2. Circuits With Multiple Sources Suppose you have the following circuit diagram: How do you determine the current throughout the circuit? How do you determine the voltage throughout the circuit? Cannot use equivalent circuits to simplify this…what to do??? - -

3. Kirchoff’s Circuit Laws Gustav Kirchoff– generalized observations of circuits Kirchoff’s Junction Rule (aka: Kirchoff’s current law) Current into node = Current leaving node I in = I out Kirchoff’s Loop Rule (aka: Kirchoff’s voltage law) Sum of voltage around any closed loop must be zero V loop = 0 V Image obtained from: http://www-history.mcs.st-and.ac.uk/BigPictures/Kirchhoff_4.jpeg

4. Applying Kirchoff’s Laws Step 1: Decide on directions for current Electron flow notation: electrons move away from -, move toward + Every node must have at least 1 current going in and 1 going out # currents = # wire lines between nodes - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3

5. Applying Kirchoff’s Laws Step 2: Sign conventions for components Electron flow notation: electrons move away from -, move toward + Every component needs a (-) end and a (+) end - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3 - - - + + +

6. Applying Kirchoff’s Laws Step 3: Perform Junction Rule for every node I in = I out Node A: I 1 + I 2 = I 3 Node B: I 3 = I 1 + I 2 Good! We’re consistent! - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3 - - - + + + A B

7. Applying Kirchoff’s Laws Step 4: Perform Loop Rule for every loop V loop = 0 V Going from - to +: negative voltage (V being lost) Going from + to -: positive voltage (V being gained) Remember: V = I R - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3 - - - + + + A B Loop 1 Loop 2 Loop 3

8. Applying Kirchoff’s Laws Loop 1: - I 1 R 2 - ε 2 - I 1 R 1 + ε 1 = 0 V Loop 2: - I 3 R 3 + ε 2 = 0 V Loop 3: - I 1 R 2 - I 3 R 3 - I 1 R 1 + ε 1 = 0 V - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3 - - - + + + A B Loop 3 Loop 1 Loop 2

9. Applying Kirchoff’s Laws Let’s say we have the following: ε 1 = 6.0 V ε 2 = 1.5 V R 1 = 2.0 Ω R 2 = 3.0 Ω R 3 = 5.0 Ω What is the current passing through the battery ε 2 ? - - I1I1 I1I1 I1I1 I2I2 I2I2 I3I3 I3I3 I3I3 - - - + + + A B Loop 3 Loop 1 Loop 2

10. Applying Kirchoff’s Laws Looking at diagram, I 2 is current passing through ε 2 How to find I 2 ? I 3 = I 1 + I 2 so I 2 = I 3 - I 1 How to find I 3 ? - I 3 R 3 + ε 2 = 0 V so I 3 = ε 2 /R 3 How to find I 1 ? - I 1 R 2 - ε 2 - I 1 R 1 + ε 1 = 0 V so I 1 = (ε 1 - ε 2 ) / (R 1 + R 2 ) What we know: I 3 = I 1 + I 2 - I 1 R 2 - ε 2 - I 1 R 1 + ε 1 = 0 V - I 3 R 3 + ε 2 = 0 V - I 1 R 2 - I 3 R 3 - I 1 R 1 + ε 1 = 0 V

11. Applying Kirchoff’s Laws I 3 = ε 2 /R 3 I 1 = (ε 1 - ε 2 ) / (R 1 + R 2 ) I 1 = (6.0 V - 1.5 V) / (2.0 Ω + 3.0 Ω) I 1 = 4.5 V / 5.0 Ω I 1 = 0.90 A I 2 = I 3 - I 1 What we know: I 3 = I 1 + I 2 - I 1 R 2 - ε 2 - I 1 R 1 + ε 1 = 0 V - I 3 R 3 + ε 2 = 0 V - I 1 R 2 - I 3 R 3 - I 1 R 1 + ε 1 = 0 V Oh no! A negative current! What does that mean? = 1.5 V / 5.0 Ω = 0.30 A = 0.30 A - 0.90 A = -0.60 A Negative current: goes in opposite direction of what you said it did

12. Try Another! Information: ε 1 = 120 V ε 2 = 120 V R 1 = 1.0 Ω R 2 = 2.0 Ω R 4 = 4.0 Ω R 5 = 5.0 Ω R 6 = 6.0 Ω Find currents across: R 1, R 4

13. Let’s Step It Up A Bit! Information: ε 1 = 120 V ε 2 = 120 V R 1 = 1.0 Ω R 5 = 5.0 Ω R 2 = 2.0 Ω R 6 = 6.0 Ω R 3 = 3.0 Ω R 8 = 8.0 Ω R 4 = 4.0 Ω Find currents across: R 3, R 6 19.0 A 15.9 A