Application of the fundamental laws (solve) | Electrical engineering | Khan Academy

So in the last video, we did our circuit analysis. We set up the four equations that we needed to solve in order to figure out all the voltages and currents in our example circuit. And so now we're going to solve it. This is a matter of doing the algebra to solve a set of four simultaneous equations.

What I want to do now is just write my equations down here and tidy them up a little bit. So the first one we're going to bring down is the voltage statement here, which I can write as V1 plus V2 equals Vs. I'm going to bring down the I equation over here; it's going to be I1 minus I2 equals Is.

Now I have to pick—I have four variables here. I'm going to start doing substitutions based on these expressions up here. These are my Ohm's law expressions, so we're going to use those to substitute for the voltages, and we're going to solve the whole thing in terms of currents.

So the first equation becomes V1 is R1 I1 plus R2 I2 equals Vs. The second equation stays the same; I1 minus I2 equals Is. The next step in the solution is going to be to figure out how to eliminate one of these currents. Now I think what I'll do is I'll eliminate I2, and I can do that if I multiply this equation by R2 and add these equations together.

So let's do the multiplication by R2 first. We'll rewrite the first equation up here: R1 I1 plus R2 I2 equals Vs, and then down the bottom, the equation is going to be R2 I1 minus R2 I2 equals R2 Is. We're going to add these equations together, and we can see these terms nicely cancel out.

So I'm going to end up with R1 plus R2 times I1 equals Vs plus R2 Is. I can get the solution in symbolic terms for I1: I1 equals Vs plus R2 Is over R1 plus R2.

What I want to do now is actually put in the original real values and solve this circuit all the way through. So we'll move up and keep going. So let me sketch the circuit again, the real circuit again. We had a voltage source with 15 volts; it went to a 4K resistor that was R1. We had a 2K resistor here and a current source, and that was 3 milliamps.

We want to discover this was I1, and this was I2. Those are the two unknown currents. Let's see if we can find those. So I1 equals Vs, which is 15 plus R2 Is, which is 2K times Is, which is 3 milliamps, divided by 4K plus 2K.

So let's solve this: that equals 15 plus 2K times 3 milliamps divided by 6K. So that equals 15 plus (2K times 3 milliamps), which is 6 volts. This is volts, and that's divided by 6K, equal to 21 volts over 6K, equal to 3.5 milliamps.

So that's this value right here; 3.5 milliamps is I1. So we have one of our unknown currents, and we want to find the other one that's sitting right here. We can figure out I2. So minus I2 equals Is; I1 equals 3 milliamps minus 3.5 milliamps equals -0.5.

Let's get rid of the minus signs; I2 equals 0.5 milliamps. That's that value right there. Let's put some boxes around our answers we have so far: here's one, and here's another one. There's I2 and I1.

Now let's finish up by figuring out the voltages. We can use Ohm's law to figure out the voltage across the 4K resistor, and that's V1. So V1 equals I, which is 3.5 milliamps times 4K ohms, and that equals (3.5 * 4) is 14 volts.

Whenever you multiply milliamps times K ohms, those units cancel, and you get volts. So V1 equals 14 volts, and our last current—or sorry, our last voltage is V2. V2 equals I2, which we decided was 0.5 milliamps times 2,000 ohms, 2K, and V2 equals 1 volt.

And now we've solved our circuit. That's a complete circuit analysis using the fundamental laws.