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.**

_____________

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) *ρ*2/*ρ*1of the resistivities and (b)
*R*2/*R*1of 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

------------------------------------------------------------------------------------------------------------------------------

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

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