Electric Circuits Legacy Problem #17 Guided Solution

One of the biggest complications which a student has when it comes to electric circuit problems is knowing what equation to use. After all, there's plenty of them. If you click the link near the bottom of this page going back to the overview page for this set of problems, you'll see the equations there. You ought to get familiar with some of the equations and the relationships between the quantities having to do with electric circuits. In this problem, knowing what equation to use is probably the most complicating thing about the problem. After all, the math's not that difficult. What you're to do is to compare the resistance of two light bulbs. One of them's in the car. You know it's a 1.5 amp bulb and a 12 volt battery. The other one's in the house. It's a 100 watt bulb and a 110 volt household circuit. Now one of the things you'll need to be able to do is to analyze the units given in the problem so that you know what the quantity is. For the car bulb, 1.5 amps refers to the current. So I write it down. I equal 1.5 amps. And it's hooked up to a 12 volt battery. And that's the so-called electric potential difference or voltage drop across the light bulb. So I write down delta V equal 12 volts. Writing the quantities down in this way, defining them in terms of the symbols used in the equations, helps me to pick the appropriate equation for finding the resistance. If I go back to that set, to the overview page for this set, I'll find the equation that goes delta V equal I times R. And knowing delta V is 12 volts and I is 1.5 amps allows me to solve for R quite easily. It comes out to be 8.0 ohms. Now I attack the bulb that's in the circuit, in the household circuit. The 110 volts refers to the delta V, the electric potential difference across this bulb. And the 100 watts refers to the power of the light bulb. So I know P equal 100 and delta V equal 110 volts. Again, writing it down in this way allows me to go back to those equations and find the one that's appropriate for solving for R. There you'll find an equation that says P equal delta V squared over R. Rearranging and substituting and then solving for R yields about 121 ohms. We can round that to two significant digits such as 120 ohms. In the end we could conclude that the light bulb that is present in the house has about 15 times greater resistance than the light bulb that's present in the car.