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Chapter 4

Chapter 4

Chapter 4

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.

1 n

1 m

The symbol for energy, W, represents work, but should not be confused with the unit for power, the watt, W.

Chapter 4

EnergyThe 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 kW20 min = 1/3 h1.2 kW X 1/3 h = 0.4 kWh

In general, energy (E) is equivalent to power (P) multiplied by time (t).

Chapter 4

PowerPower 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 2RP I2VP

R

The symbol for Power is P

Chapter 4

equals horsepower

Chapter 4

Chapter 4Power Formulas

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

P VR

2

R VP

2

V PR

P VI

I PV

V PI

P I R 2

I PR

R PI

2

Where:P = Power V = Voltage I = Current R=Resistance

Chapter 4

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

Law.

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

P = VIV = IR

RVI =

Power Formulas

Chapter 4

20 V 4

5 A P = VI = 20V × 5A = 100 W

P = I2R = 25A × 4Ω = 100 W

P = V2

R = 400V4Ω

= 100 W

Applying Power Formulas

Chapter 4Electric Power

To calculate electric cost, start with thepower:

• 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

What power is dissipated in a 27 resistor if the current is 0.135 A?

2

2(0.135 A) 27 0.49 W

P I R

Power Dissipation

Given that you know the resistance and current, substitute the values into P =I 2R.

When current flows in a resistance, heat is produced from the friction between the moving free electrons and the atoms obstructing their path.

Heat is evidence that power is used in producing current.

Chapter 4

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

Power Dissipation

12 A 110 V1320 W

P IV

The most direct solution is to substitute into P = IV.

Chapter 4

What power is dissipated in a 100 resistor with 5 V across it?

Power Dissipation

2

25 V0.25 W

100

VPR

The most direct solution is to substitute into .2VP

R

It is useful to keep in mind that 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 ampere-hours 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,

OUT

IN

Efficiency = PP

Power lost

Outputpower

Inputpower

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

Chapter 4

Ampere-hour rating

Efficiency

Energy

Joule

A number determined by multiplying the current (A) times the length of time (h) that a battery can deliver that current to a load.

The ability to do work.

The SI unit of energy.

Selected Key Terms

The ratio of output power to input power of a circuit, usually expressed as a percent.

Chapter 4

Kilowatt-hour (kWh)

Power

Watt

The rate of energy useage

The SI unit of power.

Selected Key Terms

A large unit of energy used mainly by utility companies.

Chapter 4Quiz

1. A unit of power is the

a. joule

b. kilowatt-hour

c. both of the above

d. none of the above

Chapter 4Quiz

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 4Quiz

2. The SI unit of energy is the

a. volt

b. joule

c. watt

d. kilowatt-hour

Chapter 4Quiz

2. The SI unit of energy is the

a. volt

b. joule

c. watt

d. kilowatt-hour

Chapter 4Quiz

3. If the voltage in a resistive circuit is doubled, the power will be

a. halved

b. unchanged

c. doubled

d. quadrupled

Chapter 4Quiz

3. If the voltage in a resistive circuit is doubled, the power will be

a. halved

b. unchanged

c. doubled

d. quadrupled

2

REP

Chapter 4Quiz

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 4Quiz

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

WV 44.0330

)12( 2

Chapter 4Quiz

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 4Quiz

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

)3)(12( AV

Chapter 4Quiz

6. The power rating of a resistor is determined mainly by

a. surface area

b. length

c. body color

d. applied voltage

Chapter 4Quiz

6. The power rating of a resistor is determined mainly by

a. surface area

b. length

c. body color

d. applied voltage

Chapter 4Quiz

7. The circuit with the largest power dissipation is

a. (a)

b. (b)

c. (c)

d. (d)

R 100

+10 V R 200

+15 V R 300

+20 V R

+25 V

(a) (b) (c) (d)

Chapter 4Quiz

7. The circuit with the largest power dissipation is

a. (a)

b. (b)

c. (c)

d. (d)

R 100

+10 V R 200

+15 V R 300

+20 V R

+25 V

(a) (b) (c) (d)WV 1100

)10( 2

WV 125.1200

)15( 2

WV 333.1300

)20( 2

WV 563.1400

)25( 2

Chapter 4Quiz

8. The circuit with the smallest power dissipation is

a. (a)

b. (b)

c. (c)

d. (d)

R 100

+10 V R 200

+15 V R 300

+20 V R

+25 V

(a) (b) (c) (d)

Chapter 4Quiz

8. The circuit with the smallest power dissipation is

a. (a)

b. (b)

c. (c)

d. (d)

R 100

+10 V R 200

+15 V R 300

+20 V R

+25 V

(a) (b) (c) (d)WV 1100

)10( 2

WV 125.1200

)15( 2

WV 333.1300

)20( 2

WV 563.1400

)25( 2

Chapter 4Quiz

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 4Quiz

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

AAh

220

Chapter 4Quiz

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 4Quiz

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.