Lamps and parallel circuits...

[Antoneo]

I just need some help with understanding why, in a parallel circuit with two lamps, one lamp becomes brighter when you remove the other. Shouldn't the brightness theoretically remain the same?

Sorry, I guess I should have mentioned this in the beginning but this was done/observed with a circuit on a breadboard not in the house.
Here is a diagram of the circuit.

 

[rival]

if i recal right, it will stay the same, receptacles are in parallel, when you turn something off thats on the same circuit everything else doesnt draw more, each lamp should draw whatever it needs to run properly

 

[Pikachu]

The brightness would theoretically remain the same if your power supply and conductors had adequate capacity.

 

[mastertech01]

If the load appled by both bulbs exceeds the capacity of the power source then both would glow dimmer than they are rated, and once one is removed from the circuit adequate power would be applied to the remaining bulb to allow it glow brighter. A poor ground in a DC circuit can cause a similar problem.

 

[Cyberian]

<< Shouldn't the brightness theoretically remain the same? >>


Yes. If it does not, I would strongly suggest that you have your wiring checked.

 

[woodie1]

Either the wires are too small for the load or the source isn't powerful enough.

 

[DanFungus]

woohoo, physics class at work! uh, yeah, my guess is that the power source =

 

[Antoneo]

^ and if it the link doesn't seem to work, cut and paste or drag the page onto the address bar.

 

[Harvey]

The lamp brightness changes because of the 12 ohm resistor in series with the two lamps (and their respective series resistors).

Assume both lamps are the same, and call the impedance of each lamp R-L.

With both lamps in, the current through the circuit is determined by the parallel value of (R-L + 100 ohms) and (R-L + 1000 ohms) plus 12 ohms. When you remove either lamp, the current will drop, but the 12 ohm resistor is still there, and the drop across it changes proportionally with the current. Therefore, the voltage drop across the lamp also changes.

If you built the circuit, you can determine the impedance of the lamp by measuring the voltage drop across it.