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

AI Thread Summary
Mike is struggling with a circuit problem involving Kirchhoff's laws and has shared his equations for analysis. He initially merged two 20-ohm resistors incorrectly and derived equations that led to a negative current for one variable. After clarification on his approach, he corrected his equations using loop analysis, resulting in the correct values for the currents Ia and Ib. The final calculations indicate a load current of approximately 2.67A and a voltage across the load resistor of about 26.67V. The discussion highlights the importance of clearly stating intermediate steps to avoid errors in circuit analysis.
physicsfun_12
Messages
24
Reaction score
0

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
    15.1 KB · Views: 440
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top