Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

DC Electricity Concept Question

Status
Not open for further replies.

tsa256

New Member
Hello Everyone, great forum lot of information!!

I been a long time electrical person but as I've been reading more and more into the math behind electricity and current I become stumped every once in a while.

Recently I became stuck on a scenario such as the one in the attached picture. As you can see there is a battery and two wires that have a decreasing radius. My question is simple, would the potential difference between node A and node B be greater than the 1 volt of difference in the battery. The analogy of water in a pipe tells me yes but I am not sure how to tackle this problem; specifically the contentiously increasing resistance mathematically using ohms laws or max wells equations. Any help would be greatly appreciated.
 

Attachments

  • Untitled.jpeg
    Untitled.jpeg
    30.3 KB · Views: 165
Beware of using water analogies to try to understand electricity, the equivalence is limited and in many cases just plain wrong.

In your situation there is one source of EMF in the circuit, the 1 volt battery.
So there can never be more than 1 volt in that circuit.

You have wires with varying resistance, in the circuit as drawn this does not matter, there is no load and hence no current flowing, so there can be no volt drop in the wires.
The potential difference a points A and B is exactly the same as it is at the battery terminals ie 1 volt.

However if there were to be a load at points A and B such that a current flowed between them, then the potential difference between A and B would be less than that at the battery terminals.
What that potential difference would be we cannot calculate because we do not have enough information. We would need to know the resistance of the wires and the resistance of the load.

JimB
 
One aspect of the water analogy does work here:

If the ends of the pipes (circuit) are not connected (open), there is no water flow (zero current flow), the Water Pressure (Voltage) is the same along each pipe (wire). It doesn't matter that the pipe (wire) gets skinnier (more resistive).
 
Last edited:
Beware of using water analogies to try to understand electricity, the equivalence is limited and in many cases just plain wrong.
I've noticed that a number of people really seem to dislike using a water analogy for electricity, but I think it's quite useful in demonstrating the basics of electricity to a novice, including voltage, current flow, resistance, inductance, capacitance, and requiring a complete circuit for the current to continue to flow.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top