Solve for I1, I2, I3 in Circuit w/ Loop Rule

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The discussion focuses on solving for the currents I1, I2, and I3 in a circuit with three resistors and three ideal batteries using loop rules. The user sets up two equations based on the left and right loops of the circuit, along with a third equation derived from the relationship between the currents. A method involving matrices and row reduction is employed to simplify the system of equations. Ultimately, the solution yields I1 = 0.300 Amps, I2 = 0.500 Amps, and I3 = 0.200 Amps. The user expresses gratitude for the assistance received in solving the problem.
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Homework Statement


In the problem there's a circuit pictured with 3 resistors and three ideal batteries (emf = V). I need to calculate the current through all the resistors (I1,I2,I3).
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Homework Equations


I1 - I2 + I3 = 0
V=RI

The Attempt at a Solution


left loop is 92V - 140*I1 - 210*I2 + 55 = 0
right loop is 51V -35*I3 - 210*I2 + 55v = 0

147V - 140*I - 210*I2 = 0
112V - 210*I2 - 35*I3 = 0

How do I solve for I1, I2, I3?
 
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You have three linear equations and three variables, so you just solve them the same way you would any other linear system. There aren't any rules that you have to follow; you just try to manipulate the equations to reduce the number of unknowns, hopefully reaching 1 after you've incorporated all the equations.

If you want a more mechanical but generally more tedious way of solving linear systems, try Gaussian elimination.
 
I know this is algebra but how do I set up a system of equations with only the two equations:
147V - 140*I1 - 210*I2 = 0
112V - 210*I2 - 35*I3 = 0
 
You've already established a third equation, namely I1 - I2 + I3 = 0
 
Never mind, I got it with matrices.

I1 - I2 + I3 = 0
140*I1 + 210*I2 + 0*I3 = 147
0*I1 + 210*I2 + 35*I3 = 112

matrix:
[ 1 -1 1 0 ]
[ 140 210 0 147 ]
[ 0 210 35 112 ]

row reduced echelon form
[ 1 0 0 .3 ]
[ 0 1 0 .5 ]
[ 0 0 1 .2 ]

answer:
I1 = 0.300 Amps
I2 = 0.500 Amps
I3 = 0.200 Amps

Thanks for the help though.
 
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