1. 10.13 Series & Parallel Circuits

2. Series Circuits electric circuit in which the loads are arranged one after another in series. A series circuit has only one path along which electrons can flow. If that pathway is interrupted, the whole circuit cannot function. 1 2 3

3. Series Circuits adding an extra bulb to a series string of lights makes all the bulbs dimmer. electrons use up all their potential difference going around a series circuit no matter how many loads are in the circuit. Each load will use part of the total potential difference, depending on how much it resists the flow of electrons.

4. Series Circuits the amount of current is the same in all parts of a series circuit. if more resistors or loads are added, it will increase the total resistance of the circuit. this decreases the current.

5. Parallel Circuits an electric circuit in which the loads are arranged so that electrons can flow along more than one path. the points where a circuit divides into different paths or where paths combine are called junction points An interruption or break in one pathway does not affect the other pathways in the circuit. 1 2 3

6. Parallel Circuits adding a new pathway with more resistors or loads does not affect the resistance in any of the other pathways. adding extra resistors or loads in parallel circuit decreases the total resistance of the circuit.

7. Parallel Circuits each electron has the same amount of energy, and electrons must lose all their energy on the path they are on. this is why the potential difference across parallel resistors will always be the same, even though the resistors themselves are different

8. Parallel Circuits since a parallel circuit has many pathways for current to flow, the current coming from the battery is not the same as the current along each pathway at each junction point, some electrons go one way while others go another

9. Summary of Current, potential difference, and resistance in series and parallel circuits. CircuitPotential Difference Series circuit Each load uses a portion of the total potential difference supplied by the battery. V T = V 1 + V 2 + V 3 Parallel circuit Each load uses all the potential difference supplied by the battery. V T = V 1 = V 2 = V 3

10. Circuit Current Series circuit The current is the same throughout a series circuit. I total = I 1 = I 2 = I 3 Parallel circuit The current divides into different paths. A pathway with less resistance will have a greater current. I total = I 1 + I 2 + I 3

11. CircuitResistance Series circuit The current decreases when more resistors are added. R T = R 1 + R 2 + R 3 Parallel circuit Adding resistors in parallel decreases the total resistance of the circuit.

12. Sample Question 1: In Series 1. Consider 2 and 4 resistors in series with a 12 V battery. Assuming ideal ammeters and voltmeters, what will be: I 1, I 2, V 1, V 2. Recall: For resistors in series, the total resistance, R T, is given by: R T = R 1 + R 2

13. Solution Current, I 1 First, calculate the total resistance of circuit: R T = R 1 + R 2 = 2  + 4  = 6  Now use Ohm's law to calculate current: V= I R So I = V / R = 12 V / 6  = 2 A The current in the wire to the left of the 2 resistor = 2 A Current, I 2 In a series circuit, the current is always the same, so the current in the wire to the right of the 4 resistor is also 2 A.

14. Voltage, V 1 Use Ohm's law to calculate voltage across the 2 resistor: V = IR = 2 A × 2  = 4 V The voltage V 1 = 4 V Voltage, V 2 Similarly for the 4 resistor: V = IR = 2 A × 4  = 8 V The voltage V 2 = 8 V Check: V T = V 1 + V 2 12V = 4V + 8V

15. Sample Question 2: In Parallel Now consider two resistors in parallel with a 6 V battery. For resistors in parallel, the total resistance is given by: Using the values displayed on the graph, calculate V 1, V 2, I 4, I 5.

16. Voltage, V 1 First, calculate the equivalent resistance of the two resistors: 1 / R = 1 / R 1 + 1 / R 2 = (1 / 3) + (1 / 4) = (7 / 12 So R = (12 / 7) Solution

17. Now use Ohm's law to calculate voltage V 1 : V 1 = I 1 × R = 3.5 A × (12 / 7) = 6 V i.e. the battery voltage, V 1 = 6 V The voltage V 1 = 6 V

18. Voltage, V 2 Use Ohm's law to calculate voltage across the 4 resistor: V 2 = I 3 × R 2 = 1.5 A × 4 = 6 V The voltage V 2 = 6 V Note that for parallel circuits the voltage is the same across all resistors.

19. Current, I 4 This must be identical to I 3, i.e. 1.5 A The current I 4 = 1.5 A Current, I 5 This must be identical to I 1, i.e. 3.5 A The current I 5 = 3.5 A