Electric Circuits Legacy Problem #26 Guided Solution

Success on series and parallel circuit problems require a solid understanding of the mathematical equations and also a good understanding of the conceptual ideas which underlie those equations. Here we have a series circuit that consists of three identical resistors. The 12 volt battery is pumped and charged to the circuit and the rate at which the charge flows is determined by an ammeter which reads 0.360 amperes. That's the current through the battery and that's the current through all three of the resistors since in series circuits the current is everywhere the same. So one thing that I can calculate if I know the voltage across the entire circuit and I know the current through that circuit is I can calculate the total resistance. I use the equation delta V equals I times R and I rearrange it to solve for R. It becomes R equals delta V over I, 12 divided by 0.360 gives me 33.333 repeating ohms. Now that's the total or equivalent resistance and in series circuits the total or equivalent resistance is simply the sum of the individual resistance values of those resistors in the circuit. So I have three identical resistors and together their resistance values sum up to 33.33 ohms. That is R1 plus R2 plus R3 is equal to 33.333 and each R value is the same. So I can rewrite that as 3 times R is equal to 33.333 ohms. So each one of the resistors has a resistance value that's 33.333 ohms divided by 3 also known as 11.11 repeating ohms. Now that I've determined the resistance values of the individual resistors I can determine the voltage drops across the resistors. Now there's a couple of ways to do it. One of them is the intuitive conceptual way and the other is the more mathematical way. I'm going to trace you through the more mathematical way first. To define the voltage drop across a resistor you need to use the equation again delta V equal I times R. But you have to remember that in series circuits the I happens to be the I of the circuit and the R happens to be the R of the resistor through over which you're trying to find the voltage drop. So when I go delta V equal IR to find the voltage drop across the first resistor what I would be doing is going .360 amps the current through it multiplied by 11.1111 ohms the resistance of it. I end up getting 4.00 volts exactly. Now that's the voltage drop across the first resistor but it's also the voltage drop across the other resistor since they too have the same resistance value and the same current. Well the conceptual way of doing this last question of finding the voltage drop is to simply recognize that each of the resistors has the same resistance and the same current so they must also have the same voltage drop. There's three of them so the three voltage drops together must add up to the 12 volt voltage gain as charge passes through the battery. So if I just took 12 volts and divided it by three so I'd have three equal size voltage drops but I'd end up with 4.00 volts.