1. 1 ECE 221 Electric Circuit Analysis I Chapter 5 Branch Currents Herbert G. Mayer, PSU Status 1/5/2015

2. 2 Syllabus Goal Goal KVL and KCL KVL and KCL Circuit for Sample Problem Circuit for Sample Problem Sample Problem: 3 Equations Sample Problem: 3 Equations Solution via Substitution Solution via Substitution Three Methods? Three Methods?

3. 3 Goal We’ll analyze a simple circuit, named the Sample Problem With 2 constant voltage sources, 3 resistors Computing the 3 branch currents i 1, i 2, and i 3 “Doing it the hard way” using KVL and KCL, and arithmetic substitution Branch Currents can actually be measured by an Amp-meter, inserted into a conducting line We use: R1 = 100 Ω, R2 = 200 Ω, R3 = 300 Ω, and V1 = 3 V, V2 = 4 V And Passive Sign Convention See figure: Sample Problem

4. 4 KVL and KCL Quick reminder: Kirchhoff Voltage Law (KVL): The sum of all voltages around any closed path in a circuit is = 0 V Kirchhoff Current Law (KCL): The sum of all currents at any node in a circuit is = 0 A

5. 5 Circuit for Sample Problem

6. 6 Sample Problem: 3 Equations Original Linear Equations, via KCL and KVL: (1)i1- i2- i3=0 (2)R1*i1+ R2*i2- v1=0 (3)-R2*i2+ R3*i3+ v1 + v2= 0

7. 7 Solution via Substitution Original Linear Equations: (2)’i1=v1 / R1 - i2 * R2 / R1 (3)’i3=-( v1 + v2 ) / R3 + i2 * R2 / R3 (2’) and (3’) in (1), and (1) being negated v1/R1 - i2*R2/R1 -i2 + (v1+v2)/R3 - i2*R2/R3=0 i2*(-1 - R2/R1 – R2/R3) = -v1/R1 - (v1+v2)/R3 i2=4 / 275 i2=0.0145455 A i2=14.5455 mA

8. 8 Solution via Substitution i1=v1 / R1 - i2 * R2 / R1 i1=3 / 100 - ( 4 / 275 ) * 2 i1=1 / 1100 i1=0.000909A i1=0.909 mA i3=i1 – i2= -3/220=0.909 -14.5455 i3=-13.6365 mA

9. 9 Three Methods? We’ll learn other methods for computing circuit parameters One of them using Cramer’s Rule for Matrix Operations And a third one named Mesh-Current Method Then we’ll revisit this same problem, and solve it in different ways