Physics 101 Homework Set 10 Solutions Chapter 23: ------------------ Exercises: 4: When the current is increased, the power dissipated increases and the bulb glows brighter. This is because a) there are more electrons passing through the bulb and giving up their energy there and b) each electron deposits more energy (the energy deposited per coulomb is the voltage which is V = I R. If you increase I, then you increase V.). 16: The rest of the energy is transformed to heat. Incandescent light bulbs are much less efficient than flourescent bulbs. This means that a much lower fraction of the energy is transformed into light and a much higher fraction is transformed into heat. You can feel this yourself. Flourescent bulbs are much cooler than incandescent bulbs. 18: A filament is just a thin wire. A thicker filament has less resistance than a thinner filament. If you plug it into a wall outlet (ie: into a 110 V power source), there will be more current flowing through the filament with less resistance (the thick one). Note that this means that it will glow brighter. 20: I = V / R. If you double V and R, then I(new) = 2V / 2R = V/R = I. If you halve V and R, then I(new) = 0.5 V / 0.5 R = V/R = I. I will be unchanged in both cases. 22: A 110-V appliance is designed to run when plugged into a 110 V outlet. If you plug it into a 220 V outlet, then it will draw twice as much current as it is designed to draw. Each Ampere of current will deposit twice as much as power (since P = IV and V is doubled). Thus, the 110-V appliance will use FOUR times as much power when it is plugged into a 220 V outlet as it is designed to do. If your 1500 W hair dryer is plugged into a 220 V outlet, it will use 6000 W. This can easily start a fire. On the other hand, an appliance designed to run at 220 V will draw half as much current when plugged into a 110 V outlet. Each Ampere of current will deposit half as much as power (since P = IV and V is halved). Thus, the 220-V appliance will use one-fourth times as much power when it is plugged into a 110 V outlet as it is designed to do. Therefore it probably won't work at all. But at least it won't start any fires. Therefore it will do less damage plugging a 220-V appliance into a 110 V outlet. Note that this is a serious question. American appliances run on 110 V. European ones use 220 V. You need a transformer (see chapter 25) to run appliances from the US in Europe and vice versa. 26: The high beam filament will have less resistance. Both operate on the same voltage (since both are powered by the car's electrical system). The high beams use more power, therefore they need more current, therefore they must have less resistance. (We worked this out before when we calculated the resistance of a 100 W and a 50 W lightbulb.) 28: If I connect the two resistors in series, then their total resistance will be R = R1 + R2 which will be bigger than either R1 or R2 individually. Note that 'equivalent resistance' means the 'total resistance of all the resistors together'. It does not mean that the two resistors are equal. 38: The two light bulbs will glow brighter if they are connected in parallel. I demonstrated this in class with both the car battery and large light bulbs and with the blue circuit boards. The reason for this is that if you have a current I through one light bulb, then you will have a total current 2I through 2 bulbs in parallel (two paths, each with current I) but you will only have a current 0.5*I through the 2 bulbs in series (since the resistance is doubled and there is only one path for the current to take). Therefore, the two bulbs in parallel will use four times as much power as the two bulbs in series and will glow about 4 times brighter. Since the bulbs in parallel use more power, they will use up the battery faster. 46: The more bulbs connected across a battery, the more total current the battery supplies. The more current the battery supplies, the less voltage it can supply. Therefore, if you add too many bulbs, then the brightness of each bulb will diminish. 50: If a 60 W bulb and a 100 W bulb are connected in series, the 60 W bulb will have the greater voltage drop (and hence will glow brighter). The reason for this is that the 60 W bulb has a larger resistance than the 100 W bulb. (We worked this out in class.) Therefore, when they are connected in series, both bulbs have the same current running through them. The voltage drop in each bulb is equal to the current times the resistance in that bulb (V(bulb) = I(bulb) * R(bulb)). Since they both have the same current, V(bulb) = I * R(bulb). Since the 60 W bulb has a larger resistance, it will have a larger voltage drop and it will glow brighter than the 100 W bulb. I know, this sounds strange. Just remember that we never connect them in series, only in parallel. Therefore the ratings (60 W and 100 W) refer to how they work in parallel. Problems: ---------- 6: A 4 W nightlight is plugged into a 120 V circuit. a) it draws I = P / V = 4 W / 120 V = 0.033 A b) therefore its resistance is R = V / I = 120 V / 0.033 A = 3600 ohms c) Energy = power * time. We can measure energy in either Joules or kW-hr. In 1 year = 365 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 3.1 * 10^7 seconds. Therefore the energy used in one year is 4 W * 3.1*10^7 s = 1.3*10^8 J. I can convert this to kW-hr: E = 1.2*10^8 J * (1 kW-hr / 3.6*10^6 J) = 36 kW-hr. Alternatively, I can calculate the number of kW-hr directly using: 1 year = 365 days/year * 24 hours/day = 8760 hours so E = 4 W * (1 kW / 1000 W) * 8760 hr = 35 kW-hr The answers are slightly different due to rounding errors. d) At $0.15/kW-hr, the cost of this energy is Cost = 35 kW-hr * $0.15/kW-hr = $5.25 Note that a 100 W light bulb consumes 25 times more energy than a 4 W bulb. If you leave that on for the entire year, then it will cost 25 * $5 = $125 which is MUCH more than the cost of the bulb. Estimation ----------------------- This problem requires you to estimate your input numbers. I am not looking for an exact answer, just a reasonable one. We need to estimate a) the number of households in the region and b) the power use per household. Let's just consider electrical power (ignoring oil, gas and gasoline). The population of the region is about 10^6 people. (The US population is 3 * 10^8 people [300 million] and we are a reasonably large metropolitan area.) Any population number between 0.2 million and 5 million is certainly a reasonable guess. My electric bill is about $150/month for four people. At $0.10/kW-hr, this means that we used about 1500 kW-hr per month. This is in terms of energy per time (ie: power) but I need to convert it to standard units (ie: W). There are 30 * 24 = 720 hours per month. This means that P = 1500 kW-hr / 720 hr = 2 kW. That is the average power use for my family of four. Therefore, we each use 0.5 kW. Therefore, if we are an average family, then the region uses 10^6 * 0.5 kW = 500 MW (Mega-Watt) of electrical power. We probably use more than the average (so I should lower the estimate). I also need to count business and industrial use (so I should raise the estimate). I think I'll stop here. I estimate that the area uses between 100 and 2500 MW of electrical power. These problems frequently have more than one solution. Another approach is to look at the US use of power. The US used 3.6*10^12 kW-hr in 2000 (I found that in google). This is an average power use of 3.6*10^12 kW-hr/yr / 8760 hr/yr = 4.1*10^8 kW. Hampton Roads has 1 million of the US's 300 million people so we used 1/300 of that total or 1 * 10^6 kW = 1000 MW. Any answer within a factor of 10 of this is a good answer and should probably get full credit.