Electrical Engineering: Basic Laws (10 of 31) Kirchhoff's Laws: A Medium Example 1

In this video I will use Kirchhoff's law to find the current (I=?) of a more challenging single loop circuit.

Next video in this series can be seen at:

[maxtuck]

Thank you for your time and teaching skills. Since the voltage drop across the 6 ohm resistor is 48 volts and the battery supplied voltage is 2X that voltage drop then the supplied voltage of that battery should be 96 Volts and Kirchhoff's law will work out.

[mikeymikemike]

You are a life saver! This class at my school has been a nightmare for me. Love my professor but it's been really messy and the textbook, rather than helping me, has me in fits of rage. I've probably said this before in another video, but you are a saint Professor van Biezen!

[Backflipmarine]

So just to clarify what you say at 3:10 - 3:20, you put -vo because the polarity is pre written for you on the schematic? Otherwise you would have to assume I=vo/6?

[djtfan]

For anyone still having trouble figuring out why the professor put a minus sign in the Ohm's law equation, here's how I think about it: yes, you assumed the direction of the current, and you have to calculate the drops according to the assumed direction. BUT, here's the catch, since there's a clear giveaway of the current direction on one of the resistors, and we're going against the literal flow, we're expected to get a negative current anyway. So, how can he set up an Ohm's law equation to get the negative current, since he MUST get the negative current? Well, since the resistance can't be negative, and we don't know what the current is YET, but we DO know it MUST be a negative, he has to setup the equation in such a way to make sure I=V/R turns out negative, which he does by putting a negative sign in front of the voltage.


What's actually happening is, we know the voltage is some *V*, and depending on the direction we assume we add or subtract the *V*, and resistance is *R* .
But the current is *-I*, since we have the wrong direction, and the current WOULD'VE turned out to be negative, had we calculated the *I* any other way.


All in all, it is imperative that *I=V/R* be negative, taking into consideration our wrongly assumed direction, BUT paying attention to the hint we were given.

[ansgug]

This is the reason why I believe the voltage is given a negative sign, excuse me if I'm wrong: since we are assuming that I is going clockwise and we are given the hint that the voltage actually rises over the 6 ohm resistor if we go along with that assumption, then V0 is actually the voltage drop over the 6 ohm resistor in the opposite direction of what we have assumed. Since V0 is the voltage drop in the opposite direction of our assumption, we must use -V0 to change the direction of V0, which gives us the voltage drop over the 6 ohm resistor in the direction that matches the direction that we assumed I to go in. Is this right?

[okzzvil]

Thank you Michel for your effort. God bless you more. and more power.

[explore_Finland]

very nice teaching style. One thing about which i am confused is why the polarities of 6ohm are not as same as 4ohm?

[oliverdiaz]

Hello sir I've been watching your videos and it helped me a lot.

Im a bit confused with the equation u made 12--4i--2Vnot+4--6i=0
Shouldn't it be 12--4i--2vnot+4+6i=0 since you use negative sign in the 4ohms, the 6ohms should be positive.

Please sir correct me if im wrong I really need to understand this topic.

Thank you sir.

[GilLopez-ph]

Would there be a 48v positive across from the -48v, going clockwise? Thank you so much for your videos sir!

[riqyrizqyandra]

why does the voltage between C and D is -48 V ? in the video it says 2Vo, what the reason?

[rosina]

Thanks so much, I really like your teaching. it has really helped me

[stevedasilvaferreira]

The reason for negative on the 6ohm, is because we assume direction clockwise. We then ignore, the signs on the resistors, only considering the signs for the voltages. ... Afterwards we consider the given sign, for the 1 resistor, and using V = IR results in the -V, to rectify our error initially. Solving for I, and replace to get correct answer

[makkerandy]

hi, what would happen if you put one of the Voltage source in a opposite polarity...?

[sldspaceoflearningdevelope]

Hi, I am so confused you have two possibility
First one is
12-4I-2V0+4+V0=0
The second one is
12-4I-2V0+4-V0=0

Now for nodes e and a you use VOltage drop then for node c and d you use voltage

[jeffsam]

i think he just made an error when considering the 6 ohm resistor. since theres always be a voltage drop across a resistor when its placed in the direction of current. the signes on the 6 ohm resistor must be interchanged and KVL equation must be done again. please look this video up again sir. therefore 12-4I-2vnot+4=0 . vnot = 6I . then 12-4I-2*6I+4-6I=0, I=16/22=0.7272A . Vnot = 6*0.7272 = 4.3636V therefore 2Vnot=2*4.3636 = 8.7272V
sir please correct me if im wrong

[trickytricks]

We've already established the video has a mistake in the voltage across the dependent voltage source. But my question becomes, is the correct answer -96V or +96V? With the correct direction of current in mind.

Nevertheless, thank you for the video !!!

[ahmedshaheen]

thank you very much . i am really love your way in physics

[samisiddiqi]

Wouldn't the voltage difference across Vo be negative? We determined I = -8, then corrected it's orientation so the new Current, I = 8.

According to the new current orientation, it is a voltage drop as well so wouldn't Vo = -48?

[MohammadkazimHalimy]

Sir may you explain why the resistor with 6ohm is different from other sources?

[Ching-sxnc]

Sorry sir, i just want to ask about
there are positive sign and negative sign around the 6 ohm resistor, is it refers to the terminal of resistor or the change in voltage along the path from e to a.