# Amps in a parallel circuit?

Watch
Announcements
#1
I'm slightly confused about a couple 'rules' to do with amps in a parallel circuit. I'm not sure how much sense this is going to make but I'll try to explain as simply as possible.

As stated in my revision guide the amps in each branch of a parallel circuit sums to the amps before and after the branches connect (around the battery). So, why is it that on many books and websites it is stated that amps are also higher in the branches of parallel circuits than in series circuits (where I thought the current was the same between every component -and so is surely higher?).

I am aware this is because of Ohm's law, as the current must increase when the resistance in a circuit decreases (as it does when more branches are added), but how come current increases in parallel but also decreases? I feel I am missing something obvious but I can't see it. Feel free to roast me 0
3 years ago
#2
Ohm's law states that the potential difference across a component is proportional to the resistance across that component, up until a limit of proportionality. An example of this limit may be in a filament light bulb, etc, where temperature rapidly becomes the dominant factor and resistance rapidly rises. So, we know V = IR, with I being our deciding constant up until this limit.

For components in parallel, let's use the example of two resistors, we now have two paths for our current to flow. Hence, seeing as it can flow either way, the effective resistance is halved; half the current goes either way, rather than all through one resistor, and hence overall the current has met half the resistance it would otherwise have met. This is how you add up resistances in parallel, i.e. (1/R1 + 1/R2) = (1/Rtot)

We've also established that all the current going in will end up coming out, i.e. from our parallel part of the circuit: this is Kirchoff's Current Law. There is also another law, i.e. that the sum of the potential differences across all the components equals that of our source.

So, we have the concept that any and all current going into a junction must come out of another junction, and we have the V = IR equation from Ohm's law, paired with the fact that we can handle components/resistances in parallel. The question to ask is, how do we know how much current goes across each component in parallel? This all depends on the resistances of each component. The flow of charge will always follow the easiest path, the one with the least resistance, such that it balances out in the end to minimise the total resistance to flow. The last item of interest for our entire length of awful text I've written: The potential difference across any component in parallel is equal to the potential difference of any other component in parallel for a given parallel system, i.e. for the infinite number of resistors we have in parallel at a point, they all have the same potential difference. The proof for this lies with the fact that we can consider the entire system as one effective component by summing up resistances in parallel and then considering the current across the entire effective component, or by just considering common sense on a few example setups and then deriving it based on that.

So,
V = IR
Current in = Out for a given junction
(1/R1 + 1/R2 + ... 1/RN) = (1/Rtotal)
Total voltage around a loop of the circuit = EMF of circuit (inc int. res.)

That's about all you need to know to work out any circuit that you might find. Current in parallel all adds up to the same value, the amount before it enters the parallel part of the circuit, it's just that certain branches of the parallel circuit will have larger currents than other branches.

Hope I helped... 99% sure I didn't, but hey, I tried, right?

(Original post by Saucy_muffin)
I'm slightly confused about a couple 'rules' to do with amps in a parallel circuit. I'm not sure how much sense this is going to make but I'll try to explain as simply as possible.

As stated in my revision guide the amps in each branch of a parallel circuit sums to the amps before and after the branches connect (around the battery). So, why is it that on many books and websites it is stated that amps are also higher in the branches of parallel circuits than in series circuits (where I thought the current was the same between every component -and so is surely higher?).

I am aware this is because of Ohm's law, as the current must increase when the resistance in a circuit decreases (as it does when more branches are added), but how come current increases in parallel but also decreases? I feel I am missing something obvious but I can't see it. Feel free to roast me 0
3 years ago
#3
I'm a little confused what you're asking not going to lie, The only rules you really need to know is that in serial Amps are equal all across, there is only one pathway that the current may take. In parallel you have two pathways, if they're similar then current will be more or less halved equally between the two paths. Remember that if one pathway in a parallel circuit has zero resistance then there will be more current going through it compared to one with say, 10Ohms of resistance. Think of the current as being inversely proportional to the resistance across a wire.
0
#4
(Original post by Callicious)
Ohm's law states that the potential difference across a component is proportional to the resistance across that component, up until a limit of proportionality. An example of this limit may be in a filament light bulb, etc, where temperature rapidly becomes the dominant factor and resistance rapidly rises. So, we know V = IR, with I being our deciding constant up until this limit.

For components in parallel, let's use the example of two resistors, we now have two paths for our current to flow. Hence, seeing as it can flow either way, the effective resistance is halved; half the current goes either way, rather than all through one resistor, and hence overall the current has met half the resistance it would otherwise have met. This is how you add up resistances in parallel, i.e. (1/R1 + 1/R2) = (1/Rtot)

We've also established that all the current going in will end up coming out, i.e. from our parallel part of the circuit: this is Kirchoff's Current Law. There is also another law, i.e. that the sum of the potential differences across all the components equals that of our source.

So, we have the concept that any and all current going into a junction must come out of another junction, and we have the V = IR equation from Ohm's law, paired with the fact that we can handle components/resistances in parallel. The question to ask is, how do we know how much current goes across each component in parallel? This all depends on the resistances of each component. The flow of charge will always follow the easiest path, the one with the least resistance, such that it balances out in the end to minimise the total resistance to flow. The last item of interest for our entire length of awful text I've written: The potential difference across any component in parallel is equal to the potential difference of any other component in parallel for a given parallel system, i.e. for the infinite number of resistors we have in parallel at a point, they all have the same potential difference. The proof for this lies with the fact that we can consider the entire system as one effective component by summing up resistances in parallel and then considering the current across the entire effective component, or by just considering common sense on a few example setups and then deriving it based on that.

So,
V = IR
Current in = Out for a given junction
(1/R1 + 1/R2 + ... 1/RN) = (1/Rtotal)
Total voltage around a loop of the circuit = EMF of circuit (inc int. res.)

That's about all you need to know to work out any circuit that you might find. Current in parallel all adds up to the same value, the amount before it enters the parallel part of the circuit, it's just that certain branches of the parallel circuit will have larger currents than other branches.

Hope I helped... 99% sure I didn't, but hey, I tried, right?
That was certainly not awful. So does this mean that the amps after the junction will be less than the current going through the resistor since there is resistance and thus the current must be higher as the voltage has not changed? Or am I speaking complete jibberish? I think I'm misreading my diagrams. There is one where there are two amp meters before two resistors whose amps sum to the amps before the junction. But Ohm's law states that the current must be bigger than this around a resistor. (I'm just blabbering in the hope that you spot a rookie mistake). Thanks for the help so far and good luck deciphering that 0
3 years ago
#5
(Original post by Saucy_muffin)
I'm slightly confused about a couple 'rules' to do with amps in a parallel circuit. I'm not sure how much sense this is going to make but I'll try to explain as simply as possible.

As stated in my revision guide the amps in each branch of a parallel circuit sums to the amps before and after the branches connect (around the battery). So, why is it that on many books and websites it is stated that amps are also higher in the branches of parallel circuits than in series circuits (where I thought the current was the same between every component -and so is surely higher?).

I am aware this is because of Ohm's law, as the current must increase when the resistance in a circuit decreases (as it does when more branches are added), but how come current increases in parallel but also decreases? I feel I am missing something obvious but I can't see it. Feel free to roast me I would help but I seriously am not getting what you are trying to ask xD? Also I think you mean "current" instead of amps. Amps is what current is measured in but current is the actual thing going around in a circuit.
0
3 years ago
#6
(Original post by Saucy_muffin)
That was certainly not awful. So does this mean that the amps after the junction will be less than the current going through the resistor since there is resistance and thus the current must be higher as the voltage has not changed? Or am I speaking complete jibberish? I think I'm misreading my diagrams. There is one where there are two amp meters before two resistors whose amps sum to the amps before the junction. But Ohm's law states that the current must be bigger than this around a resistor. (I'm just blabbering in the hope that you spot a rookie mistake). Thanks for the help so far and good luck deciphering that Yall got me confused xD, okay, so... two amp meters before a resistor... the way that's described, their sum should equal the amount before the junction, so, that's fine and dandy;

Remember, Ohm's law is V is proportional to R, current is just a constant introduced. V = IR is an equation derived from Ohm's Law. The current partitioning between the two resistors depends only on the resistances, and the current will be partitioned opposite to this, i.e. if one resistor is higher, it gets less than the other resistor. (higher resistance, not altitude, I mean, xD... sorry, revising and my keyboard is on my bed, and having a huge microphone literally in your face while typing is annoying)

The sum before the junction only describes what goes in, and each amp meter describes what each resistor in series with it gets.
0
3 years ago
#7
(Original post by Saucy_muffin)
I'm slightly confused about a couple 'rules' to do with amps in a parallel circuit. I'm not sure how much sense this is going to make but I'll try to explain as simply as possible.

As stated in my revision guide the amps in each branch of a parallel circuit sums to the amps before and after the branches connect (around the battery). So, why is it that on many books and websites it is stated that amps are also higher in the branches of parallel circuits than in series circuits (where I thought the current was the same between every component -and so is surely higher?).

I am aware this is because of Ohm's law, as the current must increase when the resistance in a circuit decreases (as it does when more branches are added), but how come current increases in parallel but also decreases? I feel I am missing something obvious but I can't see it. Feel free to roast me As you've said, having more branches means less resistance, so more total current; the current in the single-branch sections of a parallel circuit will therefore increase. However, this current will be equal to the total current across all parallel branches; the current might not be the same within each parallel branch (one with more resistance will have a lower current in it, as V=IR => I=V/R and the PD is the same in each parallel branch).
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How would you feel if uni students needed to be double vaccinated to start in Autumn?

I'd feel reassured about my own health (21)
16.8%
I'd feel reassured my learning may be less disrupted by isolations/lockdowns (40)
32%
I'd feel less anxious about being around large groups (12)
9.6%
I don't mind if others are vaccinated or not (13)
10.4%
I'm concerned it may disadvantage some students (6)
4.8%
I think it's an unfair expectation (31)
24.8%
Something else (tell us in the thread) (2)
1.6%