Electric Circuits Legacy Problem #27 Guided Solution

It's not too uncommon in an electric circuit problem for a student of physics to get stuck before they get started. A problem like this is a good example of a problem in which one can get stuck quite quickly. In such a situation, it's rather important that a student practice the habits of an effective problem solver, reading the problem carefully and developing a mental picture of what's going on, maybe even diagramming the situation, identifying the known and the unknown quantities, extracting that information from the verbal statement, identifying a procedure or strategy for getting from the known to the unknown quantity. Here I read of a series circuit that has two resistors and one of the first things I do when I read that statement is I start sketching a little schematic diagram. I draw a picture of a battery and the schematic symbols are used. I draw a picture of little squiggly resistors and I connect them and I connect them in series and if you have difficulty with drawing schematic diagrams, you might want to click on the link to the physics classroom where they discuss the series and parallel schematic diagrams that you can draw. So I have a 4.5 volt battery and I have a 160 mA current. So I write down delta V total equals 4.5 volts. That's the voltage drop across the entire circuit, both resistors included. And I write down I equal 160 mA. Now the little m in front of the A represents milli. It would be the same thing if it was milligrams or mg or milliliters, M capital L. Stands for milligrams or this stands for milliamps and that's not a great unit so right away I'm going to deal with that and I'm going to convert it to 0.160 milliamps because milli means one one thousandth. Now I also know that resistor A has three times the resistor of resistance B. So when I draw my little schematic diagram of my circuit, I label one of the resistors as A and the other as B and I make this statement, RA equal three times RB. That's going to be important as we go through this problem. Now the unknown quantities here are RA and RB. That's what I'm looking for. Now that I've read the problem, diagrammed it, extracted the information from the verbal statement and related it to the actual symbols used in the equation, I can begin to plot a strategy. And one of the things that I notice right away is that I have an I value and I have a delta B value and one of my equations is delta V equal IR and the big concept underlying circuits is that you can use that equation any time you have two corresponding quantities that correspond to one another that's found in this equation. Like I have the delta V across the entire circuit and I have the I through the entire circuit. And so I can find the resistance of the entire circuit, not of any individual resistor, but of the entire circuit. Sometimes called that the total resistance or equivalent resistance. So I take that equation and I rearrange it to say R of the total circuit or entire circuit equal delta V total over I total where delta V total is 4.5 and I total is .160 amps. When I do my math I get a total resistance or equivalent resistance of 28.1250 ohms. Now that's the sum of the resistance values of the two individual resistors, that's always the case on series circuits. So what I could say is that RA plus RB is equal to 28.1250 ohms. But you might recall that you wrote earlier that RA equal 3 times RB. So I could rewrite my equation 28.1250 equal RA plus RB as equal 3RB plus 1RB, which simplifies to 4RB. So 28.1250 equal 4RB if I divide each side of the equation by 4, I get RB to be 7.0313 ohms. I can round that to a decimal place and it would be 7.0 ohms. Now to get RA, it's simply a matter of recognizing that resistor A is 3 times the resistance as resistor B and I just found B. So I take the 7.0313 ohms and I multiply it by 3 and I get 21.0938 ohms for resistor A's resistance value. I can round that to one decimal place as well and it becomes 21.1 ohms.