# Electric Resistance

## Practice

### practice problem 1

- What is the light bulb's operating resistance?
- Determine the cross sectional area of the filament.
- Calculate the resistivity of tungsten using the results of part a. and b.
- How does the resistivity calculated above compare to the value quoted in standard reference tables? Why are these two values so different?
- How can a 53.3 cm filament fit into a light bulb that is only a few centimeters wide?

#### solution

Use power and voltage to determine resistance.

*R*=*V*^{2}⇐ *P*=*V*^{2}*R**P*Numbers in. Answer out.

*R*=*V*^{2}*P**R*=(120 V) ^{2}60 W *R*= 240 ΩLight 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*= π*r*^{2}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}m^{2}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}m^{2})(0.533 m) ρ = 7.48 × 10 ^{−7}ΩmThe resistivity calculated above is considerably larger than the value quoted in standard reference tables.

7.48 × 10 ^{−7}Ωm≈ 13 times larger than this source says 5.60 × 10 ^{−8}Ωm7.48 × 10 ^{−7}Ωm≈ 14 times larger than this source says 5.28 × 10 ^{−8}ΩmThat'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.

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.

### practice problem 2

#### solution

Answer it.

### practice problem 3

#### solution

Answer it.

### practice problem 4

#### solution

Answer it.