Volts and Amps. Simple question for someone who knows this stuff

[jrichrds]

What happens when you increase the Amperage of a 4.8v light bulb from 0.5A to 0.75A? (actually, replacing the light bulb with another)

 

[Chaotic42]

The power consumption goes from 2.4W to 3.6W?

 

[ScottMac]

It might get brighter - it depends on the resistance of the filiment.

The total current draw (max) is determined by dividing the resistance into the voltage (I=E/R). That would give you a maximum draw if that much current is available. Anything above that number would be unused.

(Ohm's Law: E=IR, I=E/R, R=E/I - E=Volts, I=Current(amps), R=resistance (ohms))

FWIW

Scott

 

[FoBoT]

the only way to make the current go up is to either lower the resistance or raise the voltage (see ohm's law above)

if you want to calculate power its P=IV , Watts(Power)= Amps(current)*Voltage(V)

of course this is all for a DC (direct current) system, if you are dealing with AC (alternating current), you have to include the Power Factor and it gets a bit more complicated depending on how inductive (like motors) or capacitive (like electronic circuits) or purely resistive (like a heater) the circuit is

what was the question?

 

[jrichrds]

haha, okay. It's a flashlight bulb powered by 4 AA batteries. Original was 4.8v 0.5A. Closest replacement I could find was one labeled "4D batteries / 1-6V" on the outside package, but the krypton light bulb itself was labeled 4.8v 0.75A.

No big deal, just curious what the difference is on a technical level.
So it's just not as efficient and drains the battery faster, right?

 

[NutBucket]

Yep, it'll drain the batteries faster. Hence, it spec'd D-cells instead of AA's.