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 = E2/R
P = I2 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 nu
mbers or very small numbers.

7. SEATWORK #1
Convert the following: (use only milli (10-3), kilo (103) and Mega (106)
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 103 Ω 24 kΩ
2. 0.0458 A 45 X 10-3 A 45 mA
3. 56,000 W 56 X 103 W 56 kW
4. 5,000 V 5 X 103 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 (103) and Mega (106)
WITH METRIC PREFIX SCIENTIFIC NOTATION REAL NUMBERS
E. 6 MV 6 X 106 A 6,000,000 V
1. 7.25 kΩ
2. 50 mA
3. 50 kW
4. 4.25 MV
5. 3.33 mA

10. SEATWORK #2
Answer Key:
WITH METRIC PREFIX SCIENTIFIC NOTATION REAL NUMBERS
E. 6 MV 6 X 106 A 6,000,000 V
1. 7.25 kΩ 7.25 X 103 Ω 7,250 Ω
2. 50 mA 50 X 10-3 A 0.05 A
3. 50 kW 50 X 103 W 50,000 W
4. 4.25 MV 4.25 X 106 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=V2/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 kΩ
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 kΩ
P= (220)(0.05)
P= 11 W

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