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## Electrical engineering

### Course: Electrical engineering > Unit 2

Lesson 3: DC circuit analysis- Circuit analysis overview
- Kirchhoff's current law
- Kirchhoff's voltage law
- Kirchhoff's laws
- Labeling voltages
- Application of the fundamental laws (setup)
- Application of the fundamental laws (solve)
- Application of the fundamental laws
- Node voltage method (steps 1 to 4)
- Node voltage method (step 5)
- Node voltage method
- Mesh current method (steps 1 to 3)
- Mesh current method (step 4)
- Mesh current method
- Loop current method
- Number of required equations
- Linearity
- Superposition

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# Application of the fundamental laws (solve)

We solve the equations created by direct application of the fundamental laws: Ohm's Law and Kirchhoff's Laws. (Part 2 of 2).

## Want to join the conversation?

- Is there a voltage drop across Is? There must be, right? 1 Volt in this case, the same voltage drop we have across R2. V=IR, so 1V=3mAxRs... So the resistance of the current source must be 333.33 ohms (1V/3mA).... Is this correct?(7 votes)
- You are correct that there is a 1 volt across the current source, Is. It is tempting to apply Ohm's Law to the current source. But, you have to be careful when you do this, because a current source is not a resistor. The current in this current source will be 3 mA no matter what the voltage is, so it's not very resistor-like. Voltage is not proportional to current. When you do the calculation of 1 V / 3 mA = 333 ohms, you get what seems like a valid "resistance" for the current source, but it is misleading.

It may help to look at the article "Ideal elements and sources" (https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-ideal-circuit-elements). In the section on Ideal Current Sources there is an i-v plot of a current source (a current source plots as a horizontal line). Compare that to the i-v plot for a resistor. (which plots as a sloped line). Try drawing the i-v plot for the 3 mA current source from the example circuit above, then mark the point where i = 3 mA and v = 1 volt. Then draw an i-v plot of a 333 ohm resistor and mark the same point. Can you see why its a bit awkward to talk about the resistance of a current source? The current is always the same, and the voltage can be anything, so Ohm's Law doesn't apply very well to a current source.(17 votes)

- hi , i wanted to ask that when i have the equation in the video I1-I2=Is

, how did you delete I2? why did you chose R2 in R2(I1-I2=Is)?(6 votes) - Why did you include a current source? It threw me in a little " loop".(4 votes)
- Ha! Current sources are a little mind bending. They are not as familiar in everyday life as voltage sources but they are an essential electrical concept used in every transistor circuit.(1 vote)

- 05:15why is 6 volt when 2K.3mA= 6v ?(3 votes)
- In all of the schematics drawn, it shows the current moving away from the positive side of the voltage source. Is this correct? Shouldn't the current flow into the positive side?(2 votes)
- We always point the current arrow in the direction of movement of
**positive**charge. This is a very old habit in electricity, dating back to Ben Franklin. We do this even though we all know the negative electrons carry the current (in metals).(1 vote)

- What happens if your current source is flowing against current(2 votes)
- A current source
*_defines_*the current magnitude and direction. The voltages and currents in the rest of the circuit adjust themselves to account for that definition.(2 votes)

- Two voltage sources are connected to a parallel circuit,how do I apply superposition method?(2 votes)
- Hello Varney,

This is one of my favorite techniques because it works so well for unbalanced thee phase systems...

Regarding your question. Start by turning on power supply #1 and turning off (shorting) power supply #2. Calculate the current for the circuit and set it to the side. Now turn on power supply #2 and short power supply #1. Once again calculate the current. Now all you need to do is add them together - the total current is the sum of the individual currents.

This site has a good tutorial: http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/

Regards,

APD(1 vote)

- why didn't you find the REQ of the two resistors?(1 vote)
- The two resistors are not in series and not in parallel. There is no opportunity to simplify.

Check the definitions of series and parallel,

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/a/ee-series-resistors

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/a/ee-parallel-resistors(3 votes)

- 5:30How come its 3.5mA but not 3.5A? How can one figure out that?(1 vote)
- At5:26the equation is 21V/6000ohms. That quotient is .0035 amperes, also known as 3.5 mA.(2 votes)

- How did you know that "Is" would be coming out of node B? We know that Is comes out of the current source and goes to node C and then it splits up towards node B and left towards voltage source. Why didn't part of "i1" go to the current source and add up with "Is".(1 vote)
- Each circuit element contributes its own "rule" to the circuit. For a current source the rule is, "the current through me is Is." That means current in both [edit: XXthe top and bottom wiresXX] [edit: better: the wires coming out of the current source] is Is. You can be confident that every milliamp that comes out the bottom of the current source will somehow find its way through the circuit and return to the current source through the top wire, with no extra current extra and no missing current.

Sometimes beginning students try to jump ahead and imagine the solution to a circuit before doing the formal analysis. It is really easy to get your head all confused trying to guess the currents around node b. What I suggest is for you to complete the analysis and then go back and trace out how Is flows through the circuit.(2 votes)

## Video transcript

- 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 gonna solve it. This is a matter of doing the algebra to solve a set of four
simultaneous equations. What I wanna do now is just
write my equations down here and tidy them up a little bit. So the first one we're gonna bring down is the voltage statement here, which I can write as V1 plus V2 equals Vs. I'm gonna bring down the
i equation over here. It's gonna be i1 minus i2 equals Is. And now I have to pick. I have four variables here. I'm gonna start doing substitutions based on these expressions up here. These are my Ohm's law expressions. So we're gonna use those to
substitute for the voltages, and we're gonna solve the whole
thing in terms of currents. So the first equation becomes V1 is R1 i1 plus R2 i2 equals Vs. Second equation stays the same. i1 minus i2 equals Is. The next step in the solution is gonna 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 equation is gonna be R2 i1 minus R2 i2 equals R2 Is. We're gonna add these equations together. And we can see these
terms nicely cancel out, and so I'm gonna end up with R1 plus R2 times i1 equals Vs plus R2 Is. And I can get the solution
in symbolic terms for i1. I1 equals Vs plus R2 Is over R1 plus R2. What I wanna 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 fifteen 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. And 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, 15 plus R2 is 2k times Is, three milliamps divided by 4k plus 2k. So let's solve this. That equals 15 plus 2k times three
milliamps is six volts. This is volts, and that's divided by 6k equals 21 volts over 6k equals 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 wanna find the other one. That's sitting right here. We can figure out i2. Minus i2 equals Is minus i1 equals three milliamps minus 3.5 milliamps equals minus .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. And 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. 4k ohms and that equals,
three and a half times four 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,
sorry, our last voltage is V2, is the voltage right here. And V2 equals i2 which we decided was .5 milliamp times two thousand ohms, 2k. And V2 equals one volt. And now we've solved our circuit. That's a complete circuit analysis using the fundamental laws.