Electric Circuits Legacy Problem #25 Guided Solution

If you've done electricity labs in your physics class, it's quite likely that you've used ammeters and voltmeters. These are devices which are used to measure the current in the voltage drop between two points. The ammeters in this diagram are represented by the symbol A, and they're simply wired in series in the series circuit, and they measure the current at whatever location you place it. The voltmeters, when used in lab, usually consist of two probes, which you click on metal components on the circuit, maybe on one side of the resistor and on the other side, and as such they find the electric potential difference between the two points that you tap on when you tap on the metal parts of the circuit. Here these are represented by the V with a circle around them in the diagram, and they're measuring the electric potential difference, or voltage drop, across the three resistors. In a series circuit, the total resistance is simply the sum of the individual resistance values of the three resistors in the circuit. So as we solve this problem, that's the place we're going to begin. We're going to take the 10.3, 15.2, and 12.8 ohms, and we're going to add them together and find the total resistance. It's not something that they particularly ask for, but it will be strategic in our task of trying to find the current ammeter readings and the voltmeter readings. So when I do that, I end up getting 38.3 ohms as the total, or equivalent, resistance, and the current through the battery is just going to be the battery voltage divided by this total resistance. When I take the 60.0 volt battery voltage and divide 38.3 into it, using the equation I equal delta V over R, I get 1.566 repeating amperes as the current through the battery. Though in a series circuit, that current is everywhere the same. It's the same in the battery as it is in the first resistor, as it is in the second and the third, etc. And so all the ammeter readings will read 1.566 repeating, though it would be probably to two decimal places as 1.57 amperes. Now to find the voltage drop, I can use the same equation, delta V equal I R. But when I use it, I'll have to use the right resistance value to calculate the proper delta V. For instance, for the voltmeter probe that touches on one side and the other side of the R1 resistor, I'll have to use R1's resistance value to calculate the voltage drop between those two points. So I go delta V equal I R, and for I I use 1.56 repeating, and for R I use 10.3. I get 16.1358, and that's the voltage drop across this resistor, since that's the resistance and the current between those two points. Now for R2, of course, I'm going to use the same equation, only I'm going to go I of 1.566 multiplied by 15.2 ohms for R2. I'll end up getting 23.8120 as the voltage drop across the two points being touched on opposite sides of the R2 resistor. I can round that to three significant figures as well, 23.8. Finally, for R3, I do the same thing, using as my R3 value 12.8 ohms, and I get 20.0522. I can round that to three significant digits, such that it's 20.1 volts. One thing you'll notice, that if you take these three voltage drops for going around the circuit through the three different resistors, they sum up to the 60.0 voltage gained as a charge passes through the battery. That's what we should expect.