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I am asked to find the potential difference across the points $P$ and $Q$. Using Kirchoffs second law, I calculated the 'resultant' emf as being $E$. The p.d across each resistor would then be $\frac{E}{3}$. To solve the question I went with the idea that the potential at the negative pole of the top battery(at P) would be 0. The the potential at Q would then be $E-\frac{E}{3}$ which happens to give the correct answer of $\frac{2E}{3}$.

My assumption that the p.d that at the negative pole has to be zero also implies that the p.d across the left resistor has to be $E$ which doesn't agree with the calculation.

My question is: how do I find the potential at points on a circuit with multiple batteries, when some oppose others?

  • $\begingroup$ Possible duplicate of How to determine direction of current mentally in complex electrical networks? $\endgroup$ Jun 15 '18 at 14:30
  • $\begingroup$ Using superposition, as suggested, this can be solved. I have also identified what is wrong with my method in case this question might be read by someone else. My assumption that the potential at the negative terminal in a battery is $0$ is wrong. What is correct is that the potential increases by $E$ after the battery. Let $a$ be the at P. Potential right after the battery is $a+E$. Potential at Q would be $a+E-\frac{E}{3}$. Calculating p.d across $PQ$ using this method works as well. $\endgroup$
    – whois
    Jun 20 '18 at 12:01

A reliable, and usually easy, way to do this is using superposition. The Khan Academy has an excellent article on it it here.

I urge you to take the time to read the article because it explains the procedure in detail. But in brief you take the circuit and replace all the batteries but one with a wire. Do this for each battery, so in your circuit you'd get three separate circuits. For each of the three circuits calculate the voltages and currents, then simply add the three circuits together and that will give you the correct values for the circuit with all three batteries present.

So in this case superposition gives you these three circuits to combine:


Superposition may initially seem a bit complicated, but once you get used to it the method gives you a quick and reliable method for approaching circuits like this.

  • $\begingroup$ This works because elementary circuit theory is a linear theory. $\endgroup$ Jun 15 '18 at 11:53

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