 So now we're ready to take a look at this circuit and try to analyze the entire circuit. Now in some of my previous videos, I had shown a example where we could take a chart and put all of our values into a chart and use that to help us organize. Now that's really useful if you have just a series circuit or just a parallel circuit. But in this case, we've got some series and some parallel in the same circuit. And in that case, using a single table can actually confuse you. So instead, we want to break down and identify series and parallel portions of this graph. And to start with, we want to look and find a series or parallel section. And in this case, I can look here and see that I've got a series section of those resistors. And in series, I can actually take that and figure out that the two resistors add up. 2 ohms and 3 ohms gives me 5 ohms. And for right now, I'm just going to call that resistor A. So the equivalent of this is if I replaced it with a single 5 ohm resistor. Now that 5 ohm resistor and this 4 ohm resistor are going to be in parallel. And that means we have to use the inverse law to find that set of resistances. And so instead of having 2 ohms and 3 ohms, I have 5 ohms. And then instead of having 4 ohms and 5 ohms, I now have 2.222 ohms. And you want to keep a few decimal points here, at least three significant digits. I've got four shown here. The more you round it off, the more rounding area you'll have in your final answers for your current and voltage. Now if I did replace all of this with a single 2.222 ohm resistor, which I call resistor B, then the 1 ohm resistor and that resistor B are going to be in series together, giving us our 3.222 ohms. Now the real trick here with doing the circuit analysis is not just doing the individual resistances, but to actually make little tables for each sub-part of the circuit. So here I've got my 2 ohms and 3 ohms, giving me 5 ohms. It's not the equivalent resistance of the entire circuit. It's just the equivalent resistance of that equivalent part of the circuit. But I don't know the voltage or the currents for any of those at this point. Similarly, I'm going to take another table for that parallel part of the circuit, where I now have my 5 ohm RA, my 4 ohm R4, giving me my 2.222 ohm RB. And then finally for my last series part of the circuit, I'm going to plug in my 1 ohm, my 2.22 ohm, and my 3.22 ohm. Now at this point, I can actually use the voltage of the battery, because the voltage of the battery is in a simple series circuit with R1 and the equivalent of this whole set over here. So it was if I had just a single circuit with 1 ohm, 2.2 ohms, and a 20-volt battery. Now at this point, I can work backwards through the circuits. If I wanted to find the current here, the current is the voltage over the equivalent resistance. And so this 20 volts divided by this 3.22 ohms is going to give me 6.21 amps. And I can actually put that into my table now. And because this was a series circuit, all three of these cells have to be the same. So I can actually sort of highlight them there and then fill in that it's got to be the same value for each one of these. Then I can actually work each of these rows. And just to space this out here, the first row is going to be the current times the resistance. So that's 6.21 amps times 1 ohm, which is going to give me the 6.21 volts. And then this row is 6.21 amps times 2.22 ohms, which will give me 13.79 volts. Now that means I actually now know my current and voltage going through my original 1 ohm resistor. That's been solved. What I need to do now is figure out what's going on here. And that particular part of the circuit, I know the equivalent values going through that section. That means I could take this line of my table and transfer it to this table. Because the equivalent voltage over that section of the circuit is 13.79 volts and the equivalent current through that portion of the circuit is 6.21 amps. Now this is a parallel circuit. And that means this section of the circuit is going to have all the same values for the voltages. And what that really means going back and looking at my table is that from here to here, I had a voltage drop. And it's the same voltage drop over this side as it is over this side. And that voltage drop is my 13.79 volts. Once I've done that, now I should be able to go back and find my currents. And the current for the top row here is going to be my 13.79 volts divided by my 5 ohms, which will give me a value of 2.758 amps. And then my value for the next row is going to be my 13.79 volts divided by 4 ohms, which will give me 3.448 amps. Now if you go through and add these up, you'll find out that within rounding, you get the same value here. And I could probably round these off to give me 2.76 amps. And 3.45 amps. And then that rounds off to the same value that my IB rounded off at. Now there's no problem keeping for significant digits out there. I'm just rounding things off a little bit here. Now the next step here is to realize that as I'm working through, this line here had to do with the series which went in there. So I can transfer this line of the table to this because it represented this section of the circuit. And so I know that that particular section of the circuit had a voltage of 13.79 volts and a total current of 2.76 amps. And that means that from here to here, the voltage dropped 13.79. And the current through that branch is 2.76 amps. I had volts there. It should be amps. Now this is a series section. And that means that these parts, again, have to be the same. So I can then transfer those values up to find the current through each one of those resistors. And once I have the current through each one of those resistors, I can go back and find the voltage through each one of those resistors. And again, I take the current times the resistance for each resistor. And that gives me for my 2 ohm 1, I've now got 5.52 volts. And then I've got 8.28 volts. And if you add those up, it gives you just about the 13.79 volts. It actually gives me 13.80 volts, which is some of my rounding error. If I had kept everything to four significant digits, then I would have had a more accurate value at the end. But I'm still well within the tolerances that we use in our class, which was three significant digits. So based on this, I can now come back and actually figure out what I had for each resistor. So my R1, which was 1 ohm, had a voltage. And that's my delta V. I don't have a delta symbol real handy, so I'll just write delta V. And that was what I figured out way down here at the bottom. And so that was 6.21. And it had a current one of 6.21 amps. Then my R2, which was my 2 ohm, had a delta V2, which was equal to, well, we found that here on our last chart, 5.52 volts, and an I2, which was equal to 2.76 amps. My R3, which was my 3 ohm, had a voltage of 8.28 volts and 2.76 amps. And then finally, my R4, which was my 4 ohm resistor, had a delta V voltage. And we found that one on this table. So that was 13.8 volts and an I4 of 3.45 amps. So it took a lot of steps to find the individual currents and voltages for each resistor. But by breaking it down into series and parallel subsections, I was able to work all the way back to the equivalent circuit and then work backwards through the tables to find all of my values. It's not a short process, but it is an easy process in the sense of it just uses our series and parallel rules and our strategy of breaking into a table for each subset area of the circuit and then working through in multiple, multiple steps finding the individual values. I'm going to have one more video that's going to show you a different method.