I don't understand resistors anymore?!?

Enya1998

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As far as I'm concerned, a resistors 'resists' the flow of charge, or in other words a resistor stops/reduces the current, I got it. Now, when you have 2 resistors in series the current remains the same througjt the whole circuit, but when its in series the current decreases?? I'm just so confused that I don't even know what I'm talking about. Could someone please explain how current (and voltage) are effected by resistors in series and parallel, and explain why briefly please? Thank you!

Sorry for the stupid question btw

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username2769500

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Being thinner.less free electrons per nuclei And overall electron density within the internal structure of the metal.

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TheVirtualPhoton

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(Original post by Enya1998)
As far as I'm concerned, a resistors 'resists' the flow of charge, or in other words a resistor stops/reduces the current, I got it. Now, when you have 2 resistors in series the current remains the same througjt the whole circuit, but when its in series the current decreases?? I'm just so confused that I don't even know what I'm talking about. Could someone please explain how current (and voltage) are effected by resistors in series and parallel, and explain why briefly please? Thank you!

Sorry for the stupid question btw

I dropped physics at AS but Ill give it a go and see what I remember
Voltage is essentially the input to the circuit so will be constant.
Resistors in series add together to increase the resistance of the circuit, those in parallel act differently depending on the resistance of each resistor.
If the resistance increases, using R=V/I the current will decrease

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MoonlightBoo

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(Original post by Enya1998)
As far as I'm concerned, a resistors 'resists' the flow of charge, or in other words a resistor stops/reduces the current, I got it. Now, when you have 2 resistors in series the current remains the same througjt the whole circuit, but when its in series the current decreases?? I'm just so confused that I don't even know what I'm talking about. Could someone please explain how current (and voltage) are effected by resistors in series and parallel, and explain why briefly please? Thank you!

Sorry for the stupid question btw

I don't really understand your question, but I'll try to give you the difference between 2 resistors in series vs parallel

In a series circuit, if you have 2 resistors, the current will decrease more than if you had 2 resistors in a parallel circuit. This is kind of hard to explain, so I'll give you an example.

Imagine you are at a playground. Playground 1 only has one set of slides, which is one slide right behind another. If you have 10 kids, it will take longer for them all to get to the other side of the slide because they all have to go through 2 slides. (This is a metaphor of a series circuit)
Playground 2, however, has 2 slides next to each other. If you have 10 kids, they can distribute themselves, so 5 can go on each slide, and the time it takes for all of them to get to the other side will have halved. (This is a metaphor or a parallel circuit)

Do you see what I mean? Sorry if it didn't make sense, hope I helped though!

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Joinedup

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(Original post by Enya1998)
Now, when you have 2 resistors in series the current remains the same througjt the whole circuit, but when its in series the current decreases??

The first statement is a rule - the total current in series resistors is the same at every point in that circuit.

the second statement describes what happens when you add a series resistor to an existing circuit to make a new circuit.

the current in the old circuit is the same at every point in the old circuit.
the current in the new circuit is the same at every point in the new circuit.
BUT the current in the old circuit isn't the same as the current in the new circuit, the current in the new circuit is lower than the old circuit because it's got greater total resistance.

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K-Man_PhysCheM

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(Original post by Enya1998)
As far as I'm concerned, a resistors 'resists' the flow of charge, or in other words a resistor stops/reduces the current, I got it. Now, when you have 2 resistors in series the current remains the same througjt the whole circuit, but when its in series the current decreases?? I'm just so confused that I don't even know what I'm talking about. Could someone please explain how current (and voltage) are effected by resistors in series and parallel, and explain why briefly please? Thank you!

Sorry for the stupid question btw

Let's analyse two different scenarios (note this is using A-level knowledge):

one where we have 2 fixed resistors of resistance

and

, connected in series to a dc power supply of EMF

and negligible internal resistance,

and one where the same resistors are connected in parallel.

In the first scenario (2 in series), there is a single closed loop. By Kirchhoff's 1st Law, the sum of currents entering any point is equal to the sum of currents leaving any point, so the current,

is the same at all points in the series circuit.

And by Kirchhoff's 2nd Law, the sum of the EMFs is equal to the sums of the pds in a closed loop:

So

where

and

are the pds across each resistor, 1 and 2.

By Ohm's Law,

Hence we can deduce that:

Rearranging:

$I = \dfrac{V_1}{R_1} = \dfrac{V_2}{R_2}$

Or, $I = \dfrac{V_1+V_2}{R_1+R_2} = \dfrac{V_0}{R_1+R_2}$

-------
When they are connected in series, we have 2 closed loops with the power supply. By Kirchhoff's 1st:

, where

is the current at the power supply, and

and

are the current at resistors 1 and 2 respectively.

By Kirchhoff's 2nd:

and

, therefore

By Ohm's Law:

As

in parallel:

As

:

$I = \dfrac{I_1 R_1 + I_1 R_2}{R_2} = \dfrac{I_1(R_1+R_2)}{R_2}$

And as $I_1 = \dfrac{V_1}{R_1}$ and

:

$I = \dfrac{V_0(R_1+R_2)}{R_1R_2}$ , which is a standard result.

If we compare this with our expression for current in a series circuit above, the difference is that the series circuit has $V_0 \times \dfrac{1}{R_1+R_2}$ while the parallel circuit one has $V_0 \times \dfrac{R_1+R_2}{R_1R_2}$ .

Note: $\dfrac{1}{R_1+R_2} = \dfrac{1}{R_1+R_2} \times \dfrac{R_1+R_2}{R_1+R_2}$

$= \dfrac{R_1+R_2}{{R_1}^2 + 2R_1R_2 + {R_2}^2}$

This is clearly smaller than $\dfrac{R_1+R_2}{R_1R_2}$

Therefore, the current drawn from the power supply in a circuit will always be smaller when 2 resistors are connected in series than when the same 2 resistors are connected in parallel.

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