Electric Circuits Legacy Problem #28 Guided Solution

You know, I've been teaching a number of years, and I can't tell you the number of times I've heard a statement from a student that goes something like this. You know, physics problems are very easy to solve as long as I know the formula. And I've always kind of thought to myself that, of course, as long as you know the formula, they're very easy to solve. But you have to realize that to get the formula requires a good deal of work. It requires being able to take the problem, which is packaged within a verbal statement, to sort of dissect that problem, pull out the numbers, equate them with the variables of our physics formulas, and to use your physics conceptual understanding in order to put it all in context and visualize the situation. And then you know the formula. So it takes a good deal of effort to find the formula. Once the formula is found, the problem is really no longer a physics problem, it's a math problem. So here when I read about this 9-volt battery in a series circuit, 2.5 ohm, 3.5 ohm resistor, what I'm beginning to do is I'm beginning to write down statements like delta V total equal 9 volts, R1 equal 2.5 ohms, R2 equal 3.5 ohms, and I've drawn a picture, a diagram of the circuit with a little battery schematic symbol and a couple squiggly lines for the resistor symbols, and I've connected them by wires, and I've got just a schematic diagram and that information written down. And then I list find P1, find P2, and find P total. So the problem demands that I find the total power, the power of each resistor, P1, P2, and P total. So I need a power equation, quite evidently, and if I look back on the overview page in this set of problems, I'll find several power equations. There's P equal I times delta V, P equal I times R squared, and P equal delta V squared over R. But knowing how to use the equation or what equation to use demands that you have some conceptual understanding. Regardless of how I solve this problem, I'm going to need to get the I values. I'm going to need to get I1 and I2 and I total. Now I know delta V total and I know R1 and R2. So for a series circuit, I can find R total by simply summing up the individual resistance values. The 2.5 ohms and the 3.5 ohms sums to 6 ohms and that's the total resistance. And I have the total voltage drop across the circuit, 9 volts. So if I go delta V equal IR, another one of my formulas, I can find I total. It's simply delta V total over R total, 9 volts over 6 ohms. That gives me 1.50 amps. Now I know the current. One of my equations, it's the current through the battery. But since it's a series circuit, it's current everywhere within the circuit, including I1 current and I2 current. That means I can now use P equal I squared times R in order to find the three different power values. To find P1, I use 1.50 amps and 2.5 ohms, the resistance of resistance 1. When I go I squared times R, I get 5.6250 watts. I could use for P2, 1.50 for the current and 3.5 ohms for the resistance. I'd be going I2 squared times R2 and that gives me P2. It comes out to be 7.875 watts. For P total, I can just simply sum them up and I get 13.5 watts. Or I could go I squared times R for the total battery, the total circuit. That is 1.50 squared times 9.0 volts and that gives me 13.5 watts. There's another equation you could have used though. That's the P equal I times delta V. To use it, you first have to calculate delta V. Delta V for the individual resistors. For an individual resistor, delta V is the current through that resistor times the resistance of that resistor. For resistor 1, delta V1 is equal to I1 times R1. It comes out to be 3.75 volts. Then to find its power, you take the I delta V equation, which would be 1.5 times 3.75 volts. You get 5.625 watts. Now for delta V2, you need to do the same thing. I2 times R2, 1.5 times 3.50 ohms. You get 5.25 volts. The power of that resistor is equal to 1.5 ohms, the current through it, multiplied by the voltage drop across it, 5.25 volts. It comes out to be the same values, 7.875 watts.