# Determining potentials at points in a circuit with multiple batteries 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?

• Possible duplicate of How to determine direction of current mentally in complex electrical networks? Jun 15 '18 at 14:30
• 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. Jun 20 '18 at 12:01 