1. Simple Circuits & Kirchoff’s Rules Parallel CircuitSeries Circuit

2. Simple Series Circuits  Each device occurs one after the other sequentially.  The Christmas light dilemma: If one light goes out all of them go out. R1R1 R2R2 + V R3R3

3. Simple Series Circuit - Conservation of Energy IIn a series circuit, the sum of the voltages is equal to zero. V source + V 1 + V 2 + V 3 = 0 Where we consider the source voltage to be positive and the voltage drops of each device to be negative. V source = V 1 + V 2 + V 3 Since V = IR ( from Ohm’s Law ): V source = I 1 R 1 + I 2 R 2 + I 3 R 3 V1V1 V2V2 + V V3V3

4. R1R1 R2R2 + V R3R3 Simple Series Circuit - Conservation of Charge  In a series circuit, the same amount of charge passes through each device. I T = I 1 = I 2 = I 3 I

5. Simple Series Circuit – Determining R equivalent  What it the total resistance in a series circuit? Start with conservation of energy V source = V 1 + V 2 + V 3 V source = I 1 R 1 + I 2 R 2 + I 3 R 3 Due to conservation of charge, I Total = I 1 = I 2 = I 3, we can factor out I such that V source = I Total (R 1 + R 2 + R 3 ) Since V source = I Total R Total : R Total = R Eq = R 1 + R 2 + R 3

6. Simple Parallel Circuit  A parallel circuit exists where components are connected across the same voltage source.  Parallel circuits are similar to those used in homes. R1R1 R2R2 + V R3R3

7. Simple Parallel Circuits  Since each device is connected across the same voltage source: V source = V 1 = V 2 = V 3 V1V1 V2V2 + V V3V3

8. IIn parallel circuits, the total current is equal to the sum of the currents through each individual leg. Consider your home plumbing: YYour water comes into the house under pressure. EEach faucet is like a resistor that occupies a leg in the circuit. You turn the valve and the water flows. TThe drain reconnects all the faucets before they go out to the septic tank or town sewer. AAll the water that flows through each of the faucets adds up to the total volume of water coming into the house as well as that going down the drain and into the sewer. TThis analogy is similar to current flow through a parallel circuit. Simple Parallel Circuits Analogy How Plumbing relates to current

9. Simple Parallel Circuits – Conservation of Charge & Current  The total current from the voltage source (pressurized water supply) is equal to the sum of the currents (flow of water through faucet and drain) in each of the resistors (faucets) I Total = I 1 + I 2 + I 3 + V I Total I1I1 I3I3 I2I2

10. Simple Parallel Circuit – Determining R equivalent  What it the total resistance in a parallel circuit? Using conservation of charge I Total = I 1 + I 2 + I 3 or Since V source = V 1 = V 2 = V 3 we can substitute V source in (1) as follows

11. Simple Parallel Circuit – Determining R equivalent  What it the total resistance in a parallel circuit (cont.)? However, since I Total = V source /R Total substitute in (2) as follows Since V source cancels, the relationship reduces to Note: R total has been replaced by R eq.

12. Kirchoff’s Rules  Loop Rule (Conservation of Energy): The sum of the potential drops (Resistors) equals the sum of the potential rises (Battery or cell) around a closed loop.  Junction Rule (Conservation of Electric Charge): The sum of the magnitudes of the currents going into a junction equals the sum of the magnitudes of the currents leaving a junction.

13. Rule #1: Voltage Rule (Conservation of Energy) R1R1 R2R2 + V R3R3 ΣVΣV V source – V 1 – V 2 – V 3 = 0

14. Rule #2: Current Rule (Conservation of Electric Charge) I1I1 I3I3 I2I2 I 1 + I 2 + I 3 = 0

15. Example Using Kirchoff’s Laws  Create individual loops to analyze by Kirchoff’s Voltage Rule.  Arbitrarily choose a direction for the current to flow in each loop and apply Kirchoff’s Junction Rule. + + R 3 = 5Ω  2 = 5V I1I1 I2I2 I3I3  1 = 3V R 1 = 5Ω R 2 = 10Ω

16. Ex. (cont.)  Apply Kirchoff’s Current Rule (I in = I out ): I 1 + I 2 = I 3 (1)  Apply Kirchoff’s Voltage Rule to the left loop (Σv = 0):  1 – V 1 – V 2 = 0  1 – I 1 R 1 – I 3 R 2 = 0 Substitute (1) for I 3 to obtain:  1 – I 1 R 1 – (I 1 + I 2 )R 2 = 0 (2)

17. Ex. (cont.)  Apply Kirchoff’s Voltage Rule to the right loop:  2 – V 3 – V 2 = 0  2 – I 2 R 3 – I 3 R 2 = 0 Substitute (1) for I 3 to obtain:  2 – I 2 R 3 – (I 1 + I 2 )R 2 = 0 (3)

18. Ex. (cont.)  List formulas to analyze. I 1 + I 2 = I 3 (1)  1 – I 1 R 1 – (I 1 + I 2 )R 2 = 0(2)  2 – I 2 R 3 – (I 1 + I 2 )R 2 = 0(3)  Solve 2 for I 1 and substitute into (3)  1 – I 1 R 1 – I 1 R 2 – I 2 R 2 = 0 – I 1 R 1 – I 1 R 2 = I 2 R 2 –  1 I 1 (R 1 + R 2 ) =  1 - I 2 R 2  1 - I 2 R 2 (R 1 + R 2 ) I 1 =

19. (  1 - I 2 R 2 ) (R 1 + R 2 )  Plug in known values for R 1, R 2, R 3,  1 and  2 and then solve for I 2 and then I 3. Ex. (cont.)  2 – I 2 R 3 – + I 2 R 2 = 0  2 (R 1 + R 2 ) – I 2 R 3 (R 1 + R 2 ) –  1 R 2 + I 2 R 2 2 – I 2 R 2 (R 1 + R 2 ) = 0 [ [  1 - I 2 R 2 (R 1 + R 2 ) R 2 – I 2 R 2 = 0  2 – I 2 R 3 – [ [ Multiply by (R 1 + R 2 ) to remove from denominator. 5V (5Ω+10Ω) – I 2 5Ω (5Ω+10Ω) – 3V(10Ω) + I 2 (10Ω) 2 – I 2 10Ω (5Ω+10Ω) = 0 I 2 = 0.36 A

20. Ex. (cont.)  Plug your answer for I 2 into either formula to find I 1  1 – I 1 R 1 – (I 1 + I 2 )R 2 = 0  What does the negative sign tell you about the current in loop 1? I 1 =  1 - I 2 R 2 (R 1 + R 2 ) I 1 = 3V – (0.36A)(10) (5 + 10) I 1 = -0.04A

21. Ex. (cont.)  Use formula (1) to solve for I 3 I 1 + I 2 = I 3 -0.04A + 0.36A = 0.32A

22. How to use Kirchhoff’s Laws A two loop example: Analyze the circuit and identify all circuit nodes and use KCL. (2)  1  I 1 R 1  I 2 R 2 = 0 (3)  1  I 1 R 1   2  I 3 R 3 = 0 (4) I 2 R 2   2  I 3 R 3 = 0  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 (1) I 1 = I 2 + I 3 Identify all independent loops and use KVL.

23. How to use Kirchoff’s Laws  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 Solve the equations for I 1, I 2, and I 3 : First find I 2 and I 3 in terms of I 1 :  From eqn. (2) From eqn. (3) Now solve for I 1 using eqn. (1):

24. Let’s plug in some numbers  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3  1 = 24 V  2 = 12 V R 1 = 5  R 2 =3  R 3 =4  Then, and I 1 =2.809 A I 2 = 3.319 A, I 3 = -0.511 A