Kirchhoff's rules applied to a two-loop circuit give I1 = I2 + I3, 18 = 6I1 + 3I2 and 45 = 6I1 + 2I3, with every current in amperes. What is I1?

1. Three independent equations are enough for three unknown currents, so solve by substitution. 2. From 18 = 6I1 + 3I2, rearranging gives 3I2 = 18 minus 6I1, so I2 = 6 minus 2I1. 3. From 45 = 6I1 + 2I3, rearranging gives 2I3 = 45 minus 6I1, so I3 = 22.5 minus 3I1. 4. Substituting both into the junction equation gives I1 = (6 minus 2I1) + (22.5 minus 3I1). 5. Collecting terms gives I1 = 28.5 minus 5I1, so 6I1 = 28.5 and I1 = 4.75 A. 6. Checking, I2 = 6 minus 9.50 = -3.50 A and I3 = 22.5 minus 14.25 = 8.25 A, and -3.50 A plus 8.25 A returns 4.75 A. 7. 8.25 A is the value of I3, the commonest mix-up once all three currents are on the page. 8. 5.70 A follows from dropping one of the 6I1 terms, and 2.85 A from dividing 28.5 by 10 instead of by 6. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.3 Kirchhoff's Rules_