## Physics A: Problem Set 8: Electric Power

### recommended reading

Barron's Let's Review: | 9.6 Power and Energy in Electric Circuits |

physics.info: | Electric Power |

Wikipedia: | Electric power |

HyperPhysics: | DC Electric Power |

### classwork

- Fermi National Accelerator Laboratory is the premiere high energy physics laboratory in the United States. Typical experiments at Fermilab involve producing beams of subatomic particles, accelerating them to high velocities, and them smashing them into stationary targets or other particle beams. Read the following passage from their newsletter,.
[Physicist Sergei] Nagaitsev's group has created a record-breaking electron beam with a continuous current of 500 milliamps at an energy of 3.5 MeV. To the layperson, these numbers may not seem significant. After all, half an amp is the current flowing through the wire of a typical light bulb. However, the electrons in the beam travel at a much higher energy than those in a wire, leading to a beam power that even amazes the non-experts. "We are holding a world record in DC beam power," Nagaitsev said. "About two megawatts." For comparison, this amount of power is sufficient to operate two thousand kitchen refrigerators simultaneously.

- What kind of typical light bulb would have a half an amp of current running through it? (Assume a North American household voltage of 120 V.)
- Does the beam described by Nagaitsev actually have a power of about two megawatts? Justify your answer with a calculation.
- What power does Nagaitsev assume a typical home refrigerator needs to operate?
- What make and model of household refrigerator would require the approximate power you calculated in part c.?
- Purchase the refrigerator you identified in part d.

Light bulbs used to be sold by the power they consumed. This is back in the day when light bulbs consisted of a scorching hot tungsten filament inside a partially evacuated glass bulb. Even the idea of a light "bulb" is an anachronism since the things that we use to give us light today are rarely "bulb" shaped. When Dr. Nagaitsev described his "half an amp" light bulb, he was thinking of hot filament devices that were sold by their power consumption. Given that US household voltages are around 120 V, power is an easy thing to compute. Just start with the left equation.

*P*=*VI**P*= (120 V)(0.5 A)*P*= 60 WUse the power equation from the previous part of this problem again. Use different numbers — numbers that Dr. Nagaitsev said went with the electron beam (voltage and current).

*P*=*VI**P*= (3.5 × 10^{6}V)(500 × 10^{−3}A)*P*= 1,750,000 W = 1.75 MWIs this two megawatts? Literally no, but so what? It's close enough. If I had to get my point across to the public, what would I gain by saying the power was 1.75 megawatts instead of two megawatts?

Dr. Nagaitsev says that the 2 MW beam is like 2,000 refrigerators. Divide these two numbers.

*P*=2 × 10 ^{6}W2,000 *P*= 1,000 WOr divide these two numbers.

*P*=1.75 × 10 ^{6}W2,000 *P*= 875 WFinding a refrigerator that uses 1,000 W of power (or 875 W or any other amount) is next to impossible. Your refrigerator is not on all the time even when it is plugged in. A refrigerator cycles on and off throughout the day depending on things like how many times you open the door, how long you keep the door open, how hot it is inside your kitchen, and how cold you would like your food to be. If my kitchen was outside my house and my house was in Antarctica, my fridge would never turn on. In fact, I would have to run it in reverse (an impossibility) to keep my food cold but not frozen. When someone is talking about the power of their refrigerator, they're probably talking about the power it consumes when the refrigeration cycle is on (as opposed to just resting). This info is not easy to find by slogging through consumer websites. They give their customers the more useful answer to the question, "In a typical house, under typical use, how much

*energy*would this fridge consume in a year"? Energy and power are not the same thing, so this question has no easy answer.Contact your teacher after you have purchased the refrigerator described in part d. Show all work including an equation and substitution with units.

- Which of these apparently similar rechargeable batteries is "better"? Answer this by computing the…
- charge in…
- milliamp hours [mAh]
- coulombs [C]

- potential energy in…
- milliwatt hours [mWh]
- joues [J]

Compile your findings in a table like this one.

Two apparently similar rechargeable batteries type voltage charge energy (V) (mAh) (C) (mWh) (J) NiZn 1.6 2500 NiMH 1.2 2500 There is no correct sequence for completing this table. Here's one I just thought of. Start with the definition of potential difference or voltage…

*V*=*W**q*Solve it two ways.

*q*=*W**V*⇐ *V*=*W**q**W*=*qV*⇐ *V*=*W**q*Compute the charge of the NiZn battery first.

*q*=_{NiZn}*W**V**q*=_{NiZn}2500 mWh 1.6 V *q*= 1563 mAh_{NiZn}Compute the energy of the NiMH battery (or the work it can do) using similar logic.

*W*=_{NiMH}*qV**W*= (2500 mAh)(1.2 V)_{NiMH}*W*= 3000 mWh_{NiMH}Convert the charges to coulombs using the definition of current.

*q*=*I*∆*t*⇐ *I̅*=Δ *q*Δ *t*Shift the decimal 3 places when converting from milliamps to amps. Replace an hour with 3600 s.

*q*=_{NiZn}*I*∆*t**q*= (1.563 A)(3600 s)_{NiZn}*q*= 5625 C_{NiZn}*q*=_{NiMH}*I*∆*t**q*= (2.500 A)(3600 s)_{NiMH}*q*= 9000 C_{NiMH}Convert the energies to joules using the defintion of power.

*W*=*P*∆*t*⇐ *P̅*=Δ *W*Δ *t*Shift the decimal 3 places when converting from milliwatts to watts. Replace an hour with 3600 s.

*W*=_{NiZn}*P*∆*t**W*= (2.500 W)(3600 s)_{NiZn}*W*= 9000 J_{NiZn}*W*=_{NiMH}*P*∆*t**W*= (3.000 W)(3600 s)_{NiMH}*W*= 10,800 J_{NiMH}Compile the results in a table.

Two apparently similar rechargeable batteries type voltage charge energy (V) (mAh) (C) (mWh) (J) NiZn 1.6 1563 5625 2500 09,000 NiMH 1.2 2500 9000 3000 10,800 The nickel metal hydride battery can hold more energy than the nickel zinc battery, which translates to longer periods between rechargings. That makes the NiMH the "better" battery. Higher voltage might be useful for some applications, but I don't know of any. AA, AAA, C, and D cells all should be around 1.5 V. Electrical engineers who design battery powered devices count on this.

- charge in…

### homework

- Read the following description of the execution of Kenneth Stewart in Virginia on 23 September 1998.
The electric cycle, 1,825 volts at approximately 7.5 amps for 30 seconds, then 240 volts at approximately 1.5 amps for 60 seconds… a 5 second pause intervenes, and the cycle is repeated, was designed to render the condemned brain dead within the first few moments. The function of the remainder of the cycle is to stop the body's organs so that a physician can certify that death has occurred…

Fairly straightforward. Energy is power times time. Electric power is voltage times current. Energy is a scalar, so just add up the parts of the cycle and double each to get the total.

*E*= ∑*Pt*= ∑*VIt**E*= 2[(1825 V)(7.5 A)(30 s)+ (240 V)(1.5 A)(60 s)] *E*= 864,000 J - The useful life of a battery is stated in ampere-hours. A typical 12 V car battery has a discharge time of 60 Ah. Suppose you left the headlights on when your car was parked and that they drew 3 A of current.
- How long would your battery last before it went dead?
- How much energy would it consume?

Use dimensional analysis.

60 Ah = 20 h 3 A Use the appropriate energy equation. Watch the units.

*E*=*VIt**E*= (12 V)(3 A)(20 × 3600 s)*E*= 2,600,000 J = 2.6 MJ

- A standard 60 W 120 V light bulb has a tungsten filament that is 53.3 cm long and 46 μm in diameter.
- 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?

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.