Elementary Electricity

[Mohit Pandey]

I have just been introduced to the fascinating world of electricity. :)

I have read about the series and parallel combination of resistors. .

I know that current remains the same in a series combination and P.D. remains the same in a parallel combination.

 

To understand the concept properly I have used the analogy of flow of water in a pipe and the resistors are obstructing the flow of water. I have attached the pictures of the two combinations which is being formed in my mind while talking about it. The pipes symbolise conductors.

 

In series, the resistors allow same amount to flow but the pressure of water[P.D] is not same. Imagine we are watering the plants in garden using a pipe. When we press it, the pressure increases[ more the pressing , more the pressure] but the amount of water flowing remains the same.

Now in electricity language, higher the value of resistors , more is the P.D using Ohm's law[ even when the current is same.]

 

In parallel, there are two paths for water and therefore the water pressure[P.D] is same in both.

Till now I think it is correct. But now the problem arises.

Because the resistors again obstruct the flow of water and therefore should again increase the pressure of water[P.D].

But no, P.D. is same in parallel combination of resistors. I am also unable to link current in this.

 

Can anyone help me?:cheer:

[Mohit Pandey]

One more thing.:cheer:

Work=VQ[where V=P.d and Q= charge]

Work=Fs[ where F is force applied and s is displacement]

 

From two equations,

VQ=Fs

 

Now whom is equal to whom?

[Mohit Pandey]

44 views!:)

But no replies!;)

 

Come On! Come out of your shell of laziness.:mad:

[Jay-qu]

I would avoid taking the analogy of water for electricity to far, at some point they are going to act fundamentally different so you may as well get used to thinking of electricity as the flow of electrons and not water.

 

The PD or voltage drop across a parallel part of circuit is always constant, if you have different resistances on different parts of the parallel circuit then the current will change to maintain the same PD where V=IR - Ohm's Law

 

You may wish to read this tutorial ;)

[Mohit Pandey]

I would avoid taking the analogy of water for electricity to far, at some point they are going to act fundamentally different so you may as well get used to thinking of electricity as the flow of electrons and not water.

 

Thanks Jay-qu for your reply.:hihi:

I used the analogy of water to account for the same current in series and same P.D in parallel. Is there any reason behind this? Or other expanation?

 

You may wish to read this tutorial :)

 

Good one:). Expand it with magnetism.

By the way, is your name Jayden Newstead?

[Jay-qu]

Yeah thats me I had planned on expanding the site, but had no time with uni - now that Im on summer holidays I think I will do some writting.

 

Series is one after the other - so that the current that passes through one componant must pass through the next since it has nowhere else to go.

 

To understand the constant PD thing it would be prudent to understand what PD means. Voltage is measured in Joules per coloumb (charge) so it is literally the amount of energy the electrons have. Potential difference is just the relative change in the electrons energy. When considering parallel cases the current must split - but when it rejoins it must have undergone the same voltage drop.

[Mohit Pandey]

Yeah thats me :) .

Hello Jayden ! May I ask you what 'qu' stand for in your name in Hypography?

 

Series is one after the other - so that the current that passes through one componant must pass through the next since it has nowhere else to go. .

That means coulombs of charge flowing per second would remain same(I=Q/t) but the electrons would have less energy. Am I correct?

 

When considering parallel cases the current must split - but when it rejoins it must have undergone the same voltage drop

.

I am unable to understand this. Please elaborate it. Is there any difference between P.D and voltage drop?:)

 

Also see my one more question in the second post.

[Qfwfq]

Jayden hasn't been doing too badly, but I suppose now he's either at the disco with some absolutely charming young woman or getting ready for bed. :) I suppose it's only a bit less late in India but you're a more tenacious student than him!

 

Is there any difference between P.D and voltage drop?

They're the same thing. The volt is a unit of electric potential; drop means decrease, difference. Just two different expressions. :)

 

From two equations,

VQ=Fs

 

Now whom is equal to whom?

Now electric potential is defined as potential energy per unit of charge and of course the difference in potential is just the work the field does on the charge going from one point to the other. At the same time, F is given by the electric field E times the charge (IOW the field is the force per unit of charge) so you could well write:

 

VQ = Fs = QEs

 

so you have: V = Es and of course by V we mean the difference between the two points.

 

One important thing to keep in mind: Talking about the P. D. between two points, as I defined it above, makes sense if the work does not depend on the path taken between the same two points. Of course you are really summing a whole bunch of Fs products, each one for a little stretch s along the path. This is why the voltage between the two resistors in parallel is the same. You'll probably learn later on that this is true when there isn't a changing magnetic field through the loop.

[Jay-qu]

Thanks for subbing for me Q I did have a night down at the pub after exams but now my computer has a bad virus and Im struggling to get to the internet as often as I would like..

 

I hope that answers your questions Mohit

[Mohit Pandey]

I suppose it's only a bit less late in India but you're a more tenacious student than him!

Hello Qfwfg! Thanks for your compliments. There is need to be tenacious on Hypo.

F is given by the electric field E times the charge (IOW the field is the force per unit of charge)

I couldn't understand the bolded words.

 

 

Originally Posted by Jay-qu When considering parallel cases the current must split - but when it rejoins it must have undergone the same voltage drop

Hello Jayden! I hope your computer is now OK.

You haven't elaborated this statement.

[Qfwfq]

I couldn't understand the bolded words.

They mean that the force is directly proportional to the charge.

 

[math]\vec{F}=q\vec{E}[/math]

 

So the field E tells you how much force there will be on a unit of charge; on two units there will be twice the force etc...

 

Hello Jayden! I hope your computer is now OK.

You haven't elaborated this statement.

But I did, in the last paragraph of my post, "One important thing to keep in mind". If the work wasn't the same, going through one resistor or the other between the same two points, then it wouldn't make sense to say there's a potential at each point and talk about the potential difference between them.

[Mohit Pandey]

Sorry, Qfwfg ! But I am unable to understand you.:)

First, I used the analogy of water to account for the same current in series and same P.D in parallel. But then Jayden told me that it is not possible. Then I waited for someone to come forward to make some some changes in my analogy and make it a success. But sadly, no one came forward.

 

For series, he said-

Series is one after the other - so that the current that passes through one componant must pass through the next since it has nowhere else to go.

OK. I understood this.

Now, what you say seems to be different from Jayden's. Jay said-

Originally Posted by Jay-qu When considering parallel cases the current must split - but when it rejoins it must have undergone the same voltage drop

What here undergone the same voltage drop means?

Can you explain it using a numerical problem?

For example, see the attached picture. Let the total voltage going through the circuit be 150 V. First resistor is of 10 ohms and second that of 15 ohms . Then the current flowing through first resistor will be 15 A and that through second resistor will be 10 A. Then it rejoins . The combined current flowing now 25 A. In this context, explain undergone the same voltage drop

Qfwfg-F is given by the electric field E times the charge (IOW the field is the force per unit of charge)

What does IOW means?

Cheers!

[Qfwfq]

I used the analogy of water to account for the same current in series and same P.D in parallel. But then Jayden told me that it is not possible.

He only said that analogies need a bit of caution. The analogy you made is not really a bad one but it does need a bit of care, to work it properly would require considering Bernoulli's theorem but I don't know if it's worth it for you to do so.

 

What here undergone the same voltage drop means?

It's very simple. The voltage drop going from A to B is the difference between the potential at A and that at B; this obviously is the same for the two paths from A to B. It only depends on the pair of points, not on the path between them.

 

IOW means In Other Words.

[Mohit Pandey]

Thank you.

[Mohit Pandey]

QUESTION

Welcome back!:naughty:

Magnetic Fields and Currents - Picture - MSN Encarta

At the centre of the circular loop, the arcs of the concentric circles representing magnetic field lines appear as straight line.

That means the field is uniform, doesn't it? Then what special would happen to a particle that is placed there? Would it go in the direction of the straight line at the centre would point for infinite distance?

[Turtle]

Sorry, Qfwfg ! But I am unable to understand you.

First, I used the analogy of water to account for the same current in series and same P.D in parallel. But then Jayden told me that it is not possible. Then I waited for someone to come forward to make some some changes in my analogy and make it a success. But sadly, no one came forward. ...

He only said that analogies need a bit of caution. The analogy you made is not really a bad one but it does need a bit of care, to work it properly would require considering Bernoulli's theorem but I don't know if it's worth it for you to do so.

 

I think what was not clearly stated by anyone regarding the analogy of water flow in a pipe to electric current flow in a conductor, is the consideration of the pipe's cross section area. Wouldn't it be the case that the voltage is analogous to pressure, and the amperage analogous to the volume of water in a pipe by virtue of its cross-sectional area?

 

On another note, I have read somewhere, and can't find it just now, that a system of closed water pipes with flexible membranes on the ends, can be used to transmit power in the same way as AC electrical current. A motive force is applied to one membrane to make it vibrate, and the water transfers the vibration to any and all other membranes where the motive force can be extracted to do work. :cup: :cup:

[Mohit Pandey]

Thank you Turtle!:thumbs_up

I think now I can link current with parallel analogy. At the point of seapartion of two pipes from a single pipe, the cross-section would change- I am slightly confused it would be a increase or decrease. Consequently, the amount of water flowing or current flowing would also change.

Clear up to this. But here again the problem starts.:rolleyes:

When we press the pipe acting as reistors, the water pressure or P.D would increase. But P.D remains same in a parallel connection.

Again , help! help!