Answer:
the resistance of the wire has no effect on the brightness of the bulb.
Explanation:
Let's apply ohm's law for your light bulb circuit plus wires plus switch
V = I R_{bulb} + I R_ {wire}
the current in a series circuit is constant
V = I (R_{bulb} + R_{wire})
To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.
Incandescent bulb
Power 60 W
let's use the power ratio
P = V I = V2 / R
R = V2 / P
the voltage value for this power is V = 120 V
R = 120 2/60
R_bulb = 240 Ω
Resistance of a 14 gauge copper wire (most used), we look for it on the internet
R = 8.45 Ω/ km
in a laboratory circuit approximately 2 m is used, so the resistance of our cable is
R = 8.45 10⁻³ 2
R_wire = 0.0169 Ω
let's buy the two resistors
R_{bulb} = 240
R_{wire} = 0.0169
therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.
In summary, the resistance of the wire has no effect on the brightness of the bulb.
