Potential Differences in a Series Circuit
A cell is connected to two light bulbs in series. When current flows, there is a potential difference across each resistor. The potential difference across both resistors is equal to the sum of the potential difference across each resistor.
Think of the two bulbs as two resistors R1 and R2. Suppose that the potential difference across each one is 2 V and 3 V respectively. Then the potential difference across both resistors is 2 + 3 = 5 V.
It is important to be very clear about the meaning of this. Lets label three points on the circuit as A, B and C. R1 is between A and B, and R2 is between B and C. The potential difference between A and B is 2 V. The potential difference between B and C is 3 V. Then potential difference between A and C is 2 + 3 = 5 V.
It is quite simple to undertand why. The potential difference between A and B is 2 V means that it takes 2 J of work to move 1 C of charge from A to B. Likewise, the potential difference between B and C is 3 V means that it takes 3 J of work to move 1 C of charge from B to C.
Therefore it takes 2 + 3 = 5 J of work to move 1 C of charge from A to C. This means 5 V between A and C. So the reason why we can just add the potential differences is because they are just work done on 1 C of charge.
Next, we need to think about what happens if we move point A a bit. Will the potential difference between A and B change? The answer is no. But why?
The reason is that the wire is assumed to have zero resistance. So no work is needed to move the charge along the wire. This means that you are free to move point A anywhere along the same wire. The potential difference between A and B stays the same because there is no change in the work done. Of course, you must not move A through a cell or a resistor.
This means that even if you move both A and C in the above circuit until they touch the cell, the potential difference between them is still the same.
So potential difference between A and C in these two circuits are the same.