Constant Current in a series circuit

[Celluhh]

Homework Statement

I have learned that current is constant throughout a series circuit, with or without resistors. Even if the no. of resistors in the circuit increases, the current is the same through each resistor. Resistors resist current flow, even though the current flowing through any point in the circuit remains the same as charge cannot"disappear", will the speed of the current remain the same through each resistor? An ammeter measures current, but does it measure the speed of current or the flow of current? When a circuit is complete, do the charges start flowing out from the battery cell or are there already charges in the wires in the circuit? I understand the current drawn and the current flowing through a series circuit is basically the same idea, but how does the current "see"the no. of resistors in the circuit and adjust its speed to fit the resistance in the whole circuit? Assuming voltage is constant.

 

[danielakkerma]

Hey there, welcome to the forum!
Let me just ascertain that I follow you, first:
You mentioned that:

Even if the no. of resistors in the circuit increases, the current is the same through each resistor.

But you must recall that:
$\normalsize I = \frac{V}{\sum{R}}$
Meaning that unless V grows along with the number of resistors, that total current will not remain that same.
The velocity of the charges in a conductor, to answer your query, depends on a characteristic factor, known as the conductivity. The drift velocity of the charges(which makes up the bulk of their speed), upon the application of a potential, is reliant upon the current traveling through(excepting fringe effects, and huge Vs),The amount of charge passing, per second, is what the Ammmeter measures.
For a full description on how it works, try:
http://en.wikipedia.org/wiki/Ammeter"
Hope that helps somewhat,
Daniel

 

[PeterO]

Homework Statement

I have learned that current is constant throughout a series circuit, with or without resistors. Even if the no. of resistors in the circuit increases, the current is the same through each resistor. Resistors resist current flow, even though the current flowing through any point in the circuit remains the same as charge cannot"disappear", will the speed of the current remain the same through each resistor? An ammeter measures current, but does it measure the speed of current or the flow of current? When a circuit is complete, do the charges start flowing out from the battery cell or are there already charges in the wires in the circuit? I understand the current drawn and the current flowing through a series circuit is basically the same idea, but how does the current "see"the no. of resistors in the circuit and adjust its speed to fit the resistance in the whole circuit? Assuming voltage is constant.

The battery in the circuit establishes an electric field through the wires and resistors, so charges all the way around the circuit begin to move at the same time.
The ammeter measures the flow rate - typically in coulombs per second. Given that it takes over 6 x 10¹⁸ electrons to make a whole coulomb of charge, the number of electrons flowing is probably best described as unimaginable. Even if the current is only several micro amperes, that is still over 6 x 10¹² electrons per second - still an unimaginably large number, though we are happy to say 6 trillion.

 

[Celluhh]

is reliant upon the current traveling through[/B](excepting fringe effects, and huge Vs)

Thanks Daniel for taking the trouble to answer! But I don't understand this part.Why is the drift velocity of the charges reliant on the current traveling through? In a series circuit, the current is constant throughout all resistors right?

i found this on wikipedia( i think it actually explains what i don't understand):

Electric currents in solids typically flow very slowly. For example, in a copper wire of cross-section 0.5 mm2, carrying a current of 5 A, the drift velocity of the electrons is on the order of a millimetre per second.
Any accelerating electric charge, and therefore any changing electric current, gives rise to an electromagnetic wave that propagates at very high speed outside the surface of the conductor. This speed is usually a significant fraction of the speed of light, as can be deduced from Maxwell's Equations, and is therefore many times faster than the drift velocity of the electrons. For example, in AC power lines, the waves of electromagnetic energy propagate through the space between the wires, moving from a source to a distant load, even though the electrons in the wires only move back and forth over a tiny distance.

The ratio of the speed of the electromagnetic wave to the speed of light in free space is called the velocity factor, and depends on the electromagnetic properties of the conductor and the insulating materials surrounding it, and on their shape and size.

unfortunately i do not understand this explanation.=( Does this actually also mean that the current "sees" the resistors and decides what speed to travel at because disturbances(resistors) travel at the speed of light??

 

[Celluhh]

The battery in the circuit establishes an electric field through the wires and resistors, so charges all the way around the circuit begin to move at the same time.

Thanks peter for the answer! Hmm by this do you mean that there are already charges in the wires, basically everywhere in the circuit?

 

[danielakkerma]

Hi,
Perhaps a more atomic scaling of the elucidation could help you make sense of it.
The Current, "decides" as you say, how large it should be, based on the number of resistors connected, due to the need to conserve energy. The current takes the form that is most restrictive in wasting its power on producing heat in a resistor. Each resistor R_i, carrying a current I, will produce a potential difference V_i. And as a consequence, you're right, the charges travel at an hoarsely slow speed, this is because they sweep, "drift" towards the various ends of the potential source, so, an individual electron, if we were able to monitor its exact location at any given time, could take several hours or even days, to cross some wires. But since a single electron does not produce the current passing through the circuit, but rather the entire agglomeration, so some charges which were initially closer to the source, would reach it faster(as the distance is shorter) and so forth.
What must be considered however, is that by virtue of the classical theories on conductivity, the metal sees the electric field attached to it "instantly", and uniformly across it(due to the continuity principle), and so every particle experiences roughly the same force on it.
Hope that helps,
Daniel

 

[Celluhh]

hmm may i ask what you mean by"metal"? do you mean the resistor? Ok so the current decides its form, but then my question still remains, how does it know how many resistors there are in the circuit?

hey can i ask u another question? when i conducted an experiment in the sch lab, i connected the end of the circuit to each terminal of the battery. But i realized i connected the wrong end to the wrong terminal. so i switched the ends to their correct terminals. but the bulbs for both experiments had different brightness, and even had the same order of brightness for both experiments.bulb x was the brightest, bulb y next and followed by bulb z.
although bulb x remained as bright as the previous bulb x, bulb y remained as bright as the previous and so on. can you explain what technical fault there was? I don't think there was a problem with the bulb caused we tested them before the experiment and they were of the same brightness.
Thanks a lot!

 

[danielakkerma]

Hi...
The current "does" so, mainly because the Sum of all Vs in the circuit has to equal the source, i.e Kirchhoff's Voltage law. By increasing the number of resistors, you get more V_is, and therefore the total Potential difference scatters across more hubs as it goes along.
As for the brightness, that's an interesting point, and if you can, please elaborate on it some more.
You should note that the blubs' intensity depends on the power, or P = R*I^2;
Since the current is one for the entire series circuit, assuming the resistances are the same, they should all light up with equal brightness.
Please tell us exactly what had happened,
Daniel

 

[Celluhh]

Hi...
The current "does" so, mainly because the Sum of all Vs in the circuit has to equal the source, i.e Kirchhoff's Voltage law. By increasing the number of resistors, you get more V_is, and therefore the total Potential difference scatters across more hubs as it goes along.

Daniel. I'm so sorry but i still don't get how more potential difference allows the current to "decide" its speed, (this is what you mean right?)oh and by the way i don't understand kirchhoffs voltage law either.>.<

 

[Celluhh]

As for the brightness, that's an interesting point, and if you can, please elaborate on it some more.
You should note that the blubs' intensity depends on the power, or P = R*I^2;
Since the current is one for the entire series circuit, assuming the resistances are the same, they should all light up with equal brightness.
Please tell us exactly what had happened,
Daniel

Firstly, my group tested if bulb x, bulb y and bulb z were working properly by connecting then with wire to the battery. They all lit up with about the same brightness. However when we connected them in a series circuit, with bulb x first, followed by a resistor of one ohm, then bulb y, then a resistor of 2 ohm, then bulb z. Bulb x lit up the brightest and bulb y and z were dimmer but of the same brightness.
we then switched the places of the resistors but that had no effect on the brightness of the bulbs.
After that, we realized we connected the wires to the wrong terminals and the bulbs were not in the correct order. So we changed the connection and connected the wire (which was first connected to the postive terminal)to the negative terminal and vice versa however, the brightness of the bulbs were not affected.
Can you explain why please? we came up with things like current flow from postivive terminal, electron flow from negative terminal, but do these actually affect the brightness?


oh yeah, some people say that current flows from positive terminal so in a series circuit, whichever bulb gets the current first will use up the current and then the current will travel to other bulbs and whatever current passes through other bulbs does not affect the previous bulb because the latter is located behind it. this is a wrong conception, right?

 

[danielakkerma]

Well, physically speaking, there shouldn't be any difference at all in their illumination; You better verify the equipment used and the entire setting as a whole.
The only influence on the current would arrive from the different resistors you used, as you connected them in series. Since they have different values, different power in each, that could sway the whole thing either way.
For further reading, it might avail you to see this(In case I've harangued you sufficiently, and grown wearisome :
http://wiki.answers.com/Q/What_happens_to_light_intensity_of_lamp_in_series_circuit_when_more_lamps_are_added_in_circuit"
Hope that's more edifying,
Daniel

 

[Celluhh]

If the wattage of the lamps varied, the brightest lamps would be those with the lower wattage rating (yes, this is not a mistake!).

why is this not a mistake? are you saying power decides the brightness of the lamps too?

is it cos the voltage in higher when there are more watts?

 

[Celluhh]

The only influence on the current would arrive from the different resistors you used, as you connected them in series. Since they have different values, different power in each, that could sway the whole thing either way.

i'm sorry, i don't really understand what you're saying. can u explain this more simply? thanks a lot!

 

[danielakkerma]

are you saying power decides the brightness of the lamps too?

.
Yes, that's exactly what happens.

can u explain this more simply? thanks a lot!

Sure, as you yourself had mentioned, you used,

followed by a resistor of one ohm, then bulb y, then a resistor of 2 ohm, then bulb z.

, meaning that the bulbs were interspersed by resistances of different values, that can affect their luminosity.
The power of any bulb, will be measured by: R*I^2 or V^2/R. The wattage directly depends on R, meaning, the smaller R is, the lower the wattage(Less heat produced for the same current). Since the power/intensity of each bulb, is its power, V^2/R, the smaller R is, the greater its brightness; But only should the resistances of the bulbs be different. There should be no effect if they're the same!
Daniel

 

[Celluhh]

.

The power of any bulb, will be measured by: R*I^2 or V^2/R. The wattage directly depends on R, meaning, the smaller R is, the lower the wattage(Less heat produced for the same current). Since the power/intensity of each bulb, is its power, V^2/R, the smaller R is, the greater its brightness;
Daniel

ok, so you mean the smaller the resistance, the lower the wattage, thus the brighter the bulb?

i think i get you now! so maybe the bulbs in the series circuit of my experiment were affected by the different resistors! cos my teach just said" maybe your setup is wrong". -.-/like that helps. Thanks so much!^^

 

[Celluhh]

oh gosh i just realized i have a huge problem. current flows from positive terminal of battery, electrons flow from negative terminal of battery, so exactly what causes the bulbs to light up? the charges(current right)? then what are the electrons for? are they actually the same thing? my teach said current flows from the battery, but ain't there already charges in the wires? or is that just a term?


ARGHHHHHH I'MMA TOTAL IDIOT.

 

[danielakkerma]

The current is defined as the local rate of change of the charges per unit time, or $\Large I = \frac{dq}{dt}$
By convention, The positive current was thought to originate from Positive charges, but of course, as the molecular theory of conductors, and indeed matter in general developped, it was quickly established that only electrons carry the current, in particular, those present in the so-called "conduction/valence" bands of the material.
These, start to "flow", or drift, towards the terminus, actively, when an external electric field is applied, such as with a potential difference...
The changing of the terminii, in such constructs as the light bulbs does nothing to change their luminosity, except for the sign of current due to the above convention(that is, current flows from plus to minues).
But, in Semiconductors, this is a tad more complicated. Diodes are one such characteristic beings, http://en.wikipedia.org/wiki/Diode" , that only allow a passage of current one way.

i think i get you now!

That's Great, I do hope I really did explain this thoroughly.

ARGHHHHHH I'MMA TOTAL IDIOT.

Never berate yourself.

 

[NascentOxygen]

In a series circuit, the current is constant throughout all resistors right?

In electronics, the word "constant" has a meaning different to how you are using it. A better word for what you mean is "identical", meaning that the current in each resistor is identical with that each other resistor in that series string in your arrangement.

 

[Celluhh]

The changing of the terminii, in such constructs as the light bulbs does nothing to change their luminosity, except for the sign of current due to the above convention(that is, current flows from plus to minues).

um i don't really get this line. what do you mean by the sign of current? why does it suddenly become negative? if i understand you correctly, the electrons are the one who carry the positive current right, and they flow from the negative terminal right?



oh by the way, my teacher mentioned that because the moment u close a circuit the current decides the path with the keast resistance which is the path they will take, as i mentioned in one of my earlier posts, but i have a new problem. if the current decides what path to take, then why are there charges in the circuit in the first place? won't the curren(charges) flow out from the battery only when the circuit is closed? or does it mean that the charges will simply go the reverse direction if they are in the branch with a resistor to the branch with no resistor?


thanks so much to both of you for your help!

 

[NascentOxygen]

oh by the way, my teacher mentioned that because the moment u close a circuit the current decides the path with the keast resistance which is the path they will take,

Your teacher is wrong. Current will flow in all paths; it's just that more current will flow in the path with least resistance.

why are there charges in the circuit in the first place?

Because we've deliberately built the circuit using elements/materials that have lots of available charges. The circuit comprises conductors in one form or another. All conductors have charges which are readily moved by the application of a voltage. Metals have oodles of electrons that can move and produce a current. Copper and gold are especially good for this. But in a pinch you could improvise a conductor out of a wet piece of string.

You could construct a circuit using insulating materials, but then it wouldn't do anything very spectacular, except NOT conduct electricity. That may or may not be handy. Often we really do want to stop electricity flowing from some point to another. It's useful, but not spectacular.

 

[Celluhh]

Your teacher is wrong. Current will flow in all paths; it's just that more current will flow in the path with least resistance.

heheh, i think that's what my teacher meant just that i kind of conveyed it wrongly. >.<
i'm kind of irritated though cos my teach didn't really teach us this topic, just made us do experiments. maybe he's trying to get us to do self directed learning.

hmm so the charges in the circuit are different from the charges in current? so do the charges in the circuit move along with the current as well?

thanks!

 

[NascentOxygen]

hmm so the charges in the circuit are different from the charges in current?

No. They are one and the same.

so do the charges in the circuit move along with the current as well?

The moving charges ARE the current. If they stop, then there is no current.

 

[Celluhh]

so, the current which are the charges in the circuit are already in the circuit so what happens when the circuit is closed? what happens to the charges in the branch with lower resistance and the charges in the branch with higher resistance? they can't change branches right?

 

[NascentOxygen]

When the battery pushes an electron into one end of the wire, that added electron pushes on an electron that is already in the copper wire, that electron pushes on one beside it, which pushes on one beside it, which pushes on one beside it, and so on, all the way around, until an electron at the end of your circuit is pushed into the other terminal of the battery. Each individual electron might move only a couple of mm per hour, we can't say exactly, and we don't really care. They are so tightly packed that as soon as one appears from the battery, one is pushed back into the battery's other terminal by the chain gang of electrons. You can pretend that each electron goes all the way around the circuit at breakneck speed, but that's dreaming.

So the battery pumps electrons out of one terminal, and draws in an exact equal number at the other! The "force" with which it pushes them is the voltage, the rate at which it pumps them out gives the size of the current. More electrons per second makes the current greater.

 

[Celluhh]

thank you so much nascentoxygen! you have just enlightened me!
however, if this is so, then why is the current less in the branches of a parallel circuit and why do we say that the current splits to go to different branches? why is the voltage of all the branches in a parallel circuit equal to the battery's voltage?

\thanks!

 

[NascentOxygen]

why is the current less in the branches of a parallel circuit

It isn't.

and why do we say that the current splits to go to different branches?

Some electrons go one way, some go the other. The battery voltage pushes, and they all move.

why is the voltage of all the branches in a parallel circuit equal to the battery's voltage?

Because that's how you have it connected. It wouldn't be a parallel circuit if you had it connected some other way.

 

[Celluhh]

It isn't.

Because that's how you have it connected. It wouldn't be a parallel circuit if you had it connected some other way.

it isn't?? but the textbook says that no matter how much current there is in a branch in a parallel circuit, it will be lesser than the current out from the source.\


oh so basically the theory is the same as that in a series circuit, just that the electrons go different ways when they come to a junction, is that it? less current goes to a path with more resistance because the current " decides "which path to go the moment the circuit is closed right?
um for the last point can u be more specific? i don't really get what you mean,...as in it is a parallel circuit cos we connected it that way but why is the voltage in each branch the same as that of the battery?

 

[NascentOxygen]

the textbook says that no matter how much current there is in a branch in a parallel circuit, it will be lesser than the current out from the source.\

Then the textbook is correct. The current drawn from the source = the current in one of the parallel branches + the current in the other parallel branch.

oh so basically the theory is the same as that in a series circuit, just that the electrons go different ways when they come to a junction, is that it? less current goes to a path with more resistance because the current " decides "which path to go the moment the circuit is closed right?

The voltage to the parallel branches is the same. It pushes more electrons through the branch of lower resistance because they move more easily. In the parallel branch of higher resistance, the voltage is able to push fewer electrons around. They resist it more.

um for the last point can u be more specific? i don't really get what you mean,...as in it is a parallel circuit cos we connected it that way but why is the voltage in each branch the same as that of the battery?

I pictured that you had the parallel branches connected directly to the battery. But there is no need to, of course. But to be in parallel, they must both experience the same voltage difference at the same time and place. If they don't, then they aren't in parallel--they must be in some other arrangement.

 

[Celluhh]

In the parallel branch of higher resistance, the voltage is able to push fewer electrons around. They resist it more.


I pictured that you had the parallel branches connected directly to the battery. But there is no need to, of course. But to be in parallel, they must both experience the same voltage difference at the same time and place. If they don't, then they aren't in parallel--they must be in some other arrangement.

u understand what u mean by the second statement, but why do the electrons resist it more?

why do you say that to be in parallel, they must both experience the same voltage diff at the same time and place?
please forgive me if i am lacking in my knowledge.>.<

 

[NascentOxygen]

u understand what u mean by the second statement, but why do the electrons resist it more?

The electrons don't resist, the material itself resists parting with its electrons. They are less free to move. That's the definition of having higher resistance.

why do you say that to be in parallel, they must both experience the same voltage diff at the same time and place?

Because that's the definition of parallel circuits. Can you devise an arrangement where two parallel branches DON'T connect to the same voltage points? -No, you can't.

 

[Celluhh]

The electrons don't resist, the material itself resists parting with its electrons. They are less free to move. That's the definition of having higher resistance.


Because that's the definition of parallel circuits. Can you devise an arrangement where two parallel branches DON'T connect to the same voltage points? -No, you can't.

oh yeah ur right! but what about voltage drops after every resistor? i know one of the voltage laws is that the total voltage drop in a circuit is zero, but i don't actually understand why.>.<

 

[NascentOxygen]

i know one of the voltage laws is that the total voltage drop in a circuit is zero, but i don't actually understand why.

That is an unnecessarily erudite way of looking at things, unless you are involved in electronics analysis. I don't think you need bother yourself with it.

A plainer way of expressing the same thing is: if you connect a voltage source to a series circuit, then the sum of the voltages across all the series elements must exactly equal that of the voltage source. (Not all that astonishing, really. )

 

[Celluhh]

A plainer way of expressing the same thing is: if you connect a voltage source to a series circuit, then the sum of the voltages across all the series elements must exactly equal that of the voltage source. (Not all that astonishing, really. )

which brings me to my next question, why is that so? how did they arrive at this theory?

i'm really sorry for bothering you with my questions, thanks so much for answering!

 

[NascentOxygen]

Well, if you added up all the voltages across the series branch of resistors, and found that it was LESS than the voltage of the source, where could that missing voltage have gone?

Fortunately, no one has ever had to solve a mystery of such missing voltage, because it has never happened. And never will.

 

[Celluhh]

um i know what ur trying to say but why couldn't some of it be used up by the resistor? there are voltage drops...i mean there is potential difference and this is due to different levels of electrical energy across a component which is actually the elctrical energy used up bby the component which is also tha voltage of the component right... (this is what my tuition notes define voltage as)