Answer to Question #191186 in Electric Circuits for Tessa

Question

Household wiring often contains 12-gauge copper wires, having a diameter of approx. 2 mm.

Consider a circuit of such a wire connecting a 60 W light-bulb with a standard 230 V household

outlet and a light switch 5 m away.

• What is the current drawn by the light bulb if you flick the switch?

• After what time t does the first electron from the voltage source reach the light bulb?

Assume the density of charge carriers to be n = 8 · 1028 m-3, and their charge to be

1e = 1.6·10-19 C. Does that time make sense, considering that the light bulb immediately

begins to shine, when flicking the switch?

• If i would want to collect 1 g of electrons from that wire in 1 hour, what current would

need to flow? (i.e. for 1 g of electrons passing the cross-section of the wire in 1 hour) The

mass of one electron is given as me = 9.1 · 10-31 kg.

• How big is the electric field in the copper wire, if we assume the wire to have a resistivity

ρ = 1.68 · 10-8 Ωm?

Answer 1

Given, $P=60W, V=230volts,d=2mm,l=5m$

(a) Current $I=\dfrac{P}{V}=\dfrac{60}{230}=0.269A$

(b) time $t=\dfrac{\text{ Total charge on electron}}{\text{ total current}}$

$=\dfrac{nq}{I}=\dfrac{8.1028\times 1.6 \times 10^{-19}}{0.269}=48.19\times 10^{-19} s$

(c) Using faraday first law of electrolysis-

$W=ZIt$

$\Rightarrow 1= \dfrac{i\times 60\times 24\times 24}{96500} \\[9pt] \Rightarrow i=2.792A$

(d) Resistivity, $\Rho=1.68\times 10^{-8} \Omega m$

As we know,

conductivity $\sigma=\dfrac{1}{\Rho}=\dfrac{1}{1.68\times 10^{-8}}=5.95\times 10^7 s m^{-1}$

Current density

$J=\dfrac{i}{A}=\dfrac{i}{\pi r^2}=\dfrac{4i}{\pi d^2}=\dfrac{4\times 0.269}{3.14\times (2\times 10^{-3})^2}=\dfrac{1.076}{12.56\times 10^{-6}}=8.56\times 10^4 A/m^2$

Also, $J=\sigma\times E$

$\Rightarrow E=\dfrac{J}{\sigma}=\dfrac{8.56\times 10^4}{5.95\times 10^7}=1.43\times 10^{-3} V/m$