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?

[Mohit Pandey]

One more thing.
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.

[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]

Originally Posted by 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.

Thanks Jay-qu for your reply.
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?

Originally Posted by Jay-qu

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]

Originally Posted by Jay-qu

Yeah thats me .

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

Originally Posted by Jay-qu

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?

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
.

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!

Originally Posted by Mohit Pandey

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.

Originally Posted by Mohit Pandey

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.