Answer to Ohm's Law (Kirchhoff's Law) 2nd Question

ericmark · 22 Feb 2009

Untitled

ericmark
22 Feb 2009
2

So people can answer without seeing my answer I have posted as separate post. Until the first post is read this will make little sense.

Sorry I1 and I2 have not got the 1's and 2's as a sub text too late once I realised I could not post sub text.

Eric

Dividing the circuit into two loops
E1=I1xR1+(I1+I2)/R3
E2=I2xR3+(I1+I2)/R2
Replacing with values
8=I1x1+(I1+I2)/5
3=I2x2+( I1+I2)/5
We now have the makings of a simultaneous equation
6I1 + 5I2 = 8
5I1 +7I2 = 3
To get I2 to counsel out we multiply by 7 and 5 so
42I1 + 35I2 = 56
25I1 + 35I2 = 15
Therefore 17I1 = 41
I1 therefore = 2.411765 Amp
Substituting I1 with 2.411765
6 x 2.411765 + 5I2 = 8
14.470588 + 5I2 = 8
5I2 = 6.470588
I2 = 1.294118 Amp

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Answer to Ohm's Law (Kirchhoff's Law) 2nd Question

ericmark

Untitled

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