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Physics: Electricity & Circuits (GCSE)

These are some concepts that are really confusing me, if someone could clarify that'd be really helpful :smile:


VOLTAGE in parallel:
So you know how when identical cells are in parallel with each other, the total voltage supplied to the circuit is equal to the voltage of just one of the cells. (The explanation is that in parallel the electrons only do one "lap" aka choose a branch and go through it in a series)
Say you have 3 branches in parallel, the first branch has a 2v battery, second branch has a 2V battery, third branch has a 2V battery, fourth branch has a lightbulb. In a conventional current electricity flows from positive to negative right, so it'll flow FROM the first battery into either one of the other branches right?
So surely, the electrons get an initial "push" from the first 2 volt battery, and say they choose the first branch they get another 2V push from the battery in that branch and then go back to the first branch in series. Surely that's a 4V total voltage??? So how come the rule is that the total voltage of the circuit is only equivalent to the voltage of just one of the cells?

RESISTORS in parallel:
so the fact that the more resistors there are in parallel, the more possible routes there are for electrons to go through is explained as follows: if there's only one lane open the resistance is high and traffic slow, if we open up more lanes, we expect the traffic to speed up so resistance decreases.

I honestly don't understand this? Surely the above correlates to just having more branches not necessarily more resistors? So I don't understand how having more resistors DECREASES overall resistance to that point that overall resistance is LESS than the resistance of any of the individual resistors?

CURRENT in series:
so current doesn't get "used up" and is apparently the same everywhere in a series circuit, but how come when you add resistors current decreases? Surely the resistors is "consuming" current or like using up voltage and therefore decreasing current - isn't that the same as being used up?
(edited 7 years ago)
Original post by medhelp
These are some concepts that are really confusing me, if someone could clarify that'd be really helpful :smile:


VOLTAGE in parallel:

Say you have 3 branches in parallel, the first branch has a 2v battery, second branch has a 2V battery, third branch has a 2V battery, fourth branch has a lightbulb. In a conventional current electricity flows from positive to negative right, so it'll flow FROM the first battery into either one of the other branches right?


How many branches?
This is why we have circuit diagrams, I'm not sure which way the batteries are pointed but the convention is current flows from +ve to -ve as you say...
If the identical batteries are all connected +ve to +ve and -ve to -ve then there will not be a flow of current out of the +ve of one battery into the +ve of the next - it would be going against the potential gradient, A river can't decide to flow against the gravitational gradient and start going back uphill.

RESISTORS in parallel:
so the fact that the more resistors there are in parallel, the more possible routes there are for electrons to go through is explained as follows: if there's only one lane open the resistance is high and traffic slow, if we open up more lanes, we expect the traffic to speed up so resistance decreases.

I honestly don't understand this? Surely the above correlates to just having more branches not necessarily more resistors? So I don't understand how having more resistors DECREASES overall resistance to that point that overall resistance is LESS than the resistance of any of the individual resistors?

I wouldn't think in terms of 'speed' I'd think in terms of number of units per second - if you have one conveyor belt capable of carrying 5 cans of baked beans per second and you stick another identical conveyor belt next to it the number of baked bean cans per second going through your baked bean factory has gone up from 5 per second to 10 per second - the cans are still going at the same speed as before but after 1 second twice as many of them have been moved.


CURRENT in series:
so current doesn't get "used up" and is apparently the same everywhere in a series circuit, but how come when you add resistors current decreases? Surely the resistors is "consuming" current or like using up voltage and therefore decreasing current - isn't that the same as being used up?


voltage is divided up between resistances in series - every time you add a series resistor there's less voltage across the other resistors in the circuit... and as a consequence less current. you are correct that the current is the same in all parts of a series circuit.
Reply 2
Original post by Joinedup
How many branches?
This is why we have circuit diagrams, I'm not sure which way the batteries are pointed but the convention is current flows from +ve to -ve as you say...
If the identical batteries are all connected +ve to +ve and -ve to -ve then there will not be a flow of current out of the +ve of one battery into the +ve of the next - it would be going against the potential gradient, A river can't decide to flow against the gravitational gradient and start going back uphill.

ill draw the one I meant ( sorry its a bit sh*t lol)


Screen Shot 2016-10-29 at 00.25.37.png


I wouldn't think in terms of 'speed' I'd think in terms of number of units per second - if you have one conveyor belt capable of carrying 5 cans of baked beans per second and you stick another identical conveyor belt next to it the number of baked bean cans per second going through your baked bean factory has gone up from 5 per second to 10 per second - the cans are still going at the same speed as before but after 1 second twice as many of them have been moved.


right so how does resistance affect the conveyor belt in parallel? I don't get how adding more resistors makes the overall resistance less?


voltage is divided up between resistances in series - every time you add a series resistor there's less voltage across the other resistors in the circuit... and as a consequence less current. you are correct that the current is the same in all parts of a series circuit.


thanks
Original post by medhelp
ill draw the one I meant ( sorry its a bit sh*t lol)


Screen Shot 2016-10-29 at 00.25.37.png

right so how does resistance affect the conveyor belt in parallel? I don't get how adding more resistors makes the overall resistance less?



thanks


The batteries are a bit like raised tanks of water connected to pipes - if the batteries have the same voltage it's like tanks being at the same height. the water won't flow out of one tank and back up into another and out again because it's being forced to flow from a position of high potential energy to a position of low potential energy. adding more tanks in parallel at the same height doesn't increase the pressure in the pipes - that's just given by the height of the water towers which is equal.

---
The conveyor belts were supposed to be showing that you get a greater throughput when you have two current paths in parallel conducting current*

The resistance of a resistor is the potential difference (in volts) needed to force 1 amp of current to flow through it. when you put resistors in parallel the potential difference across them both is the same.

if you had 2 resistors in parallel with a potential difference of 1V and they both had a resistance of 100 ohms you'd have 0.01 Amps going through the left resistor and 0.01 Amps going through the right resistor - so the currrent after you join the wires together woul be 0.01 + 0.01 Amps = 0.02 Amps.
So if you calculate the effective resistance of your 2 resistor parallel network you've got 0.02 Amps flowing with 1V pushing it, R=V/I so the effective resistance is 1/0.02 = 50 ohms

---
* the general rule is that putting a resistor in parallel can only ever reduce the effective resistance of your network - even putting a rubbish conveyor belt that moved 1 can per second in parallel with your original conveyor belt would increase the rate at which cans moved through the factory, even if only by a small amount.
Reply 4
Original post by Joinedup
The batteries are a bit like raised tanks of water connected to pipes etc


thank you so much! I understand it alot better now. I would rep you again but I can't.


also, I'm really confused by part c of this question

Screen Shot 2016-10-29 at 19.22.48.png

I got 12v / 3ohms = 4A overall because i thought you only count the bit in series right? since I thought that whatever current goes into the parallel part is the same coming out?

but the answers say:
4 + 3 = 7
so current = 12 / 7 or 1.7 A
Original post by medhelp
thank you so much! I understand it alot better now. I would rep you again but I can't.


also, I'm really confused by part c of this question

Screen Shot 2016-10-29 at 19.22.48.png

I got 12v / 3ohms = 4A overall because i thought you only count the bit in series right? since I thought that whatever current goes into the parallel part is the same coming out?

but the answers say:
4 + 3 = 7
so current = 12 / 7 or 1.7 A


TBH I'm not sure what you've done.

On the left of the diagram
because S1 is closed there's a path for current through the 12 Ohm
the parallel 12 Ohm and 6 Ohm have an effective resistance of
1/(1/12+1/6)
=1/0.25
=4 Ohms

On the right of the diagram
because S2 is open there *isn't* a path for current through the 6 ohm... so there's just a single 3 Ohm here

So we've got a 4 Ohm and a 3 Ohm resistance in series - we can calculate the current coming out of the battery from that.

since all the current has to go through the 3 Ohm (there isn't another current path) the current through the 3 Ohm has to be the same as the current coming out of the battery.
Reply 6
Original post by Joinedup
TBH I'm not sure what you've done.

On the left of the diagram
because S1 is closed there's a path for current through the 12 Ohm
the parallel 12 Ohm and 6 Ohm have an effective resistance of
1/(1/12+1/6)
=1/0.25
=4 Ohms

On the right of the diagram
because S2 is open there *isn't* a path for current through the 6 ohm... so there's just a single 3 Ohm here

So we've got a 4 Ohm and a 3 Ohm resistance in series - we can calculate the current coming out of the battery from that.

since all the current has to go through the 3 Ohm (there isn't another current path) the current through the 3 Ohm has to be the same as the current coming out of the battery.


Ooooh so the resistance of the parallel part affects current through the whole circuit? Is that always the case or is just in this one since current has to go through the parallel part in this to get to the 3ohms resistor??
(edited 7 years ago)
Original post by medhelp
Ooooh so the resistance of the parallel part affects current through the whole circuit? Is that always the case or is just in this one since current has to go through the parallel part in this to get to the 3ohms resistor??

I'd look at each circuit individually tbh

all the batteries have the same voltage, it's 1.5V - so when they're in series like that you start at one of the voltmeter terminals, add 1.5 for every battery pointing left to right and subtract 1.5V for every battery pointing right to left and stop when you get to the other voltmeter terminal.

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