☑️14 - Mesh Analysis with Voltage Sources

[@ahmadwehbi6233]

what is the name of the book so i can solve more problems ?

[@SkanCityAcademy_SirJohn]

Fundamentals of electric circuit, Alexander and Sadiku

[@bartukacar5629]

Amazing explanation

[@SkanCityAcademy_SirJohn]

Thanks so so much.

[@Mcsfastgaming12]

sir can i ask if why does 2(i1) is positive if the polarity is + to - = negative? Can you explain me why sir?

[@SkanCityAcademy_SirJohn]

It depends on the convention you may be using. For conversational flow, + to - is positive.

[@hamzaatheb]

Could you please tell me why in mesh no 2 You but (-12 ×(i1-i2) I think should +12×(i2-i1) Thanks

[@SkanCityAcademy_SirJohn]

This is because I to took the clockwise direction which means that the direction opposes current in 12ohms. Hence it's supposed to be -12(i1-i2)

[@NokulungaMacu-uq8qu]

I understand the -12. But I thought it's going to be i2-i1 because it's the 2nd loop. Guys this is confusing because other tutors says i2 take the lead on loop 2. Please clarify thanks 1st example

[@SkanCityAcademy_SirJohn]

@NokulungaMacu-uq8qu kindly state the time in the video for easy reference

[@BlessingsKanyika-v4x]

I didn't know you also have electrical analysis tutorials oh 🥹 this channel is enough to save a semester 😅🥹

[@SkanCityAcademy_SirJohn]

Aww great

[@yahboi8377]

may i ask in example 2 why did you on mesh 1 20=voltage drops isnt it 20-4(i1-i3)-2(i1-i2)=0 ??? please reply

[@SkanCityAcademy_SirJohn]

Yes it's actually the same thing. Transpose voltage drops to the right hand side, and it's exactly what you see in the video.

[@thatschemistry9264]

How do i know when to divide the equation and what no number to use

[@SkanCityAcademy_SirJohn]

We divide the equation through by a constant when the coefficients of the variables a large. We usually don't want to be dealing with large values. For the number, divide by the coefficient of the variable with the highest power.

[@LeandrieBermudo]

In the example 1? Where did you get the 6 to divide 36?

[@SkanCityAcademy_SirJohn]

From line 3, I just divided the the greatest common factor between 36, 18 and 12, which is 6.

[@CL-sq1bs]

In example 1, how do you solve for the voltages of each resistors then?

[@SkanCityAcademy_SirJohn]

After finding the mesh currents, you just use ohms law, V = iR, to find the voltages across each resistor. Multiply the current in the resistor by the resistance of the resistor.

[@RayanSahib-z2h]

in second example 10 i0 shouldn't be negative at second mesh?

[@SkanCityAcademy_SirJohn]

No please, it's supposed to be negative, we assume conventional flow of current as this since current in leaving the positive terminal, and in line with i2 it is positive.

[@PrettyCharm_MFAHH]

What difference does it make if I° is given and we have to find v° ???

[@SkanCityAcademy_SirJohn]

It depends on the question anyways

[@Nansomiraclechurch]

Mesh analysis on dependent sources

[@nitiponsupanitinun8091]

Could you please explain the 1st example , current is zero in second loop, how's it possible. (Sorry for my English)

[@SkanCityAcademy_SirJohn]

That is actually what it means, I don't know how to further explain to you. I guess you don't understand why we should get a zero in our calculation.

[@nitiponsupanitinun8091]

@@SkanCityAcademy_SirJohnok, thanks for your answer.

[@SkanCityAcademy_SirJohn]

You are most welcome. Where do you watch this video from?

[@nitiponsupanitinun8091]

@@SkanCityAcademy_SirJohn Thailand.

[@SkanCityAcademy_SirJohn]

@@nitiponsupanitinun8091 thatsgreat, nice to hear from you from Thailand, which program do you read in school?

[@rickp9714]

in example 2 how did you determine i3= d3/d ?

[@SkanCityAcademy_SirJohn]

okay, so that is actually a formula being provided from the cramer's rule, since we want to find io which is equal to i3, we dont need to do i1 and i2 first, we can just skip to i3 = d3/d to get our solution, in the lesson i explained how to get i1,i2 and i3 from cramer's rule. Hope you got it? thank you.

[@Kidvini2025]

There is a mistake in example 2.....in mesh 2, you did KVL with the current 10io......kvl is for voltage

[@SkanCityAcademy_SirJohn]

It's not wrong. That is a dependent voltage source, which depends on current value io. So it's not a current source. But a current controlled voltage source.

[@timdavid5638]

Wonderful

[@SkanCityAcademy_SirJohn]

Thanks so much

[@ДіанаПастушенко]

but in example 2 you have i3=I_0, not i3=i0, how can you substitute it? uppercase I 0 is not the same as lowercase i0, isn't it?

[@SkanCityAcademy_SirJohn]

I understand you, but in this case, it is the same.

[@idrisabdullahi9348]

but the first example its seems to be 36= -2i1 -12(i1-i2) - 4i1 since resistance always resist the flow of current. Furthermore the current is entering from the positive terminal of the resistance and move out to the negative terminal thats why it should be negative

[@SkanCityAcademy_SirJohn]

No please, look at the direction of the current flow of i1, it is in the clockwise direction, which is the same as the direction we are taking and mind you, we consider the conventional flow of current.

[@idrisabdullahi9348]

@@SkanCityAcademy_SirJohn okay, noted

[@SkanCityAcademy_SirJohn]

Thanks for watching...

[@jerrychen7310]

Why I0 = i3 and equal i0, I can’t figure out 16:41

[@SkanCityAcademy_SirJohn]

Yes because Io is the same current in mesh 3 which is i3

[@timdavid5638]

Nice ❤️❤️❤️❤️

[@SkanCityAcademy_SirJohn]

Thank you

[@KafuKemeh]

For the first example the mesh 1 how did u get the 4 ohms to have the current of I1 while the 12 ohms the current is (I1-I2)

[@SkanCityAcademy_SirJohn]

It's simply because 12 ohms is in the middle branch, and in the middle branch, the total current is as a results of i1 and i2. Since i2 oppose i1 The total current is i1-i2

[@ndzalamankwinika2999]

Hy sir where can i send a question that i have im writing soon😢

[@SkanCityAcademy_SirJohn]

Kindly join my telegram group for tertiary students and text me directly

[@ndzalamankwinika2999]

@@SkanCityAcademy_SirJohn thanks a lot sir

[@duniasy]

​@@SkanCityAcademy_SirJohnWhere can I find the group link?

[@realmafia8611]

you made fail bro, the questions there were literally the same smh...

[@SkanCityAcademy_SirJohn]

Can you explain further?

[@MichealMadiri]

u just failed on ur own .... this is very correct

[@SkanCityAcademy_SirJohn]

@MichealMadiri great 😃