Question

Build a second circuit with a battery and a light bulb but this time add a switch. Your circuit might look something like the one at right. When a switch is open in a circuit, it means the two ends are disconnected and current cannot flow between them. When a switch is closed in a circuit, it means the two ends are connected and current can flow between them. Play with the switch to check how it affects the flow of current. With the switch closed, compare the brightness of the bulb and the flow of current in this circuit, with that of your first circuit. Did increasing the length of wire in the circuit change the brightness of the bulb or the curr

Answer 1

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

            $\frac{R_{bulb} }{R_{wire} } = \frac{240 }{ 0.0169}$

              $\frac{ R_{bulb} }{ R_{wire} } = 1.4 \ 10^4$

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.