Question

1. The current in a 100 watt lightbulb is 0.760 A . The filament inside the bulb is 0.170 mm in diameter.

a. What is the current density in the filament?

Express your answer to three significant figures and include the appropriate units.

b. What is the electron current in the filament?

Express your answer using three significant figures.

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2. Wires 1 and 2 are made of the same metal. Wire 2 has twice the length and twice the diameter of wire 1. What are the ratios (a) ρ2/ρ1of the resistivities and (b) R2/R1of the resistances of the two wires?

Answer 1

1

(a)

formula for current density

J = I/A

= 0.760/ pi d^2/4

= 0.760/ pi ( 0.170* 10^-3)^2/4

= 33483116 A/m^2

rounding to three significant J = 33500000 A/m^2

(b)

formula for electron current

i_e = I/e = 0.760/1.6* 10^-19= 4.75 * 10^18 A/C

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2)

formula for resistance is

R = rho L/A= rho L/ pi d^2/4

R1 = rho 1 L1/ pi d1^2/4

R2 = rho 2 L2 / pi d2^2/4

since L2 = 2L1 , d2 = 2 d1

R2 = rho 2 ( 2L1) / pi ( 2 d1)^2/4

R2/R1 = rho 2 ( 2L1) / pi ( 2 d1)^2/4/ rho 1 L1/ pi d1^2/4

= rho 2 ( 2L1)/ pi d1^2/ rho 1 L1/ pi d1^2/4

= rho 2 ( 2 )/ rho 1 * 4

R2/R1 =0.5

(a)

rho2/rho1 = 1

because they are made with same material