POWER LAW

Power Law Essentials and basics

1. EMPOWERMENT TECHNOLOGIES POWER LAW

2. LEARNING OBJECTIVES At the end of the lesson, the students should be able to: Define and identify POWER LAW. Review scientific notation and metric prefixes. Calculate current, voltage, resistance and power using Ohm’s Law and Power Law.

3. POWER LAW The amount of current times the voltage level at a given point measured in wattage or watts. It states that the power dissipated in a device is inversely proportional to the squared value of the voltage across it. It can also be stated as the power dissipated in a device is directly proportional to the squared value of the current going through it.

4. POWER LAW FORMULAS/FORMULAE P = I X E P = E 2 /R P = I 2 X R POWER LAW IS NAMED AFTER SCOTTISH MATHEMATICIAN JAMES WATT

5. Derivation

6. SCIENTIFIC NOTATION AND METRIC PREFIXES Scientific notation is the way that scientists easily handle very large numbers or very small numbers.

7. SEATWORK #1 Convert the following: (use only milli (10 -3 ), kilo (10 3 ) and Mega (10 6 ) REAL NUMBER SCIENTIFIC NOTATION WITH METRIC PREFIX E. 0.00725 A 7.25 X 10 -3 A 7.25 mA 1. 24, 000 Ω 2. 0.0458 A 3. 56,000 W 4. 5,000 V 5. 0.01525 A

8. SEATWORK #1 Answer Key: REAL NUMBER SCIENTIFIC NOTATION WITH METRIC PREFIX E. 0.00725 A 7.25 X 10 -3 A 7.25 mA 1. 24, 000 Ω 24 X 10 3 Ω 24 k Ω 2. 0.0458 A 45 X 10 -3 A 45 mA 3. 56,000 W 56 X 10 3 W 56 kW 4. 5,000 V 5 X 10 3 V 5 kV 5. 0.01525 A 15.25 X 10 -3 A 15.25 mA

9. SEATWORK #2 Convert the following: (use only milli (10 -3 ), kilo (10 3 ) and Mega (10 6 ) WITH METRIC PREFIX SCIENTIFIC NOTATION REAL NUMBERS E. 6 MV 6 X 10 6 A 6,000,000 V 1. 7.25 k Ω 2. 50 mA 3. 50 kW 4. 4.25 M V 5. 3.33 mA

10. SEATWORK #2 Answer Key: WITH METRIC PREFIX SCIENTIFIC NOTATION REAL NUMBERS E. 6 MV 6 X 10 6 A 6,000,000 V 1. 7.25 k Ω 7.25 X 10 3 Ω 7,250 Ω 2. 50 mA 50 X 10 -3 A 0.05 A 3. 50 kW 50 X 10 3 W 50,000 W 4. 4.25 M V 4.25 X 10 6 V 4,250,000 V 5. 3.33 mA 3.33 X 10 -3 A 0.00333 A

11. Example V= 220 V P= 2.5 kW I=? R=? Given: V= 220 V P= 2.5 kW R=? I=? Formula: R=V 2 /P I =V/R or I=P/V Solution: R= (220 V) 2 /2.5 kW R= 48,400/2500 R= 19.36 Ω I= 220 V/ 19.36 Ω or I= 11.36 A I= 2.5 kW/ 220 V I= 2500/220 I = 11.36 A

12. V= 220 V P=? I= 50 mA R=? V=? P=? I= 7.25 mA R=2.2 k Ω 1 Seatwork (1/2 yellow pad) 2

13. V= 220 V P=? I= 50 mA R=? 1 Seatwork (1/2 yellow pad) Given: V= 220 V P=? R=? I= 50 mA Formula: R=V/I P= V X I Solution: R= 220/0.05 R= 4,400 R= 4.4 k Ω P= (220)(0.05) P = 11 W

14. V=? P=? I= 7.25 mA R=2.2 k Ω Seatwork (1/2 yellow pad) 2 Given: V=? P=? R= 2.2 k Ω I= 7.25 mA Formula: V=I X R P= V X I Solution: V= (7.25 mA) (2.2 k Ω ) V= (0.00725)(2200) V= 15.95 V P= (15.95)(0.00725) P = 0.12 W