Simple Circuits & Kirchoff’s Rules

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

2. R1 + V R2 R3 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.

3. V1 + V V2 V3 Simple Series Circuit - Conservation of Energy • In a series circuit, the sum of the voltages is equal to zero. Vsource + V1 + V2 + V3 = 0 Where we consider the source voltage to be positive and the voltage drops of each device to be negative. Vsource = V1 + V2 + V3 Since V = IR (from Ohm’s Law): Vsource = I1R1 + I2R2 + I3R3

4. R1 + V R2 I R3 Simple Series Circuit - Conservation of Charge • In a series circuit, the same amount of charge passes through each device. IT = I1 = I2 = I3

5. Simple Series Circuit – Determining Requivalent • What it the total resistance in a series circuit? • Start with conservation of energy Vsource = V1 + V2 + V3 Vsource = I1R1 + I2R2 + I3R3 • Due to conservation of charge, ITotal = I1 = I2 = I3, we can factor out I such that Vsource = ITotal (R1 + R2 + R3) • Since Vsource = ITotalRTotal: RTotal = REq = R1 + R2 + R3

6. + V R2 R3 R1 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.

7. + V V2 V3 V1 Simple Parallel Circuits • Since each device is connected across the same voltage source: Vsource = V1 = V2 = V3

8. Simple Parallel Circuits AnalogyHow Plumbing relates to current • In parallel circuits, the total current is equal to the sum of the currents through each individual leg. • Consider your home plumbing: • Your water comes into the house under pressure. • Each faucet is like a resistor that occupies a leg in the circuit. You turn the valve and the water flows. • The drain reconnects all the faucets before they go out to the septic tank or town sewer. • 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. • This analogy is similar to current flow through a parallel circuit.

9. ITotal I2 I3 I1 + V ITotal 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) ITotal = I1 + I2 + I3

10. Simple Parallel Circuit – Determining Requivalent • What it the total resistance in a parallel circuit? • Using conservation of charge ITotal = I1 + I2 + I3 or • Since Vsource = V1 = V2 = V3 we can substitute Vsource in (1) as follows

11. Simple Parallel Circuit – Determining Requivalent • What it the total resistance in a parallel circuit (cont.)? • However, since ITotal = Vsource/RTotal substitute in (2) as follows • Since Vsource cancels, the relationship reduces to Note: Rtotal has been replaced by Req.

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. ΣV R1 + V R2 R3 Rule #1: Voltage Rule (Conservation of Energy) Vsource – V1 – V2 – V3 = 0

14. Rule #2: Current Rule (Conservation of Electric Charge) I1 I2 I3 I1 + I2 + I3 = 0

15. I1 I2 R1 = 5Ω R3 = 5Ω R2 = 10Ω + + 1 = 3V 2 = 5V I3 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.

16. Ex. (cont.) • Apply Kirchoff’s Current Rule (Iin = Iout): I1 + I2 = I3 (1) • Apply Kirchoff’s Voltage Rule to the left loop (Σv = 0): 1 – V1 – V2 = 0 1– I1R1 – I3R2 = 0 • Substitute (1) for I3 to obtain: 1 – I1R1 – (I1 + I2)R2 = 0 (2)

17. Ex. (cont.) • Apply Kirchoff’s Voltage Rule to the right loop: 2 – V3 – V2 = 0 2– I2R3 – I3R2 = 0 • Substitute (1) for I3 to obtain: 2 – I2R3 – (I1 + I2)R2 = 0 (3)

18. I1 = Ex. (cont.) • List formulas to analyze. I1 + I2 = I3 (1) 1 – I1R1 – (I1 + I2)R2 = 0 (2) 2 – I2R3 – (I1 + I2)R2 = 0 (3) • Solve 2 for I1 and substitute into (3) 1 – I1R1 – I1R2– I2R2 = 0 – I1R1 – I1R2 = I2R2 – 1 I1 (R1 + R2) = 1 - I2R2 1 - I2R2 (R1 + R2)

19. 2 – I2R3 – + I2 R2 = 0 [ 1 - I2R2 (R1 + R2) [ 2 – I2R3 – R2 – I2R2 = 0 Ex. (cont.) [ (1 - I2R2) (R1 + R2) • Plug in known values for R1, R2, R3, 1 and 2 and then solve for I2 and then I3. [ Multiply by (R1 + R2) to remove from denominator. 2(R1 + R2) – I2R3 (R1 + R2) – 1R2 + I2R22– I2R2 (R1 + R2) = 0 5V(5Ω+10Ω) – I25Ω(5Ω+10Ω) – 3V(10Ω) + I2(10Ω)2– I210Ω(5Ω+10Ω) = 0 I2 = 0.36 A

20. 1 - I2R2 (R1 + R2) 3V –(0.36A)(10) (5 + 10) I1 = I1 = Ex. (cont.) • Plug your answer for I2 into either formula to find I1 • 1 – I1R1 – (I1 + I2)R2 = 0 • What does the negative sign tell you about the current in loop 1? I1 = -0.04A

21. Ex. (cont.) • Use formula (1) to solve for I3 • I1 + I2 = I3 • -0.04A + 0.36A = 0.32A

22. R1 I3 e 2 I2 I1 R2 R3 e 1 How to use Kirchhoff’s Laws A two loop example: • Analyze the circuit and identify all circuit nodes and use KCL. • (1)I1 = I2 + I3 • Identify all independent loops and use KVL. (2)e1-I1R1-I2R2 = 0 (3) e1-I1R1-e2-I3R3 = 0 (4) I2R2-e2-I3R3 = 0

23. R1 I3 I2 I1 e 2 R2 R3 e 1 From eqn. (2) From eqn. (3) Þ How to use Kirchoff’s Laws • Solve the equations for I1, I2, and I3: • Firstfind I2 and I3 in terms of I1: Now solve for I1using eqn. (1):

24. R1 I3 I2 I1 e 2 R2 R3 e 1 Let’s plug in some numbers e1 = 24 V e 2 = 12 V R1= 5W R2=3W R3=4W Then,and I1=2.809 A I2= 3.319 A, I3= -0.511 A