Electric Resistance

1. Use power and voltage to determine resistance.

   R =	V2	⇐	P =	V2
   	R			P

   Numbers in. Answer out.

   R =  V2 P

   R =  (120 V)2 60 W

   R = 240 Ω

2. Light bulb filaments, like most wires, are basically long thin cylinders. Their cross sections are circles. Use the equation for the area of a circle to get the cross sectional area of the filament.

   A = πr²

   This equation has an r in it — r for radius. The problem gives us diameter, because it's much easier to measure than radius. Divide the diameter in half before using this equation and pay attention to the units. The unit μm (a micrometer) is one millionth of a meter.

   A = π(23 × 10−6 m)2 A = 1.66 × 10−9 m2

3. There's an equation that relates some of the properties of a wire to its resistance.

   R =	ρℓ
   	A

   Solve it for resistivity (ρ, "rho").

   ρ =	RA
   	ℓ

   Watch the units again. The length is given in centimeters, but the SI unit of choice is the meter.

   ρ =	(240 Ω)(1.66 × 10−9 m2)
   	(0.533 m)
   ρ = 7.48 × 10−7 Ω m

4. The resistivity calculated above is considerably larger than the value quoted in standard reference tables.

   7.48 × 10−7 Ω m	≈	13 times larger than this source says
   5.60 × 10−8 Ω m
   7.48 × 10−7 Ω m	≈	14 times larger than this source says
   5.28 × 10−8 Ω m

   That's because most references tables report the resistivity for room temperature (typically 20 °C or 300 K). An operating light bulb is going to have a temperature closer to 3,500 K. That's about 12 times hotter than room temperature.

   3,500 K	≈ 12 times hotter
   300 K

   Resistance goes up when temperature goes up. For many materials, they're basically directly proportional, which is what these numbers show.

5. How can a 53.3 cm filament fit into a light bulb that is only a few centimeters wide? It's coiled and then coiled again, kind of like DNA. It's a coiled coil.

Answer it.

Answer it.

Answer it.