Chapter 4. Chapter 4 - PDF Free Download

Chapter 4

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 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. 1 m

In general, energy (E) is equivalent to power (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 10 6 J in a kwh. The kwh is convenient for electrical appliances. Energy 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

Power The symbol for Power is P 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 2 R P V R 2

equals horsepower

Power Formulas There are three basic power formulas, but each can be in three forms for nine combinations. P VI P I 2 R P V R 2 I P V R P I 2 R V P 2 V P I I P R V PR Where: P = Power V = Voltage I = Current R=Resistance

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 * = I 2 R * = V2 / R

Applying Power Formulas 5 A P = VI = 20V 5A = 100 W 20 V 4 P = I 2 R = 25A 4Ω = 100 W P = V2 R = 400V 4Ω = 100 W

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

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 2 R. P 2 I R 2 (0.135 A) 27 0.49 W Heat is evidence that power is used in producing current.

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 A110 V 1320 W

Power Dissipation What power is dissipated in a 100 resistor with 5 V across it? 2 V The most direct solution is to substitute into P. 2 R V P R 5 V 100 2 0.25 W It is useful to keep in mind that small resistors operating in low voltage systems need to be sized for the anticipated power.

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

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? Battery 20 h

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, Efficiency = P P IN OUT Power lost 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

Ampere-hour rating Efficiency Energy Joule Selected Key Terms 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 ratio of output power to input power of a circuit, usually expressed as a percent. The ability to do work. The SI unit of energy.

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

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

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)

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

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

3. If the voltage in a resistive circuit is doubled, the power will be a. halved b. unchanged c. doubled d. quadrupled P E R 2

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

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 (12V ) 330 2 0.44W d. 2 W

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

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 )(3A)

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

7. The circuit with the largest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V R +15 V R +20 V R +25 V R 100 200 300 (a) (b) (c) (d)

7. The circuit with the largest power dissipation is a. (a) b. (b) c. (c) d. (d) 2 (25V ) 400 1.563W +10 V R +15 V R +20 V R +25 V R 100 200 300 2 2 2 (a) (10V ) (b) (15V ) (c) (20V ) (d) 100 1W 200 1.125W 300 1.333W

8. The circuit with the smallest power dissipation is a. (a) b. (b) c. (c) d. (d) +10 V R +15 V R +20 V R +25 V R 100 200 300 (a) (b) (c) (d)

8. The circuit with the smallest power dissipation is a. (a) b. (b) c. (c) d. (d) 2 (25V ) 400 1.563W +10 V R +15 V R +20 V R +25 V R 100 200 300 2 2 2 (a) (10V ) (b) (15V ) (c) (20V ) (d) 100 1W 200 1.125W 300 1.333W

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

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 20Ah 2A

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.