Resistance In Electric Circuit

1. Oct 13, 2011

san203

I have 2 questions

1st: How does the resistance decrease Current? (Apart from converting the electrical Energy into other forms of energy)

2nd: Why is the presence of Resistance felt in the entire circuit i.e. Current reduced in the entire circuit?

Suppose there is a circuit with a cell A and a resistor B.
The path of electrons from A to B should be without resistance(Apart from the resistance offered by the wire which is Negligible).
But after that from B back to A,there should be less current flowing.
Then Why is the current in the entire circuit decreased?

Thank You!

2. Oct 13, 2011

Studiot

Good morning san, welcome to Physics Forums.

Rather than thinking that resistance 'decreases the current' - perhaps because of its name - think about it another way.

The current in a circuit or part of one is proportional to the voltage across that circuit or part of one. (Ohm's Law)

The resistance is simply the constant of proportionality.

So in your example the resistance in the wire is very low, but the voltage across the wire is also very low.
So the current in the wire is the ratio of two small quantities. (Voltage /Resistance)

The voltage across the resistor is much higher so although the resistor has a higher resistance the ratio is the same as for the wire so the current is the same.

Does this help?

Last edited: Oct 13, 2011
3. Oct 13, 2011

sophiecentaur

The Resistance of a circuit is a number that tells you how many Volts you need to apply in order to make 1 Amp flow around the circuit.
This definition gives you the useful formula
V=IR (V = volts, I = current and R is resistance)
(and the other two versions you can get by re-arranging it)
Any other, hand waving definition of Resistance is likely to confuse you. It is not helpful, for example, to say it's 'how hard it is to push electrons through'. That sort of description doesn't really tie in with the above equation - so you should avoid it. That equation works!! and it's sometimes described as Ohm's Law (not strictly accurately aamof).

4. Oct 13, 2011

Staff: Mentor

Total resistance (together with the voltage) sets the current.

You can see a good analogy in water flow through pipes. Suppose you have a wide pipe carrying water, but somewhere along the way a constriction has formed in the pipe, so that 80% of the bore is blocked. Water finds it difficult to progress through this narrow opening, and this high resistance section limits the flow through the whole pipe. It makes no difference whether the partial blockage has formed in the first section of piping, or near the centre, or near the end, it limits the current just the same.

5. Oct 13, 2011

sophiecentaur

The problem with the old, simplified (/jaded), analogy of water flow is that it doesn't include the idea that electricity actually transfers Energy. The Resistance of the wire in a heater or the 'Effective Resistance' of any electrical appliance or device whilst it is working involves Energy Transfer. Just talking about constrictions in water pipes really doesn't involve any concept of Energy. Does anyone ever mention that there may be very little energy lost along the journey down a pipe -( just a bit of turbulence, in fact, for low velocities) That, in itself, is sufficient reason to tread very carefully with a water flow analogy. Likening electrons to drops of water makes things worse.
If you really want to have a good water model of electricity, then you need to have pumps and turbines with wide bore pipes connecting everything. Then you need to put your model on a mountainside, talk in terms of raising water, to give it Potential Energy and using that energy to power devices on the way down. Well, dang me- you just got yourself an Electric circuit with Potential, Current and Resistance. Why bother to get your feet wet at all?

I agree that there is an elementary analogy where circuit continuity is concerned but none of the noddy water models help with a real understanding of how electricity 'works'. I thought that people got as far as asking questions on PF in order to get a better view of things than they got in early years at School.

6. Oct 13, 2011

san203

Exactly!!!

Thanks

7. Oct 13, 2011

sophiecentaur

Energy is transferred when the current passes through one resistance - so there is less energy available for the second resistance. The two energies add up to the total supply energy (per coulomb).
Voltage is joules per coulomb. This yields Kirchofff's Second Law.
Btw, don't worry about how the volts 'know' how to sort themselves out so they're shared appropriately. It does take a finite time to settle down as the electromagnetic wave propagates through the circuit.
The same current must flow around the whole loop (if it's just a single loop). If there are multiple paths then the total current leaving the Power Supply + terminal will go into the - terminal and, when it splits at a junction, the input current is the sum of the current into the branches. (Kirchoff 1)

8. Oct 13, 2011

davenn

The current isnt decreased ( thats an incorrect assumption) its just limited by the resistance of the circuit.
In this 3 resistor circuit below, the current flowing at any point in the circuit is the same. it DOESNT change. The current flowing in the circuit is determined
by the total resistance of the circuit I = V / R.
It doesnt matter where you measure the current you will get the same result. between the battery and R1, between R1 and R2, between R2 and R3, or
finally between R3 and the battery

cheers
Dave

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9. Oct 13, 2011

davenn

Further to the previous post....

What is decreasing as you go around the circuit is the voltage.
There is a voltage drop across each resistance.
firstly work out the current flowing in that circuit I = V/R. Once you have that current
then you can work out the voltage drop across each resistor V(R1) = I x R1, V(R2) = I x R2, V(R3) = I x R3.
those 3 voltage drops will add up to the total voltage supplied by the battery (or other voltage source)

Dave

EDIT....
think of this real world situation .... I have a 12V, 5 amp plugpack, my MP3 player needs/uses 12V and 750mA (milliAmp). so what happened to the 4.25A ?
is it decreased ? NO, just because the plugpack can supply a maximum of 5A doesnt mean thats its always supplying that amount.
The internal resistance (load) of the MP3 player (across the power supply) is such that only 750mA (.75A) is flowing from the 5A rated plugpack.

does that help ?

Last edited: Oct 13, 2011
10. Oct 14, 2011

san203

Okay but the Limited.Why?

Why???
And Can you explain from the Electron Point of View,you know regarding all those Cross section area and Drift Speed.

11. Oct 14, 2011

sophiecentaur

If you can bear it, try to avoid thinking in terms of an individual electron. It's a much more statistical business. A huge number of electrons are moving randomly with a huge net RMS velocity. All your battery is doing is giving them all a small amount of extra KE in a particular direction. Get a better model in your head and you should find fewer contradictions.