According to my rudimentary understanding, I feel that a 1000 ohm resistor will draw lesser current and hence drain the battery more slowly than a 100 ohm resistor. After all a 100 ohm resistor will provide a low resistance path and allow all the current to flow through it quickly.
But this doesn't make sense practically. It's like if a connect a 12V LED strip to a 12 volt battery, it will drain all its charge faster than connecting a bigger load such as a 12 volt table fan to it. So it looks as though it's harder to drive a lower load (as the battery would drain quickly). Very confusing indeed!

 

You are confused because you have not understood mathematically the meaning of load.

What is a light load vs a heavy load?
Load is a measure of conductance, not resistance.

If you define load as equal to 1/R
then a 100Ω resistor is a heavier load than a 1000Ω resistor.

btw, the unit of conductance is mho.

 

You are confused because you have not understood mathematically the meaning of load.

What is a light load vs a heavy load?
Load is a measure of conductivity, not resistance.

If you define load as equal to 1/R
then a 100Ω resistor is a heavier load than a 1000Ω resistor.

Oh! I am really surprised. To me a 'heavy' load is something that uses more power - like an immersion coil. A 'light' load would be a cell phone. And in this case an immersion coil offers much more resistance than a cell phone.

And anyway, a 100 ohm load drains the battery faster than a 1000 ohm load, right?

 

Sometimes you get it right and sometimes you get it wrong.

And anyway, a 100 ohm load drains the battery faster than a 1000 ohm load, right?

That is correct. So which is the heavier load? Which one consumes more power?

Oh! I am really surprised. To me a 'heavy' load is something that uses more power - like an immersion coil. A 'light' load would be a cell phone. And in this case an immersion coil offers much more resistance than a cell phone.

Now work through this one again.

 

So it looks as though it's harder to drive a lower load (as the battery would drain quickly). Very confusing indeed!

So, language is confusing you.
Avoid words such as "harder to drive" and "lower load".

Focus on resistance, current and power.
I = V/R
P = IV = V * V/R

A lower resistance will draw more current and dissipate more power than a higher resistance given the same applied voltage.

 

So, language is confusing you.
Avoid words such as "harder to drive" and "lower load".

Focus on resistance, current and power.
I = V/R
P = IV = V * V/R

A lower resistance will draw more current and dissipate more power than a higher resistance given the same applied voltage.

Thank you very much. I got the point now

 

I can see where the confusion arises.

A higher "resistance" is harder to push against.

Electrical resistance resists the flow of current. You would need a higher voltage in order to push the same amount of current through a resistor of greater resistance.

However, as stated in previous posts, a higher resistance consumes less electrical power for the same applied voltage and hence is a lighter load.

This is meant for other readers who might be having the same problems.