1. Ohm’s Law V = I R + _ V I R

2. Ohm’s Law: V = IR A fundamental relationship in electric circuits. Describes how much potential difference is required to move charges through a resistance at a given current. Materials that have constant resistances are said to obey Ohm’s Law

3. Power and Energy in Electric Circuits Rate at which energy is supplied to a circuit is Power P = VI measured in Watts Can also be stated as: P = VI = (IR) I = I 2 R or P = VI = V (V/R) = V 2 /R

4. EXAMPLES 1)Calculate the rate at which energy is supplied by a 120 V source to a circuit if the current in the circuit is 5.5 A Solution:P=VI = (120V)(5.5 A) = 660 W 2) A 150 Ω resistor carries a current of 2.0 A. Calculate the rate at which heat energy is produced in resistor Solution:P heat = I 2 R = (2.0 A) 2 (150 Ω) = 600 W

5. Energy (W) Recall: Power = Work / time Therefore Work (energy) = Power x time W = Pt = VIt = I 2 RT = V 2 t / R Unit is joules

6. Example How much energy is produced by a 50 V source that generates a current of 5.0 A for 2 minutes? Solution: Don’t forget time must be in seconds! W = VIt = (50 V)(5.0 A)(120 s) = 30,000 J = 30 kJ

7. Series Circuits Or why old Christmas lights used to all go out when only one bulb was broken

8. Series Circuit Has only one current path and if that path is interrupted, the entire circuit ceases to operate.

9. The diagram represents a circuit containing three resistors in series with meters placed to measure various characteristics of the circuit - A - represents an ammeter, a very low resistance device that measures current in a circuit. - V - represents a voltmeter, a very high resistance device that measures potential difference across a circuit

10. In Series circuits… Current throughout the circuit is constant; therefore ammeter can be placed at any position. Potential Difference is equal to the SUM of the potential differences across all resistances – Known as Kirchhoff’s first rule (or simply the loop rule) Ohm’s law holds for each resistance

11. For series circuits I = I 1 = I 2 = I 3 = … V = V 1 + V 2 + V 3 + … R eq = R 1 + R 2 + R 3 + … - this is the equivalent resistance of the circuit

12. Example – calculate the meter readings 24 V 3Ω3Ω6Ω6Ω9Ω9Ω VtVt

13. Solution First find equivalent resistance R eq = R 1 + R 2 + R 3 + … = 3Ω + 6Ω + 9Ω = 18Ω The total potential difference, V T = 24 V since the source supplies the entire circuit The current through circuit (I) is V=I R eq 24 V = I (18Ω) I = 1.33 A

14. Solution cont’d Potential difference across each resistance can be found using Ohm’s Law V=IR V 1 = (1.33 A) (3Ω) = 4 V V 2 = (1.33 A) (6Ω) = 8 V V 3 = (1.33 A) (9Ω) = 12 V

15. Important Fact! As the number of resistances in a series circuit increases, the equivalent resistance, of the circuit increases and the current through the circuit decreases.

16. Example Suppose a fourth resistance of 18Ω is added to the series circuit. Calculate (a) equivalent resistance of circuit R eq = R 1 + R 2 + R 3 + … = 3Ω + 6Ω + 9Ω + 18Ω = 36Ω (b) the current through the circuit V=I R eq 24 V = I (36Ω) I = 0.67 A

17. Parallel Circuits

19. Have more than one current path. If a segment of a // circuit is interrupted, the result will not necessarily be that the entire circuit ceases to operate. House wiring is in //.

20. Parallel circuits Current separates into more than one path. The point where separation occurs is known as a junction The sum of the currents entering a junction must equal the sum of the currents leaving the junction This is Kirchoff’s second rule (or simply the junction rule)

21. example In the diagram below, what are the magnitude and the direction of the current in wire X? 1A 2A 4A X

22. For any // circuit V = V 1 = V 2 = V 3 = … = V n voltage is constant I = I 1 + I 2 + I 3 + … + I n current through entire circuit is equal to the sum of the currents through all resistances (Kirchoff’s 2 nd rule) V n = I n R n

23. example Calculate (a) the equivalent resistance (b) Currents I 1 and I 2 (c) Total current of the following circuit. 3 Ω 6 Ω 24 V I1I1 ITIT I2I2

24. Solution a) To find equivalent resistance:

25. Solution (cont’d) b) To find I 1 and I 2 and c) I T 3 Ω 6 Ω 24 V I1I1 ITIT I2I2

26. Important Note The equivalent resistance is less than any single resistance in the circuit. If more resistance is added in //, the equivalent resistance decreases and the total current increases. – The result is roughly equivalent to increasing the cross- sectional area of a conductor. This is why overloading a household circuit by connecting too many electrical appliances is dangerous. – As current increases, the amount of heat energy also increases. This can lead to fires – Fuses and circuit breakers are designed to prevent such fires from occurring.

27. example A 2 – ohm resistor is added in parallel to the previous circuit. Calculate (a) the equivalent resistance and (b) the total current of the altered circuit. 3 Ω 6 Ω 24 V I1I1 ITIT I2I2 2 Ω I (2Ω)