Physics 101 Homework Set 10 Lawrence Weinstein All problems are from Chapter 23. One steps: ----------- 23.4: V = IR -> I = V/R = 120 V / 240 Ohm = 0.5 A 23.6: P = IV = 10 A * 120 V = 1200 W Exercises: ---------- 23.6: The battery does not supply the electrons. The electrons are already in the wires. The battery merely provides the energy to make them move. 23.10: No. The same current flows into and out of a battery and into and out of a lightbulb. If more current flowed into a battery or into a lightbulb, then the battery or bulb would develop a static electric charge and you would get shocked when you touched it (that's what a van de Graaff machine does). The battery provides energy to the electrons that pass through it and the lightbulb uses the energy from the electrons that pass through it. The electrons are neither created nor destroyed. 23.14: The shorter the Mean Free Path, the higher the resistance of the material. This is because, the shorter the mean free path, the more collisions the electron will have with the atoms in the wire, the harder it will be for the electron to get through and the more resistance there will be. (Think about riding in the bumper cars at the amusement park. If you have a collision every three feet, it will be much more difficult for you to drive to the other side than if you only have one collision every 30 feet.) If you make the metal colder, then its resistance will decrease, and the mean free path will increase. 23.18: The rest of the energy goes to heat. Incandescent light bulbs get VERY hot! 23.20: The filament of a lightbulb glows because the resistance of the filament is much greater than the resistance of the connecting wire. The electrons deposit their energy where the resistance is greatest. 23.22: a) If you cut the wire in half so that it is half as long, then the resistance will be halved (ie: 5 Ohms). b) if you double the wire over, this is the same as cutting the wire in half and then placing the halves together. You will now have a wire that is half as long and twice as thick. Therefore the resistance of the doubled-over wire is 2.5 Ohms. 23.24: I = V/R. The resistance of the lightbulb does not change. Therefore, if you increase the voltage, you increase the current. The lightbulb will have a greater current when connected to a 220-V source. 23.30: a) joule/coulomb = volt b) coulomb/second = ampere c) watt-second = joule 23.40: THe lamp turns on almost instantly because all of the electrons in the circuit start moving almost simultaneously. 23.44: Bulb A is in parallel with bulb B. Therefore, closing the switch to turn on bulb B will not change the potential difference (voltage) across bulb A and therefore will not change the brightness of bulb A. 23.60: The labels 60-W and 100-W refer to the power consumed by the bulb if it is attached to a 110-V potential difference. Therefore, since P = IV, the 60-W bulb draws less current than the 100-W bulb. Since V = IR, and the 60-W bulb draws less current, the 60-W bulb has a higher resistance than the 100-W bulb. (We calculated this in class for 110-W and 55-W bulbs.) When you put two bulbs in series, the bulb with the higher resistance will use most of the electrons' energy and will therefore have the greater voltage drop. brighter. The 60-W bulb has more resistance, therefore it will have a greater voltage drop than the 100-W bulb when placed in series with the 100-W bulb. Another way to calculate this is to use V = IR. Both bulbs have the same current (since they are in series), therefore the bulb with the larger resistance has the larger voltage drop across it. When the bulbs are placed in parallel, both bulbs have the potential difference across them (this is the same as having the same voltage drop). Problems: ------------------ 23.6: a) I = P/V = 4 W / 120 V = 0.033 A b) R = V/I = 120 V / 0.033 A = 3600 Ohms = 3.6 kilo-Ohm c) One year = 1 year * (365 days / 1 year) * (24 hours / 1 day) = 8760 hours Energy = P*t = 4 W * 8.76*10^3 hr = 3.5*10^4 W-hr = 35 kWh (kiloWatt*hours) alternatively: 1 year = 8760 hours * (60 min/1 hr) * (60 s/1 min) = 3.15*10^7 s Energy = 4 W * 3.15*10^7 s = 1.2*10^8 J Either set of units (J or kWh) is acceptable. d) cost = 35 kW-hr/yr * $0.15/kWh = $5.25 per year Estimation: ------------ We need two numbers, the average current used per household and the total number of households in the US. We'll do the number of households first. The population of the US is 3*10^8. There are between 2 and 3 people per household so the total number of households = 3*10^8 people * (1 household / 3 people) = 10^8 households. Now, let's estimate the current in use at 7:00 PM on Thursday. I will start by estimating the power used and then convert that to current (since I know the voltage is 110 V). There are 5 - 10 lights on at 100 W each. The refrigerator is using about 500 W. The heating system is using several kW (since a 1-room space heater is 1500 W). Perhaps the stove or oven is on also (using about 2 kW). Therefore, the total power used is about 5 kW or 5*10^3 W (anywhere between 1 and 25 kW is close enough). Since P = IV, I = P/V = 5*10^3 W / 110 V = 50 A. Therefore, each household uses about 50 A. Thus, the total current used by US households is 10^8 households * (50 A / household) = 5*10^9 Amps. Wow, that's a lot of current!