Tuesday, October 25, 2011

Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?

More than one answer is possible:

a) very little energy is dissipated in the thick connecting wires

b) the electric field in connecting wires is very small, so emf is about equal to Ebulb * Lbulb

c) electric field in the connecting wires is zero, so emf = Ebulb*Lbulb

d) current in the connecting wires is smaller than current in the bulb

e) all the current is used up in the bulb, so the connecting wires don't matter

Thank you!Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?a

of the answers given. You would have to add a lot of wire to see it dimmer Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?a, %26amp; b.

'c' is close but it can't really be zero.Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?None of the above?

Unless the wires are supeconducting, there will be a change in brightness. It may just be insignificant.

A goes closest.Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?The best answer is probably %26quot;a%26quot;.



Generally when designing a lighting circuit, you use a thick enough wire so the resistance of the wire is very small compared to the light bulb. The resistance is finite, however, and given the diameter and length of the wires, and the current in the bulb, you can calculate the voltage drop in the wires, but most of the voltage applied appears across the bulb. With small wire resistance (Rw), and since the power loss in the wires is I^2*Rw, very little energy would be dissipated in the wires.



%26quot;b%26quot; could also be considered correct, for the same reason. With a small voltage drop, the E field would be the voltage drop divided by the length of the wire, which again would be small.



%26quot;c%26quot; is troublesome, since the E field along the wire is very small, but actually non-zero.



%26quot;d%26quot; is incorrect since the wires and the light bulb are in series, the current is identical in both the wires and the light bulb. This wouldn't be necessarily so if the circuit was driven by an AC source, and the length of the wires was not kept short compared to the wavelength of the driving source. Some radiation of energy (just like an antenna) could make the current in the wires different than in the light bulb, but usually this wouldn't be done on purpose.



%26quot;e%26quot; is incorrect since the current isn't %26quot;used up%26quot;, the charge is conserved, it just circulates through the circuit. It is %26quot;power%26quot; or %26quot;energy%26quot; that is used up. Again... the wires and bulb are in series, so the current should be equal in both wires and bulb.Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?a and b are correct assumptions.Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?a -



The thicker the wire per a given length, the less resistance, thus the less energy dissipated by the wire and more dissipated by the bulb.



b -



the resistance of the wires is low enough to not cause a sufficient drop in voltage across the wires, so there is not a noticeable difference in voltage across the bulb.Why does the brightness of a bulb not change when you use longer copper wires to connect it to the battery?I would say %26quot;A%26quot; is true until the wires used are very skinny and very long.



The other answers are gibberish.