/sci/ - Science & Math » Thread #12305704

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What do I even do here?

Anonymous Wed 04 Nov 21:21:58 2020 No.12305704 View Reply Original Report

Quoted By: >>12305715 >>12305821

Do I need to use Lagrange Multipliers?

Anonymous

Anonymous Wed 04 Nov 2020 21:25:47 No.12305715 Report

Quoted By: >>12305750

>>12305704
Looks like it

Anonymous

Anonymous Wed 04 Nov 2020 21:36:51 No.12305750 Report

Quoted By: >>12305775 >>12305783 >>12305810

>>12305715
TA is being a troll, I could have done it if it just gave a clear cut equation with zero paragraphs.

Do you do $i_1+i_2=i$ , plug that into energy loss equation, then do partials? I'm confused on what happens next.

Anonymous

Anonymous Wed 04 Nov 2020 21:45:21 No.12305775 Report

Quoted By:

>>12305750
If you know how to set up lagrange multipliers then you also know what objective function f subject to constraint equation g look like, the rest is just plugging in.

Anonymous

Anonymous Wed 04 Nov 2020 21:47:07 No.12305783 Report

Quoted By:

>>12305750
Looks like it, the partials should be equal to zero.

Anonymous

Anonymous Wed 04 Nov 2020 22:02:27 No.12305810 Report

Quoted By:

>>12305750
L = I1^2*R1 + I2^2*R2 + h*(I1+I2-I)

dL/dI1 = 2 I1 R1 + h = 0
dL/dI2 = 2 I2 R2 + h = 0
dL/dh = I1+I2-I

I1 = I2 R2/R1
I2 = I1 R1/R2

I2 = I/(1+R2/R1)
I1 = I/(1+R1/R2)

Anonymous

Anonymous Wed 04 Nov 2020 22:05:40 No.12305821 Report

Quoted By:

>>12305704
Total energy loss is
$I_1^2 R_1 + I_2^2 R_2 = I_1^2 R_1 + (I - I_1)^2 R_2$
Now differentiate it with respect to $I_1$ and set it equal to 0.
$0 = 2 I_1 R_1 - 2 (I - I_1) R_2 \\<br /> I R_2 = I_1 (R_1 + R_2)\\<br />I_1 = \frac{I}{1 + \frac{R_1}{R_2}} \\<br />I_2 =\frac{I}{1 + \frac{R_2}{R_1}}$

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