Calculating Current Supplied by Voltage Source (Complex algebra)

AI Thread Summary
The discussion focuses on calculating the total resistance (Rt) in a circuit using complex algebra, specifically through the equation 1/Rt = 1/(6+j8)Ω + 1/(9-j12)Ω. Participants confirm that the approach is correct but suggest using the term 1/Z for admittance instead of 1/Rt. They recommend formatting equations with LaTeX for clarity and emphasize the importance of considering phase shifts in current and voltage when calculating power dissipation. Additionally, there is a mention of potential typographical errors in the initial equation. The conversation aims to clarify the algebra involved in the calculations.
jackrichie
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I've attached the question that I am referring to.

I believe I'm heading in the right direction with this one by stating that:

1/Rt = 1/(6+j8)Ω + 1/(9-j12)Ω

But I am confusing myself with my algebra.

Any help is appreciated
 

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jackrichie said:
I've attached the question that I am referring to.

I believe I'm heading in the right direction with this one by stating that:

1/Rt = 1/(6+j8)Ω + 1/(9-j12)Ω

But I am confusing myself with my algebra.

Any help is appreciated

Welcome to the PF.

[STRIKE]Looks good so far[/STRIKE]. (Oops, I missed the typo that jambaugh caught below) Go ahead and keep working though the algebra, and post your work so we can check it.

BTW, it's best if you format the equations with LaTeX, instead of having to carry a bunch of parenthesis along with plain text. There is an introduction to LaTeX in the PF FAQ thread in the Feedback Forum here:

https://www.physicsforums.com/showthread.php?t=617567

:smile:
 
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If your 6 is a typo and should be an 8, then fine but I'd call it 1/Z not 1/Rt. 1/Z is the Admittance = 1/Imped.
Then apply impedance to Ohm's law (ac form)
V = IZ

where V = V_0 exp(j \omega t), I = I_0 exp(j \omega t + j \phi ).

You get I as a complex multiple of V and so can get its magnitude to find peak current. For power dissipation you will need to take into account the phase shift \phi since current and voltage won't peak at the same time.
 
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