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More Electronics Help! Kirchhoff's law!

FarmerGiles97
By FarmerGiles97
in Hobby Electronics
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Posted

Im at the end of a college assignment, and im working on Kirchhoff's law.  Ive ended up at a simultaneous equation and im stuck on what to do next. The last time i worked with algebra was when i left school ( 5 years ago)

Would anyone be able to point me in the right direction and help me? 

 

Sorry for the bad hand writing.

 

https://imgur.com/a/LLXqbZT

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I hated that friggin law. If this hasn't been answered by the time I get home I'll help you out as best I can.

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Posted

Oh dear... this gives me some pretty traumatizing flashbacks :P

 

Either way you're almost there:

10 = 28I1 + 9I2

15 = 9I1 + 16I2

 

9l2 = 10 - 28I1

9l1 = 15 - 16l2

 

I2 = 10/9 - 28I1/9

I1 = 15/9 - 16I2/9

 

Now I substitute I2 in the second equation with the first equation:

 

I1 = 15/9 - 160/81 + 16*28I1/81 = -25/81 + 448I1/81

 

From here I can work out the value of I1:

 

367I1/81 = 25/81

367I1 = 25

I1 = 25/367 ~= 0.07

 

Using this result I can find the value of I2 as well by using the first equation:

 

I2 = 10/9 - 28(0.07)/9

I2 = (10 - 28 * 0.07) / 9 ~= 0.9

 

From here, I3 is just:

 

I3 = I1 + I2 ~= 0.97

 

And there you go. CHECK MY CALCULATIONS, I am merely human and I can press the wrong buttons on the calculator - but it should be correct.

 

-EDIT-

it was not correct. I fixed it :P

Don't ask to ask, just ask... please 🤨

sudo chmod -R 000 /*

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28 minutes ago, Sauron said:

And there you go. CHECK MY CALCULATIONS, I am merely human and I can press the wrong buttons on the calculator - but it should be correct.

Your reply came up while I was working it out myself. But we have different answers.

This is what I did.

I rewrote I1 as A and I2 as B, so I wouldn't mix them up.

 

I got I1 = 25/367 and I2 = 330/367

 

I checked it using an online tool to be sure and I seem to be right.

 

Also I wouldn't have thought that I1/2 could be negative? I remember the Simultaneous Equations a lot better than Kirchhoff's law, probably because we got the formulas on a sheet in physics!

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6 minutes ago, Madgemade said:

Your reply came up while I was working it out myself. But we have different answers.

There you go, I did it in a bit of a rush :S your answer is probably correct.

6 minutes ago, Madgemade said:

Also I wouldn't have thought that I1/2 could be negative? I remember the Simultaneous Equations a lot better than Kirchhoff's law, probably because we got the formulas on a sheet in physics!

Current can absolutely be negative, it just means it runs in the opposite direction to the frame of reference.

Don't ask to ask, just ask... please 🤨

sudo chmod -R 000 /*

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20 minutes ago, Sauron said:

Current can absolutely be negative, it just means it runs in the opposite direction to the frame of reference.

Ahh, because of the resisters I was think of resistance and I don't think you can have negative resistance resister. Negative current yes, I is current so yeah should have realised that one. Hopefully OP sees how to do the simulations equations, in math class they were much harder than that, often with indices or quadratics...

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24 minutes ago, Madgemade said:

Ahh, because of the resisters I was think of resistance and I don't think you can have negative resistance resister. Negative current yes, I is current so yeah should have realised that one. Hopefully OP sees how to do the simulations equations, in math class they were much harder than that, often with indices or quadratics...

Im doing all the maths stuff along aide my course. I just havent done it, in so long i find it hard to remember lmao. 

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30 minutes ago, Madgemade said:

Ahh, because of the resisters I was think of resistance and I don't think you can have negative resistance resister. Negative current yes, I is current so yeah should have realised that one. Hopefully OP sees how to do the simulations equations, in math class they were much harder than that, often with indices or quadratics...

I rechecked my stuff, you got the right answer. I swapped a - for a +. Typical...

Don't ask to ask, just ask... please 🤨

sudo chmod -R 000 /*

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Ngl i really dont get how i get these answers and how i can answer the question? Is there any youtube videos to follow? 

 

 

I used this one, but i dont believe i can solve the equation the way he did it. 

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To answer my assignment questions, 

 

Task A) 

The current in the resistors would be as follows yes? 

 

R1= 0.07

R2= 0.9

RL=0.97

 

And Task B? 

 

Power dissapated would be 25 volt? Or no? 

 

If i am wrong how would i answer this question? 

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Right, i just read about the second question. 

 

P = I2R? 

 

So 

 

0.972 x 9 =8.47W? 

 

Am i right or no? 

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  • 2 weeks later...
Posted
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On 11/7/2018 at 6:19 PM, Sauron said:

Oh dear... this gives me some pretty traumatizing flashbacks :P

 

Either way you're almost there:

10 = 28I1 + 9I2

15 = 9I1 + 16I2

 

9l2 = 10 - 28I1

9l1 = 15 - 16l2

 

I2 = 10/9 - 28I1/9

I1 = 15/9 - 16I2/9

 

Now I substitute I2 in the second equation with the first equation:

 

I1 = 15/9 - 160/81 + 16*28I1/81 = -25/81 + 448I1/81

 

From here I can work out the value of I1:

 

367I1/81 = 25/81

367I1 = 25

I1 = 25/367 ~= 0.07

 

Using this result I can find the value of I2 as well by using the first equation:

 

I2 = 10/9 - 28(0.07)/9

I2 = (10 - 28 * 0.07) / 9 ~= 0.9

 

From here, I3 is just:

 

I3 = I1 + I2 ~= 0.97

 

And there you go. CHECK MY CALCULATIONS, I am merely human and I can press the wrong buttons on the calculator - but it should be correct.

 

-EDIT-

it was not correct. I fixed it :P

I made a mistake in my working out, which in turn fucked up your working out. 

 

'Either way you're almost there:

10 = 28I1 + 9I2

15 = 9I1 + 16I2'

This is the part i fucked up. 

 

It should have been 

 

15= 9i1 + 23i2. 

 

Now im confused on how i work out the equation. 

 

 

 

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1 hour ago, JordanHopkins97 said:

Now im confused on how i work out the equation.

You just need to get the two variables on opposite sides of the equation, like I did here:

Quote

10 = 28I1 + 9I2

15 = 9I1 + 16I2

 

9l2 = 10 - 28I1

9l1 = 15 - 16l2

 

I2 = 10/9 - 28I1/9

I1 = 15/9 - 16I2/9

the method works regardless of the numbers you use. In your case it's:

Quote

10 = 28I1 + 9I2

15 = 9I1 + 23I2

 

9l2 = 10 - 28I1

9l1 = 15 - 23l2

 

I2 = 10/9 - 28I1/9

I1 = 15/9 - 23I2/9

Then you can proceed just like I did, you just need to change the numbers to the correct ones.

Don't ask to ask, just ask... please 🤨

sudo chmod -R 000 /*

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6 hours ago, Sauron said:

You just need to get the two variables on opposite sides of the equation, like I did here:

the method works regardless of the numbers you use. In your case it's:

Then you can proceed just like I did, you just need to change the numbers to the correct ones.

367I1/81 = 25/81

367I1 = 25

I1 = 25/367 ~= 0.07

 

I just dont get where you got the 367 from? 

 

Thats the part i am confused on?

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3 minutes ago, JordanHopkins97 said:

367I1/81 = 25/81

367I1 = 25

I1 = 25/367 ~= 0.07

 

I just dont get where you got the 367 from? 

 

Thats the part i am confused on?

Oh, ok - I might have gone too fast there.

 

(with the old numbers:)

I1 = 15/9 - 160/81 + 16*28I1/81 = -25/81 + 448I1/81

I1 = -25/81 + 448I1/81

81I1/81 = -25/81 + 448I1/81 (81/81 is the same as 1)

448I1/81 - 81I1/81 = 25/81

(448 - 81)I1/81 = 25/81

367I1/81 = 25/81

 

In the end all I'm doing is adding 25/81 and subtracting I1 from both sides.

Don't ask to ask, just ask... please 🤨

sudo chmod -R 000 /*

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