Electric current is reduced by the resistance?

[Gajan1234]

How does the current is reduced by the resistance. How is the rate of charge flow decrease in the whole circuit if the variable resistor is placed in the middle of the circuit?

 

[anorlunda]

Resistance is defined as the ratio of voltage to current.

How is the rate of charge flow decrease in the whole circuit if the variable resistor is placed in the middle of the circuit?

That depends on the circuit.If you want better answers, you must learn to ask more specific questions.

 

[Gajan1234]

Resistance is defined as the ratio of voltage to current.
That depends on the circuit.If you want better answers, you must learn to ask more specific questions.

It is very hard to explain my confusion. What I mean is how will the current know to change its rate if the the variable resistor attached in the middle of the circuit. Does this mean tha the current is different before and the variable resistor. If it is then it contradict the fact that current is same in a series circuit.

Hope this question is clear to you now.

 

[anorlunda]

No, you need a picture to show the circuit.

But things in series do have the same current. If you have an idea that contradicts that, the idea must be wrong.

 

[ProfuselyQuarky]

Yeah, a resistor is actually a pretty boring component when it comes to electricity. It doesn’t really do anything actively, which makes its function difficult to understand. Actually, a lot of resistors are nothing more but a piece of bent metal wire. Resistors ultimately just allow you to acquire the proper voltage and current that you want in you circuit. If you have, say, a 9 volt battery, a resistor can decrease the voltage to 6 volts or 4 volts, for example.

It gives you control and power! Mwahahaha

 

[Gajan1234]

Yeah, a resistor is actually a pretty boring component when it comes to electricity. It doesn’t really do anything actively, which makes its function difficult to understand. Actually, a lot of resistors are nothing more but a piece of bent metal wire. Resistors ultimately just allow you to acquire the proper voltage and current that you want in you circuit. If you have, say, a 9 volt battery, a resistor can decrease the voltage to 6 volts or 4 volts, for example.

It gives you control and power! Mwahahaha

This is a good reply. Thanks for the help.

But will the current be different if it has gone through the variable resistor.

 

[Gajan1234]

No, you need a picture to show the circuit.

But things in series do have the same current. If you have an idea that contradicts that, the idea must be wrong.

Will the current reading on A1 be different to A2, the function of the resistor is to decrease the current so for this to happen, doesn't the current need to flow through the resistor?

 

[ProfuselyQuarky]

But will the current be different if it has gone through the variable resistor.

The only difference between a normal resistor and a variable resistor is that the variable resistor allows you to adjust the resistance between specific points within the circuit. With a normal resistor, you would have to add two or more resistors to acquire the voltage you want. A variable resistor does it all for you.

 

[anorlunda]

Will the current reading on A1 be different to A2, the function of the resistor is to decrease the current so for this to happen, doesn't the current need to flow through the resistor?

In that circuit, the current is the same everywhere. Increase the resistance and the current decreases everywhere.

 

[davenn]

Will the current reading on A1 be different to A2, the function of the resistor is to decrease the current so for this to happen, doesn't the current need to flow through the resistor?

you have a slightly incorrect view of the concept, the other responders haven't picked up on this

The resistance doesn't decrease the flow of current. It only allows a certain amount of current to flow according to the value of the resistor and the voltage across it according to Ohms Law ... I = V / R

look at it this way ... a common misconception ...
A person asks ... I have a 12V 10A power supply, can I use it to power my 12V 1A project ? will it burn it out?

The answer is no
The load resistance of the project determines the current that will flow in the circuit

They then ask " what happens to the other 9A?"

Nothing, those other 9A are not generated in the power supply
Just because a PSU is rated at 10A, doesn't mean it's pushing out 10A all the time
That is just a maximum it can produce before it overheats and dies
The amount of current that leaves the PSU is set by the load resistanceDave

 

[sophiecentaur]

@ Gajan: I can sense the old question of "how does the current know what to be, in a circuit" rearing its head. Fact is that, when you switch on any circuit, there is an initial pulse that runs through the circuit and things settle down to a steady sate after a vey short time (as short as a few nanoseconds perhaps). All the DC theory that we start with refers to the steady state.
Dave has pointed out that your language to describe things as you see them, is serving to confuse you. An ideal circuit doesn't exist but you can start by considering the sort of ideal circuit you have drawn. The current flowing will be the same all the way round that circuit. Current (A1) out of the battery (+)will be the same as the current (A2) into the battery (-) . Current does not 'fizzle out' as it makes its way round a circuit. But Voltage (the Potential difference between a point on the circuit and the negative terminal) does - it starts at the full value (Potential Difference) at the + terminal and as the current passes through all the series elements (say a chain of series resistors) the PD will be less and less as you go down the chain until it ends up at zero at the - terminal.
A resistor may not be as sexy as a transistor or an OP AMP but it still demands a lot more understanding that most people admit and it's well worth while getting it's behaviour well sorted before moving on.

 

[sophiecentaur]

I absolutely hate water analogies but here is a good one to illustrate the 'startup thing'.
You have a set of reservoirs at different levels on a hill. They are all empty except the top one, which is kept full (your battery). They are connected in series by a set of pipes and they have water turbines which drive machinery at their inlets (the resistances in series). When you connect the top reservoir, water starts of flow and starts to fill the second reservoir, which soon starts to overflow into the third (but starting with only a trickle, of course). After an interval, the reservoirs are more of less full and water starts to flow out of the bottom reservoir - a trickle and then full on. Only when this whole process has completed can you consider it a steady state and the turbines will then all 'know' how much power to transfer to the machines. Tens of minutes instead of less than 1ns but the principle is the same. At the end, you can say that the gravitational potential energy (your voltage) is shared out according to the drops in height between reservoirs and the water current is the same all the way down - but not before.

 

[Gajan1234]

I absolutely hate water analogies but here is a good one to illustrate the 'startup thing'.
You have a set of reservoirs at different levels on a hill. They are all empty except the top one, which is kept full (your battery). They are connected in series by a set of pipes and they have water turbines which drive machinery at their inlets (the resistances in series). When you connect the top reservoir, water starts of flow and starts to fill the second reservoir, which soon starts to overflow into the third (but starting with only a trickle, of course). After an interval, the reservoirs are more of less full and water starts to flow out of the bottom reservoir - a trickle and then full on. Only when this whole process has completed can you consider it a steady state and the turbines will then all 'know' how much power to transfer to the machines. Tens of minutes instead of less than 1ns but the principle is the same. At the end, you can say that the gravitational potential energy (your voltage) is shared out according to the drops in height between reservoirs and the water current is the same all the way down - but not before.

Thanks indeed for your help

 

[sophiecentaur]

Thanks indeed for your help

Any time - just put a cheque in the post. haha

 

[Averagesupernova]

Sophie used a water analogy. OMG! This seems like one of those moments where I should raise a flag and say welcome aboard.

 

[sophiecentaur]

I did give a disclaimer first, you'll notice and my analogy does involve a fairly good 'correspondence' with the relevant parts of the electrical system. (Please don't mention this to anyone else. {I can't find a "Blush" in the emoticons})
Next time I'll have to use a thermal example, I guess.

 

[phinds]

Actually, a lot of resistors are nothing more but a piece of bent metal wire.

Uh, seriously?

 

[ProfuselyQuarky]

Uh, seriously?

I was just trying to dumb everything a bit down, phinds.

 

[phinds]

I was just trying to dumb everything a bit down, phinds.

I think you carried it way too far

 

[ProfuselyQuarky]

I think you carried it way too far

Okay, you're right

 

[sophiecentaur]

Actually, a lot of resistors are nothing more but a piece of bent metal wire.

Uh, seriously?

I was just trying to dumb everything a bit down, phinds.

The converse is true at least.

 

[berkeman]

Any time - just put a cheque in the post. haha

That could be interpreted two ways...

 

[sophiecentaur]

That could be interpreted two ways...

Mmm. I see what you mean. But the risk of an actual cheque arriving is pretty low. In that unlikely event, it would have to be diverted to PF funds.

 

[jim hardy]

@Gajan1234
It may be helpful to oversimplify.
Have you had high school physics ? Remember atoms and electron shells ? spfd etc ?

It helps a beginner to imagine yourself very small and inside the wires.
Envision yourself in a lattice of metal atoms, each atom with its electron cloud the size of a basketball and a nucleus smaller than the ball of a ball point pen.

In fact, when just beginning you might find it worthwhile to imagine yourself a free charge hopping along from atom to atom along its outermost electrons..

In wires those hops are easy because most metal atoms don't hold tightly to charge in their outer electron shells.
Inside a resistor the atoms are not so friendly and it takes work to get from one to the next. That work shows up as heat. Resistance is in that sense analogous to friction.
Charge has to be pushed through resistive material and heat results. Rub your hands together and feel the heat of friction.

Back to our hopping charges...
Current flows in a closed loop.
By the time you have hopped around the whole loop you've made some easy hops and some difficult ones.
How hard were you pushed by the voltage?
How hard were you resisted by the resistor?
The guy behind you was pushed by same voltage, so was guy behind him...
the quicker your hops the more charges can move through in a given time.

Ratio of available voltage to resistance determines how many charges can push through every second
and charges per second is the definition of current.

Volts
__________ = current
resistance

that is Ohm's Law and it really is that simple.

Believe in algebra and make mental models for yourself.
Keep tweaking your models until they lead you intuitively to the formula. It beats memorization...

Imagination is your friend , provided you put it to work...

old jim

 

[Gajan1234]

@Gajan1234
It may be helpful to oversimplify.
Have you had high school physics ? Remember atoms and electron shells ? spfd etc ?

It helps a beginner to imagine yourself very small and inside the wires.
Envision yourself in a lattice of metal atoms, each atom with its electron cloud the size of a basketball and a nucleus smaller than the ball of a ball point pen.

In fact, when just beginning you might find it worthwhile to imagine yourself a free charge hopping along from atom to atom along its outermost electrons..

In wires those hops are easy because most metal atoms don't hold tightly to charge in their outer electron shells.
Inside a resistor the atoms are not so friendly and it takes work to get from one to the next. That work shows up as heat. Resistance is in that sense analogous to friction.
Charge has to be pushed through resistive material and heat results. Rub your hands together and feel the heat of friction.

Back to our hopping charges...
Current flows in a closed loop.
By the time you have hopped around the whole loop you've made some easy hops and some difficult ones.
How hard were you pushed by the voltage?
How hard were you resisted by the resistor?
The guy behind you was pushed by same voltage, so was guy behind him...
the quicker your hops the more charges can move through in a given time.

Ratio of available voltage to resistance determines how many charges can push through every second
and charges per second is the definition of current.

Volts
__________ = current
resistance

that is Ohm's Law and it really is that simple.

Believe in algebra and make mental models for yourself.
Keep tweaking your models until they lead you intuitively to the formula. It beats memorization...

Imagination is your friend , provided you put it to work...

old jim

THANK YOU SIR

 

[LvW]

How does the current is reduced by the resistance. How is the rate of charge flow decrease in the whole circuit if the variable resistor is placed in the middle of the circuit?

So many good answers...however, I like to add another one which is based on some fundamental electrical rules.
At first - the "current" is not "reduced" by a resistance - instead, it is determined by the driving voltage and the connected resistance (Ohm`s law).
But this is only the RESULT of some considerations and not yet an EXPLANATION.

Therefore:
1.) Question: What causes the movement of charged carriers (electrons) in a resistance? Answer: A mechanical force F=E*q caused by an electrical field E within the body of the resistor. Fot the following it is helpful to replace the resistive body by a piece of wire with a length L and the cross-sectional area A.
2.) The electrical field within this piece of wire is defined as applied "voltage V divided by the length L of the field": E=V/L.
3.) Now the material of the wire comes into play. The number and the mobility of free electrons (available to form a current) are characterized by a factor κ which is the "specific conductivity" of the material. With this factor we define a quantity called "current density" S=κE. This parameter S is a quantitative measure of the electrons ability to move.
4.) If we multiply the current density S with the real cross-sectional area A of the wire, we get a quantity called "electrical current": I=κ*E*A,
5.) Inserting the results of 2) we get I=κ*E*A=V*(κ*A/L).
6.) Defining R=L/κ*A we arrive at the ohmic law in the classical form: I=V/R.
(It apprears logical and evident that R is proportional to L and invers proportional to κ and A).

 

[davenn]

At first - the "current" is not "reduced" by a resistance - instead, it is determined by the driving voltage and the connected resistance (Ohm`s law).
But this is only the RESULT of some considerations and not yet an EXPLANATION.

yup, covered that many posts ago

D

 

[Ario Barzan]

@Gajan1234
If the current at both sides of the variable resistor is not the same, the difference must accumulate inside the resistor. Now, I ask you how is this possible?

 

[jim hardy]

If the current at both sides of the variable resistor is not the same, the difference must accumulate inside the resistor. Now, I ask you how is this possible?

When your logic leads to an impossible conclusion then one of two things is wrong:
your logic
or
the assumption from which you started.