# Why is there a potential difference? The question then asks for the potential difference between $X$ and $Y$, which is claimed to be $3.6\text{ volts}%$.

Why would there be a potential difference in this case? If I connect a lightbulb on $X$, and another on $Y$, it seems very obvious that the brightness would be the same.

Or is the standard direction of current important? So the current "encounters" the 60 ohm "first" when flowing to $X$, and the 30 ohm "second" when flowing to $Y$? That seems a little crackpot to me. So if we change the resistor to the left of $X$ to 100 ohms, and the resistor to the right of $Y$ to be 0 ohms, the differing definitions of current flow would actually cause a difference in measurement of potential?!?!

The correct answer is worded: • Current flows from the positive to the negative terminal of the battery. That provides a direction in itself. I'll leave you to figure out the rest. :) – Kitchi Feb 27 '13 at 14:00
• However, electrons flow from negative to positive. Conventional current is only a convention, so it shouldn't make a difference. The way the question is currently worded, there wouldn't be a difference (whether current flows negative to positive or the other way around), but if I change the resistors, would it now have? Would Benjamin Franklin (who first claimed current flows from positive) be able to find the "true" direction of current flow? – ithisa Feb 27 '13 at 14:02
• Well, okay - Look at it this way. Conventional current sets up a sign convention. Which one is + and which is -. So if you take the direction of the electron's motion, then the potential difference between X and Y would be the same, except with a minus sign. So it doesn't really make a difference. – Kitchi Feb 27 '13 at 14:06
• "So if we change the resistor to the left of X to 100 ohms, and the resistor to the right of Y to be 0 ohms, the differing definitions of current flow would actually cause a difference in measurement of potential?!?!" That's what I meant. The resistors are set up "symmetrically" in the original question but what if they are not? – ithisa Feb 27 '13 at 14:07
• Setting the resistor to the right of Y to zero, and then measuring what voltage is at Y is the same as measuring the voltage to the right of the 30 $\Omega$ resistor to the right of X (at the junction of the two wires). Now you can use both conventions and see what happens. – Kitchi Feb 27 '13 at 14:09

Try the following two thought experiments.

1. Eliminate the branch of the circuit containing Y. You now have three resistors: the internal resistance of the battery ($5\Omega$), and the two given resistors. Whatever current you determine to flow through the circuit, it must be the same through all three of those resistors. Because of that, the voltage drop across the $60\Omega$ resistor has to be exactly twice the resistance drop across the $30 \Omega$ resistor.

2. Eliminate the branch of the circuit containing X. All of the above still holds, except that the ordering of the resistors is different.

Now reassemble the circuit. Because the two branches have the same total resistance ($30 \Omega + 60 \Omega = 60 \Omega + 30 \Omega = 90 \Omega$), they have the same current running through them. If you change the values of the resistors, then the current might not be split equally between the two branches. But the voltage drop across the two branches will always be equal, regardless of what the total current is or how it is split between the two branches. And the voltage drop within a given branch will always be divided between that branch's resistors proportionally to their resistance values.

Consider a story in lieu of explanation. Two backpackers set out from camp (the battery) with just enough supplies to cross a mountain range. The first backpacker has to climb a shallow mountain (a $30 \Omega$ resistor) followed by a steep mountain (a $60 \Omega$ resistor). The second backpacker first climbs a steep mountain ($60 \Omega$) and then a shallow mountain ($30 \Omega$). Climbing a steep mountain takes more supplies than climbing a shallow mountain. They both end up climbing to the same height, and using up the same quantity of supplies. However, at the top of the first mountain, the first backpacker has only climbed the shallow mountain, while the second backpacker has already climbed the steep mountain. The second backpacker has less work to do, but fewer supplies. The first backpacker has more supplies, but still more work to do. Supplies, in this case, represent the voltage, and having crossed only the first mountain, the first backpacker has more supplies than the second backpacker.

You are correct in saying that the direction of current flow (conventional or electron current) does not affect the voltage difference. However, there is a fundamental difference between the two parallel branches of the circuit: one has the 60 ohm resistor on the left and the other has the 30 ohm resistor on the left.

Consider solving the circuit using conventional current. There is a larger voltage drop across the 60 ohm resistor in the top branch than the 30 ohm resistor in the bottom branch, causing the voltage at X to be lower than Y. Notice how we considered only the first resistor encountered.

Now, solve the circuit with electron current. This time, the first resistor encountered is a 30 ohm resistor on the top branch and a 60 ohm resistor on the bottom branch. Again, the resistors encountered are different, and the voltage at X is the same amount lower than Y.

While switching between conventional/electron for the entire circuit does not make a difference, doing so only for one branch (i.e. letting the first resistor encountered by current be 30 ohms for both branches) does modify the circuit.

• What if I change the 30 ohm resistor to the left of Y to 15 ohms? Now if we solve with the conventional current, we will obtain a different result (the Y would encounter a 15 ohm rather than a 30 ohm). With electron current, we obtain the same result as the previous one, as the first resistor encountered is a 30 ohm on the top, and the first resistor encountered is a 60 ohm on the bottom. Isn't this a contradiction? – ithisa Feb 27 '13 at 17:05
• With electron current, we actually do not obtain the same result as before for the voltage in the branch containing Y since decreasing the resistance of one of the resistors increased the current in that branch. Since the resistor was changed to 15 ohms, the voltage drop in electron current is $60/(15+60) = 0.8$ times the original 12 V, which is 9.6 V, leaving 2.4 V at node Y. Using a similar process to find the voltages at node X and Y for both electron and conventional current finds that the voltage difference is consistently 5.6 V in this new circuit. – Shivam Sarodia Feb 28 '13 at 1:18

Actually, what seems very obvious won't be the case, try lighting the light bulb. That was an experiment. Here's what's happening in theory.

The answer is Yes, since Electric Field that is setup in the wire has a direction, current has a direction and yeah..the Potential Energy associated with the current will drop for 60 ohm first and 30 ohm later for first resistance and in reverse for the second resistance.

So, yeah. Direction does matter.

This is a nice diagram from University Physics which makes the concept of Potential fall clear. You are correct about inserting lightbulbs at X and Y. The same current flows through each branch, so the lightbulbs will be equally bright.

However, the question is not whether there is a difference in the currents flowing through X and Y, but whether, if you connect a wire between them, any current will flow through it.

Your objection, of course, will be that if the currents through X and Y are the same, there is no reason why any current should flow either way when they are connected.

In the spirit of KDN's answer (which I applaud), think of the currents through X and Y as two identical parallel rivers flowing down a 90m hill from the LHS (top)and the RHS (bottom) of the battery, carrying the same volume per second. The route through X is initially steep and drops 60 m, whereas the route through Y is initially shallow and drops only 30m. If we opened a channel between X and Y, would any water flow between them? Of course - because X is 30m below Y, and we naturally expect water to flow downhill.

It is not the rate at which water flows through X and Y (the current) which matters but whether there is a difference in height (potential) between them.

If we switched the battery so that the top of the hill was on the RHS and the bottom on the LHS, this would reverse the direction of current flow. Now the route through X is initially shallow while that through Y is initially steep. So Y is below X and this time current would flow from X to Y.