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After combining terms and simplifying, we now have three equations, three unknowns; the rest is "just algebra." Junction a: I3 - I - I, = 0 --eq. 1 The "green" loop - 30 I, + 45 - 41 I3 = 0 --eq. 2 The "blue" loop 41 Iz -130 + 21 Iz = 0 --eq. 3 skip the algebra! Make sure to use voltages in V and resistances in 2. Then currents will be in A. You can see the solution in part 7 of today's lecture notes (which will not be presented in class). It's a pain. Generally, with the homework problems, you can easily solve directly for one variable, leaving you with two equations and two unknowns.

After combining terms and simplifying, we now have three equations, three unknowns; the rest is "just algebra." Junction a: I3 - I - I, = 0 --eq. 1 The "green" loop - 30 I, + 45 - 41 I3 = 0 --eq. 2 The "blue" loop 41 Iz -130 + 21 Iz = 0 --eq. 3 skip the algebra! Make sure to use voltages in V and resistances in 2. Then currents will be in A. You can see the solution in part 7 of today's lecture notes (which will not be presented in class). It's a pain. Generally, with the homework problems, you can easily solve directly for one variable, leaving you with two equations and two unknowns.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Find the values of I3, I2, and I1 using this 3 equations. Thank You!

After combining terms and simplifying, we now have three equations, three unknowns; the rest is "just algebra." Junction a: I3 - I, - I, = 0 --еq. 1 The "green" loop - 30 I, + 45 - 41 I, = 0 --eq. 2 The "blue" loop 41 I3 -130 + 21 I, = 0 --eq. 3 skip the algebra! Make sure to use voltages in V and resistances in Q. Then currents will be in A. You can see the solution in part 7 of today's lecture notes (which will not be presented in class). It's a pain. Generally, with the homework problems, you can easily solve directly for one variable, leaving you with two equations and two unknowns. еen
Transcribed Image Text:After combining terms and simplifying, we now have three equations, three unknowns; the rest is "just algebra." Junction a: I3 - I, - I, = 0 --еq. 1 The "green" loop - 30 I, + 45 - 41 I, = 0 --eq. 2 The "blue" loop 41 I3 -130 + 21 I, = 0 --eq. 3 skip the algebra! Make sure to use voltages in V and resistances in Q. Then currents will be in A. You can see the solution in part 7 of today's lecture notes (which will not be presented in class). It's a pain. Generally, with the homework problems, you can easily solve directly for one variable, leaving you with two equations and two unknowns. еen
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