Finding load current using kirchoffs

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matthew187

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Hi there attached is a task where i had to calculate the value of the load using kirchoffs law I would be grateful if someone can check this to see if its correct, its part of an assignment question.
 

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  • Circuit Two Kirchoffs Law.pdf
    457.7 KB · Views: 104

By simply adding eq. 1 & 2, you get

5v = 0ΩI1 + 6ΩI2

6ΩI2 = 5v

I2 = 5v/6Ω = 0.833 A

From here, Eq. 1 becomes,
3v = 4ΩI1 - 1Ωx0.833 A

4ΩI1 = 3v + 0.833v

I1 = 3.833v/4Ω = 0.958 A

And,
I1 - I2 = 0.958 A - 0.833 A = ~0.125 A



I think there is a mistake in Eq. 2 muitiplying with 4...
Correct me if I am wrong...
 

I have done it again and this is what I got the first attachment had information missing.
 

Attachments

  • Circuit Two Kirchoffs Law.pdf
    459.3 KB · Views: 116

Are you adding these two linear equations????



Then 4ΩI1 + (-16ΩI1) = -12ΩI1
I think you missed that part...
 

I've done it again can you please check if this is correct please many thanks.
 

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  • Circuit Two Kirchoffs Law.pdf
    308.1 KB · Views: 90

I have rewritten the answer. Your answer is correct, but has some mistakes in equations....

 

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