Potential at a point in a circuit

hi vadevalor!

conventionally, the current flows from the positive terminal (long line) to the negative terminal (short line)

(the electrons, of course go the other way, towards the positive!)

so the short line is at zero potential, and you have to start from the short line to get to X to find the potential at X, ie the potential difference from zero

(see eg "Electric Potential Diagrams" at http://www.physicsclassroom.com/class/circuits/u9l1c.cfm)

ie you go through the 7.6Ω, so 7.6 is in the numerator

 

The values 4.8V and 1.6V shown on the circuit diagram are misleading. Should not the unit be Ω instead of V?

The equivalent circuit is a voltage supply of 8V connected to 4.8Ω in series with 7.6Ω as shown correctly in your diagram.

Hence 8V is the voltage on (4.8 + 7.6)Ω.

And the voltage on 4.8Ω can be found be using direct proportion.

 

hi vadevalor!

start by replacing the three batteries by one 3V battery …

what is the potential at Y? and where is that measured from?

and now adjust that to start from X instead ​

 

hi vadevalor!

you must always always ALWAYS use current flow not electron flow!

X is somewhere in the middle of the 3V battery, isn't it?

 

yes, i like your diagram …

you've split the batteries into two: a 2V and a 2 minus 1 = 1V battery going the same way,

so it's easy to see whether to add or subtract

i don't understand that …

it's always current flow​

what exactly did they say?

 

ah, i see, you're interpreting the flow direction as the direction for measuring the potential

no, the flow direction (= the current direction) is the direction the charge is moving (which we assume is a positive charge, so the opposite of the electron flow direction)

since the charge moves from high to low potential, the potential decreases as the charge moves, so the potential difference is measured in the opposite direction:

ie we use a positive charge (ie not an electron), and so it will flow outside the battery from the long line (+ve) to the short line (-ve), and the potential will be highest at the long line, and decrease to zero at the short line