Electric Circuits Legacy Problem #33 Guided Solution

This problem reminds me of a common experience I have as an instructor within a physics classroom. When coming to the part of the electric circuits unit, when we're calculating the power of resistors for series and parallel circuits, I'll go through an example problem and get an answer. A student will invariably raise their hand and say, I did it a different way and got the same answer. What's up with that? And what is up with that is that there are three different equations for calculating electric power. And the equations are derived one from another. And as such, you would expect that using either one of the three equations would allow us to calculate the power of a resistor within a circuit. As I often say, my apologies to cat lovers, that there's many different ways to skin a cat. Now what I'm going to do is calculate the power here of our individual resistors and of the entire circuit three different ways. So follow along. I'm going to begin by reading the problem and diagramming a parallel circuit and remembering my big concepts of the delta V across a branch of a parallel circuit is equal to the delta V of the battery. So as I diagram this circuit, I begin to write down what I know. And what I know is that delta V total equal 9 volts. And given the concept just mentioned, I also know that delta V1, that is a resistor 1, is equal to 9 volts. And delta V2 is equal to 9 volts as well. I'm calling R1 2.5 ohms. That's the value of resistor 1. And I'm calling R2 3.5 ohms. And I want to calculate the power. I'm looking for P1 and for P2 and for total power. Now I'm going to focus on the branch resistors first, the 2.5 ohms, which I've called R1, and the 3.5 ohms, which I've called R2. And I'm thinking about the equations for power. There's three of them. And they are P equal I times delta V, P equal I squared times R, and P equal delta V squared over R. So what I need to do to calculate P for the individual resistors over the total circuit is I need to know two of the three quantities, I, R, and V. Now when I think about resistor 1, the 2.5 ohms resistor, what I know is I know that delta V is equal to 9 volts and R equal to 2.5 ohms. So perhaps the most direct way of calculating the P1 value is to say P1 equal delta V1 over R1, delta V squared over R1. That is the 9 squared divided by the 2.5 ohms. And when you do that that way, what you get for P1 is 32.400 watts. You can do the same thing for R2. It's just the 9 squared divided by the 3.5 ohms. And when you do that, you get 23.1429 watts. And you can round that to three significant digits. Now here's some alternative ways, if you wish, for calculating P1 and P2. I could use the equation P equal I times delta V, but first I need to calculate I1. In calculating I1, I need to use the equation delta V equal I times R applied to resistor 1. Rearrange that would be I1 equal delta V1 over R1. I can substitute in 9 volts for delta V1, 2.5 ohms for R1, and I would calculate 3.6 amps for I1. Now P1 equal I1 times delta V1, so I simply take the 3.6 ohms and multiply by the 9 volts. And I get 32.400 watts, the same thing I just had. I can repeat the process for first finding I2 using resistor 2 value. I2 equal 9 volts over 3.5 ohms, that's 2.5714 amps. And then I can say P2 equal I2 times delta V2, the 2.5714 amps times the 3.50 ohms. And I get the same answer as before, 23.1429 watts. Once I calculate I1, I2, of course I could have used the third equation, P equal I squared times R. I could have squared my I and calculated, and then multiplied by the R. I would have gotten the same answer. Now for P total, one way to calculate it is to simply recognize that each of the resistors are using up energy at a given rate. That's what we call power, the rate of energy consumption. And they're doing it simultaneously, so as you might expect, the total rate of energy consumption is simply the sum of the individual values for energy consumption rates. That is P total equal P1 plus P2. So we've calculated three different ways, P1 and P2 values of 32.400 watts and 23.1429 watts. Simply sum them up and you get 55.429 watts. Now here's two other ways to calculate P total. You could begin by calculating the equivalent resistance. For parallel circuits, 1 over REQ equal 1 over 2.5 plus 1 over 3.5. By the time you solve that one for R equivalent resistance, you get 1.4583. Once you get that, one thing you could quickly do is P equal delta V squared over that R value. That is 9 squared divided by 1.483 ohms. That gives you the same power value of 55.5429 watts. Or, once you got that R equivalent, you could have calculated I total, the current outside the branches, as delta V total, 9 volts, divided by the R equivalent value, 1.4583. You would have gotten 6.1714 amps. And then you could have said P total equal this I total value of 6.1714 amps squared multiplied by the R total, which is the 1.4583. That I squared R product ends up giving you the same 55.5429 watts of power.