tellmesomething said:
I was definitely glitching when i thought about this sum, i assumed the current to go from both paths of terminal A to terminal B but that doesnt make sense as the current needs to go back to A ofcourse or theres gonna be an accumalation of charges at B which would stop the current due to repulsion
Yes. You have a loop consisting of components in series. The current simply flows (clockwise or anticlockwise) around the loop.
tellmesomething said:
if a circuit has a battery with internal resistance how can it be short circuited? The whole point of short circuits is that they have 0 potential difference between the terminals of the cell right? So if it has internal resistance then that means it undergoes a voltage drop so yeah can you please explain where im going wrong?
If you short-circuit a
single source (emf = ##\mathscr E##, internal resistance = ##r##) then you connect its terminals with a zero-resistance connection. The voltage between the source’s terminals is then zero and the current through the source is ##\frac {\mathscr E}r##. Edit: And the voltage across the internal resistance is ##Ir =\frac {\mathscr E}r r =\mathscr E##; the 'volts gained' from the 'pure emf' equals the 'volts lost' from the current flowing through the internal resistance.
In the Post #1 question, the sources are connected in a loop so I’d agree with you that
each separate source should not be described as ‘short-circuited’.
However, it's worth noting that because ##\mathscr E = \alpha r##, it turns out that each cell is behaving just the same as if it were short-circuited! Each source has 0 volts between it terminals and carries a current of ##\frac {\mathscr E}r##! See below.
tellmesomething said:
If we take your values
@Steve4Physics I start at the 3V battery my current loses 6 ohms then gains 6 ohms then loses 1 ohm then gains 1 ohm. Till now voltage is 0. Now it loses 3 volts because of the internal resistance of the 3V battery. Voltage drop=3V, how is it short circuiting?
Forget the short-circuiting thing. And note that you don't 'gain' an 'lose' ohms as you move around a circuit.
Consider the Post #3 example. I'd explain it like this...
Total emf = 10V, total resistance = 20 ##\Omega## and current, I = 0.5A.
Remember that the voltage between a source’s terminals is the source’s emf minus the voltage-across the internal resistor (the so-called 'lost volts').
For the 1V emf source, the voltage-drop across its internal resistance = Ir = 0.5*2 = 1V. Voltage between source’s terminals is 1V – 1V = 0V.
For the 3V emf source, voltage-drop across its internal resistance = Ir = 0.5*6 = 3V. Voltage between source’s terminals is 3V – 3V = 0V.
Similarly the 6V emf source has 0V between its terminals.
I’d suggest experimenting with two things:
a) See what happens if ##\mathscr E = \alpha r## is not true. For example, just change any one value in the Post #3 example and work out the voltage between each source's terminals.
b) See if you can prove (using some algebra) that the emf* across each source (when ##\mathscr E = \alpha r##) is always 0 for any number of sources in a loop.
*Edit: 'emf' is wrong. Careless of me. Should be 'voltage' or, even better, 'terminal pd'. Thanks
@tellmesomething.