Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery? a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

Options B & E are correct

Explanation:

Looking at all the options, B & E are the correct ones.

Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.

Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.

Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.

Option C is not correct because although the Electric field along the wire is not zero, it is very small.

Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.

the answer is to check all of them! they all apply well at least i think so!

that would be because of both gravity and friction. so     c