Lightbulb and mini hydro generator

[samirgaliz]

I am using a 1.5 V incandescent light bulb. I know that if I connect a 9V battery to it or any higher voltage, it will burn it due to the large current flowing through it (Ohm's law if V goes up then the current goes up as well assuming constant R).

Now I a have a 12 V mini hydro generator showing 7.0 V when water runs through it.
When I connect the generator directly to the light bulb, I would expect that the light bulb will burn. But it doesn't!
I know it has to do with the power output of the generator at that voltage which might not be enough to cause damage.
But looking at it from Ohm's law, that 7.0 voltage should cause a large current through the light bulb and damage it.

Any clarification would be appreciated.
Thanks

 

[anorlunda]

The voltage changes when you put a load on the generator.

Ohm's law is not very useful for describing a generator-bulb system. Open circuit (i.e. no load) voltage of a generator is also not useful.

 

[davenn]

...

Now I a have a 12 V mini hydro generator showing 7.0 V when water runs through it.
When I connect the generator directly to the light bulb, I would expect that the light bulb will burn. But it doesn't!
I know it has to do with the power output of the generator at that voltage which might not be enough to cause damage.
But looking at it from Ohm's law, that 7.0 voltage should cause a large current through the light bulb and damage it.

Any clarification would be appreciated.
Thanks

As anorlunda said the voltage will drop under load

if you measure the generator output with the bulb attached, I would bet you will finds it a very low voltage
what this is telling you is that the generator cannot supply the current required by the bulb to light upDave

 

[samirgaliz]

Yes the voltage changes

The voltage changes when you put a load on the generator.

Ohm's law is not very useful for describing a generator-bulb system. Open circuit (i.e. no load) voltage of a generator is also not useful.

As anorlunda said the voltage will drop under load

if you measure the generator output with the bulb attached, I would bet you will finds it a very low voltage
what this is telling you is that the generator cannot supply the current required by the bulb to light upDave

Thanks to both of you. That makes sense.
But why would there be a drop in potential? Is there an internal resistance within the generator itself so when a current is driven through the load the potential drops? I am assuming it is the resistance of the coil inside the generator. Correct?

 

[berkeman]

Is there an internal resistance within the generator itself so when a current is driven through the load the potential drops?

Your intuition is good, young Skywalker...

(Google "internal series resistance of voltage source")

 

[berkeman]

https://www.electronicshub.org/wp-c...tance-connected-in-series-with-the-source.jpg

 

[CWatters]

Yes it will have some internal resistance but its also possible the generator slows down on load?

With no load the generator is easier to turn so it spins faster producing a higher voltage. On load it's harder to turn so it goes slower and the voltage falls.

This will have the effect of increasing the effective internal resistance.

 

[jartsa]

But why would there be a drop in potential? Is there an internal resistance within the generator itself so when a current is driven through the load the potential drops? I am assuming it is the resistance of the coil inside the generator. Correct?

There is a permanent magnet in the generator, right? And a coil. So I would say that the surface current on the magnet induces an opposite surface current on the coil, when the coil gets close to the magnet.

https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

Do I need to explain more?