1 Chapter 42 Energy 1 n Energy, W, is the ability to do work and is measured in joules. One joule is the work done when a force of one newton is appli...

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Chapter 4

Chapter 4 1n

Energy Energy, W, is the ability to do work and is measured in joules. One joule is the work done when a force of one newton is applied through a distance of one meter. The symbol for energy, W, represents work, but should not be confused with the unit for power, the watt, W.

1m

Chapter 4 In general, energy (E) is equivalent to power Energy (P) multiplied by time (t). The kilowatt-hour (kWh) is a unit of energy equivalent to one kilowatt (1 KW) of power expended for one hour (1h) of time. It is a much larger unit of energy than the joule. There are 3.6 x 106 J in a kWh. The kWh is convenient for electrical appliances. What is the energy used in operating a 1200 W heater for 20 minutes? 1200 W = 1.2 kW 20 min = 1/3 h 1.2 kW X 1/3 h = 0.4 kWh

Chapter 4 The symbol for Power is P

Power

Power is the rate energy is “used” (actually converted to heat or another form). Power is measured in watts (or kilowatts). Notice that rate always involves time. One watt = one joule/second Three equations for power in circuits that are collectively known as Watt’s law are:

P IV

P I 2R

V2 P R

Chapter 4

equals horsepower

Chapter 4

Chapter 4

Power Formulas

There are three basic power formulas, but each can be in three forms for nine combinations.

P VI

P I 2R

V2 P R

P I V

P R 2 I

V2 R P

P V I

P I R

V PR

Where: P = Power

V = Voltage

I = Current

R=Resistance

Chapter 4

Power Formulas

• Combining Ohm’s Law and the Power Formula – All nine power formulas are based on Ohm’s Law.

V = IR I=V R

P = VI

– Substitute IR for V or V/R for I to obtain: * P = VI * P = VI * = (IR)I * = V x V/R * = V2 / R * = I2R

Chapter 4 Applying Power Formulas

5A 20 V

P = VI = 20V × 5A = 100 W 4

2

P = I R = 25A × 4Ω = 100 W 2

400V V = 100 W = P= 4Ω R

Chapter 4

Electric Power

To calculate electric cost, start with the power: • An air conditioner operates at 240 volts and 20 amperes. • The power is P = V × I = 240 × 20 = 4800 watts. – Convert to kilowatts: 4800 watts = 4.8 kilowatts – Multiply by hours: (Assume it runs half the day) energy = 4.8 kW × 12 hours = 57.6 kWh – Multiply by rate: (Assume a rate of $0.08/ kWh) cost = 57.6 × $0.08 = $4.61 per day

Chapter 4

Power Dissipation

When current flows in a resistance, heat is produced from the friction between the moving free electrons and the atoms obstructing their path. What power is dissipated in a 27 resistor if the current is 0.135 A?

Given that you know the resistance and current, substitute the values into P =I 2R. Heat is evidence P I 2R that power is used (0.135 A) 2 27 in producing current. 0.49 W

Chapter 4 Power Dissipation What power is dissipated by a heater that draws 12 A of current from a 110 V supply?

The most direct solution is to substitute into P = IV. P IV 12 A 110 V 1320 W

Chapter 4 Power Dissipation What power is dissipated in a 100 resistor with 5 V across it?

V2 The most direct solution is to substitute into P . R 2 V P It is useful to keep in mind that R 5 V

2

100

0.25 W

small resistors operating in low voltage systems need to be sized for the anticipated power.

Chapter 4 Resistor failures Resistor failures are unusual except when they have been subjected to excessive heat. Look for discoloration (sometimes the color bands appear burned). Test with an ohmmeter by disconnecting one end from the circuit to isolate it and verify the resistance. Correct the cause of the heating problem (larger resistor?, wrong value?).

Normal

Overheated

Chapter 4 Ampere-hour Rating of Batteries Expected battery life of batteries is given as the amperehours specification. Various factors affect this, so it is an approximation. (Factors include rate of current withdrawal, age of battery, temperature, etc.)

How many hours can you expect to have a battery deliver 0.5 A if it is rated at 10 Ah? 20 h

Battery

Chapter 4 Power Supply Efficiency Efficiency of a power supply is a measure of how well it converts ac to dc. For all power supplies, some of the input power is wasted in the form of heat. As an equation, Power lost

POUT Efficiency = PIN Input power

What is the efficiency of a power supply that converts 20 W of input power to 17 W of output power? 85%

Output power

Chapter 4

Selected Key Terms

Ampere-hour A number determined by multiplying the rating current (A) times the length of time (h) that a battery can deliver that current to a load. Efficiency The ratio of output power to input power of a circuit, usually expressed as a percent. Energy The ability to do work. Joule The SI unit of energy.

Chapter 4

Selected Key Terms

Kilowatt-hour A large unit of energy used mainly by utility (kWh) companies. Power The rate of energy useage Watt The SI unit of power.

Chapter 4

Quiz

1. A unit of power is the a. joule b. kilowatt-hour c. both of the above d. none of the above

Chapter 4

Quiz

1. A unit of power is the a. joule b. kilowatt-hour c. both of the above d. none of the above It is the Watt (W)

Chapter 4

Quiz

2. The SI unit of energy is the a. volt b. joule c. watt d. kilowatt-hour

Chapter 4

Quiz

2. The SI unit of energy is the a. volt b. joule c. watt d. kilowatt-hour

Chapter 4

Quiz

3. If the voltage in a resistive circuit is doubled, the power will be a. halved b. unchanged c. doubled d. quadrupled

Chapter 4

Quiz

3. If the voltage in a resistive circuit is doubled, the power will be a. halved b. unchanged c. doubled d. quadrupled

E P R

2

Chapter 4

Quiz

4. The smallest power rating you should use for a resistor that is 330 with 12 V across it is a. ¼ W b. ½ W c. 1 W d. 2 W

Chapter 4

Quiz

4. The smallest power rating you should use for a resistor that is 330 with 12 V across it is a. ¼ W b. ½ W c. 1 W d. 2 W

(12V ) 2 0.44W 330

Chapter 4

Quiz

5. The power dissipated by a light operating on 12 V that has 3 A of current is a. 4 W b. 12 W c. 36 W d. 48 W

Chapter 4

Quiz

5. The power dissipated by a light operating on 12 V that has 3 A of current is a. 4 W b. 12 W c. 36 W d. 48 W

(12V )( 3 A)

Chapter 4

Quiz

6. The power rating of a resistor is determined mainly by a. surface area b. length c. body color d. applied voltage

Chapter 4

Quiz

6. The power rating of a resistor is determined mainly by a. surface area b. length c. body color d. applied voltage

Chapter 4

Quiz

7. The circuit with the largest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V

R 100

(a)

+15 V

R 200

(b)

+20 V

R 300

(c)

+25 V

R

(d)

Chapter 4

Quiz

7. The circuit with the largest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V

R 100

(a)

(25V ) 2 1.563W 400 +15 V

(10V ) 2 1W 100

R 200

(b)

+20 V

(15V ) 2 1.125W 200

R 300

(c)

+25 V

(20V ) 2 1.333W 300

R

(d)

Chapter 4

Quiz

8. The circuit with the smallest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V

R 100

(a)

+15 V

R 200

(b)

+20 V

R 300

(c)

+25 V

R

(d)

Chapter 4

Quiz

8. The circuit with the smallest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V

R 100

(a)

(25V ) 2 1.563W 400 +15 V

(10V ) 2 1W 100

R 200

(b) (15V )

2

200

+20 V

1.125W

R 300

(c)

+25 V

(20V ) 2 1.333W 300

R

(d)

Chapter 4

Quiz

9. A battery rated for 20 Ah can supply 2 A for a minimum of a. 0.1 h b. 2 h c. 10 h d. 40 h

Chapter 4

Quiz

9. A battery rated for 20 Ah can supply 2 A for a minimum of a. 0.1 h b. 2 h c. 10 h d. 40 h

20 Ah 2A

Chapter 4

Quiz

10. The efficiency of a power supply is determined by a. Dividing the output power by the input power. b. Dividing the output voltage by the input voltage. c. Dividing the input power by the output power. d. Dividing the input voltage by the output voltage.

Chapter 4

Quiz

10. The efficiency of a power supply is determined by a. Dividing the output power by the input power. b. Dividing the output voltage by the input voltage. c. Dividing the input power by the output power. d. Dividing the input voltage by the output voltage.

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