwise it looks ok to me.
Thank you Dave! wow... how did I mess that one up badly. Would you say I should lose 3 marks though? I did properly apply the laws and solve but just made a calculation error.

#### sophiecentaur

Gold Member
OK so far but they also want to know how the 2.613A is shared between the other resistors. One way to find the remaining current: you can easily find the volts across the triangle of resistors and then the current through each leg.

#### DaveE

Thank you Dave! wow... how did I mess that one up badly. Would you say I should lose 3 marks though? I did properly apply the laws and solve but just made a calculation error.
If I was focused on teaching concepts, I wouldn't mark off that much. However, I would also want answer in more general terms like I1 = V/(R1+R2||(R3+R4)). Then the numbers would be found last, after the equations.
But, there is a school of thought in engineering classes that the final results matter. If your bridge collapses or your circuit electrocutes someone, no one cares why or how you were wrong. This is more like the real world, where you need to check your results for errors (there will be errors, we all do that). Sometimes instructors want to make that clear with harsh penalties.

"Diagonal Series/Parallel Circuit"

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