Elementary Electricity

[Qfwfq]

Originally Posted by Mohit Pandey

I couldn't understand the bolded words.

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



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...

Originally Posted by Mohit Pandey

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]

Originally Posted by Mohit Pandey

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.

Originally Posted by Mohit Pandey

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!
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]

Originally Posted by Qfwfq

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.

[Mohit Pandey]

Thank you Turtle!
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.
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!

[Turtle]

Originally Posted by Mohit Pandey

Thank you Turtle!
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.
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!

This is where the analogy starts breaking down, because electricity is not a fluid. I think in a pipe with water the resistance is found in the roughness of the pipe interior, and the rougher a pipe interior, the higher the resistance and there is actually a drop in pressure.

In Ohm's law, I = V/R, if the voltage V remains constant, the current/amperage I drops with an increase in resistance R.

I better stop there before I blow a fitting. Hope I didn't make things clear as mud.

[Mohit Pandey]

Originally Posted by Turtle

This is where the analogy starts breaking down, because electricity is not a fluid. I think in a pipe with water the resistance is found in the roughness of the pipe interior, and the rougher a pipe interior, the higher the resistance and there is actually a drop in pressure.

No, the pressure increases. That is why in kidney, the water is filtered by increasing pressure. This is done by transporting blood from a thick pipe to a thin pipe. It was told by my teacher . However,I could not found it on Google and therefore, I can't give you the link.

Originally Posted by Turtle

In Ohm's law, I = V/R, if the voltage V remains constant, the current/amperage I drops with an increase in resistance R.

I better stop there before I blow a fitting. Hope I didn't make things clear as mud.

I could not understand the bold words as Ampere is the unit of current.
I think you need to blow not a fitting, but many fittings.

[Mohit Pandey]

Even though my second question has not answered yet, I asking the third.
In Indian domestic electric circuit, there are three types of wires- earth, live and nuetral.
In India, the potential difference between the nuetral and live wire is 220 V. However, P.D is needed for the current to flow in an electric circuit. And here, isn't the wires different and therefore, shouldn't the P.D be between the beginning and the end of ciruit?
Electrical wiring (UK) - Wikipedia, the free encyclopedia