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Edited - could someone explain how you treat a circuit with two cells in respect to circuit laws and also whether the section in the middle is in parallel or in series.

Second Edit: I’ve researched the topic some more and have come to the conclusion that I shouldn’t have been attempting this question in the first place as we haven’t covered it as of yet. This is why there was no attempt to answer the question. I had assumed it was just a slightly different version of topics we had covered and there was a very simple and easily covered answer to the question that I just wasn’t aware of. This is not the case. Thank you for the answer anyway as it was helpful. Sorry my mistake.

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closed as off-topic by Buzz, Jon Custer, ZeroTheHero, John Rennie, Kyle Kanos Dec 23 '18 at 12:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Buzz, Jon Custer, ZeroTheHero, John Rennie, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Hi welcome to PSE. The general consensus of this community is that it's okay to ask and answer homework questions. But simply providing an answer in a way that the asker doesn't learn anything is not generally accepted. In other words, we'll be your physics tutor, but not your answer guide. So, please to try to find the concept that you don't understand (here, it's something to do with double loop circuits) and ask a question about that. $\endgroup$ – psitae Dec 22 '18 at 17:35
  • $\begingroup$ "Didn’t really have a clue how to work around the problem" - do you know how to solve the problem if $R_1$ were a fixed resistance? $\endgroup$ – Alfred Centauri Dec 22 '18 at 17:46
  • $\begingroup$ @AlfredCentauri well it is fixed at 0 ohms in parts b and c so no the main issue I’m having is how to treat the two cells I’ve edited the original question now to be more specific. $\endgroup$ – Tommy Dec 22 '18 at 18:01
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    $\begingroup$ Look up Kirchhoff's circuit laws and you should be able to answer all questions. Basically the sum of all voltages on the resistor in each individual cell has to be equal to the applied voltage at each cell. $\endgroup$ – Halberd Rejoyceth Dec 22 '18 at 18:11
  • $\begingroup$ Related meta post: physics.meta.stackexchange.com/q/10951/2451 $\endgroup$ – Qmechanic Dec 23 '18 at 16:25
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could someone explain how you treat a circuit with two cells in respect to circuit laws.

I can explain how you treat a linear circuit with two voltage sources. Your circuit is a linear circuit because it contains nothing but linear components.

With a linear circuit, you can solve the currents caused by each individual source by "nulling out" all of the others. If you do that for each source, then the total current in each resistor will simply be the sum of the currents caused by each individual source.

The "cells" in your circuit are voltage sources. They force a particular voltage between two points without restricting the current at all. So, to "null out" a voltage source, you want force 0V between the same two points without preventing current. In other words, replace the cell with a wire.

So, first, replace the left cell with a straight wire, and figure out the current that will be forced through each resistor by the right hand cell. Then replace the right hand cell with a straight wire, and figure out the currents that are forced by the left hand cell. Then, compute the total current in each resistor by adding the current due to the left cell and the current due to the right cell.

Pay attention to the signs! When a source causes current to flow in the same direction as the arrow on a resistor symbol, you'll want to record a positive number. If the source forces current to flow opposite of the direction that the arrow is pointing, then you'll write down a negative number.

Finally, knowing the current in each resistor, you can use Ohm's Law to compute the voltage across each resistor.

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  • $\begingroup$ P.S., If you are given a circuit where the arrows were not already drawn for you, Just draw your own. You can try to guess in advance which way each arrow should point, but It doesn't really matter which direction you draw them. If the final answer for any given resistor is negative, then you can re-draw the arrow in the opposite direction, and change the sign of the current for that resistor from negative to positive. $\endgroup$ – Solomon Slow Dec 22 '18 at 18:44
  • $\begingroup$ P.P.S.: A current source forces a particular current, without limiting the voltage. If you ever have to null out a current source, you replace it with something that allows no current, and any voltage: In other words, you simply disconnect it. $\endgroup$ – Solomon Slow Dec 22 '18 at 18:51
  • $\begingroup$ thank you very much for the explanation think I’m about there now. Just one more misunderstanding, why when working out the value of I2 can you not eliminate the left source, use I=V/R to find the current I=12/40 = 0.3A and then take that as the value? $\endgroup$ – Tommy Dec 22 '18 at 20:16

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