Solving a Complex Kirchoff's Law Problem - Mike's Request for Help

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Homework Help Overview

The discussion revolves around a problem involving Kirchhoff's laws in a circuit analysis context. The original poster, Mike, is attempting to solve equations related to current in a circuit but is encountering difficulties, particularly with negative current values.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Mike attempts to simplify the circuit by merging resistances and sets up equations based on Kirchhoff's laws. Participants question the setup of the equations, the direction of currents, and the origin of certain values in the equations.

Discussion Status

Participants are actively engaging with Mike's approach, providing clarifications and checking the validity of his equations. There is a recognition of the need for clear intermediate steps in the calculations, and Mike acknowledges the importance of this feedback.

Contextual Notes

Mike has shared a circuit diagram and is working under the constraints of homework rules, which may limit the level of assistance he can receive. The discussion reflects an ongoing exploration of the problem rather than a definitive solution.

physicsfun_12
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Homework Statement


Hello there, hope you are well.

I am having trouble solving this problem involving Kirchoff's laws.

I have attached a copy of the circuit and problem to this post in jpeg format.

Thanks again in advance for any help.

Mike


Homework Equations





The Attempt at a Solution


Well I merged to the two 20 ohms into one 10ohm and said that the total voltage in any loop equals zero.

so 60=20Ia + 10Ib

and 20=20Ib + 10Ia

I have tried solving this many times and keep getting a negative current for Ib. My original equation must be wrong.

Thanks a lot for any help or input,

Mike
 

Attachments

  • Kirchhoff's Laws_question.jpg
    Kirchhoff's Laws_question.jpg
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Hi Mike! :smile:
physicsfun_12 said:
… 60=20Ia + 10Ib

and 20=20Ib + 10Ia

Which loop has Ia, and which has Ib, and in which direction? :confused:

And what about Ic ?

(and where do those 20s on the RHS come from?)
 
I think I've got it now.

I was using loop analysis so that's why there is only two currents.

The 20s came from 60=10Ia + 10(Ia+Ib) which expands to give 20Ia + 10Ib

You can work out the currents using simultaneous equations, giving a Ia=3.333A and Ib=-0.6666A

I of the load resistor, Il=Ia+Ib = 2.6667A

So Vl=2.6667*10 = 26.667V

Does this look about right? Thanks a lot for your help

Mike
 
Last edited:
physicsfun_12 said:
The 20s came from 60=10Ia + 10(Ia+Ib) which expands to give 20Ia + 10Ib

ohhh! you should always state an intermediate step like that :rolleyes:

(apart from anything else, you stand a good chance of making a mistake with ±s if you don't write this stuff out clearly)

Yes, the answer looks ok now. :smile:
 
Sorry about that, you are absolutely right... that's most probably why I was getting it wrong in the first place!

Thanks for your help and take care,

Mike
 

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