science universe

Series Circuits

10–1 CHARACTERISTICS OF SERIES CIRCUITS

When any number of devices are connected so that there is only a single circuit path for electrons, the devices are in series. Each device has the same amount of current in it; see Figure 10–1.

Christmas tree lights are generally connected in series. For many years such an arrangement could be bothersome because all lamps would fail when any lamp burned out. Recent innovations have eliminated this nuisance problem by “shunting” a bypass around the defective filament. This desirable improvement was necessary because when any one lamp burns out, its filament is removed from the circuit, thereby interrupting the current. Each device in a series circuit has the same current through it as every other device in the circuit, since there is only one path for the electrons to travel.

10–2 THE VOLTAGE DROP

The eight lamps shown in Figure 10–1 share the 120 volts supplied and, assuming all lamps in the circuit to be of equal size, they will share alike. In other words, each lamp has a potential difference of 15 volts (120/8) across its terminals. These individual voltages appearing across each resistance of a series circuit are known as voltage drops.

To reinforce this idea, let us consider the circuit of Figure 10–2. Let us first examine the nomenclature used in this schematic diagram. Note that all resistors have been labeled with subscripts: R1, R2, and R3. Since all the resistors are equal in size, we may assume that they will share the supply voltage of 24 volts equally among themselves. These voltage drops of 8 volts each are labeled correspondingly: E1, E2, and E3. After 8 volts are dropped at R1, only 16 volts are left for the remainder of the circuit.

These examples should confirm that

The total voltage of a series circuit is equal to the sum of all the voltage drops.

Mathematically stated,

ET = E1 + E2 + E3 + . . . + En

This equation is sometimes explained by saying that the sum of the voltage drops equals the sum of the voltage rises. In this case, E1, E2, E3, etc., represent the voltage drops, and ET the voltage rise. A voltage rise can thus be explained as a voltage source, as shown in Figure 10–2.

10–3 RESISTANCE AND CURRENT IN SERIES CIRCUITS

Compare the circuit in Figure 10–2 with that of Figure 10–3. Can you detect the similarities and the differences? Both circuits consist of three resistors attached in series to a 24-volt supply. Both circuits have 2 amperes flowing through the resistors. By Ohm’s law, we compute the total resistance (RT) that is connected to the battery.

This may have been obvious to you right along. After all, if you add all the resistances in each circuit, you obtain the same result for RT, namely, 12 ohms. In other words,

The total resistance of a series circuit is equal to the sum of all individual resistors.

Mathematically stated,

This idea of adding individual values does not apply to the current. You may remember our earlier statement, in Section 10–1, that the current in a series circuit has the same value everywhere. The current in these two circuits is 2 amperes at any given point. Yet it is customary to label the current with different subscripts as it flows through different components. We refer to it as I1 when it flows through resistor R1 and, correspondingly, I2 when it flows through R2. The current in the supply line is named IT to correspond to the supply voltage ET. Some people prefer to use the subscript S for the word supply, and write IS and ES. Just remember, by whatever name you call it,

The current in a series circuit is the same everywhere.

Mathematically stated,

So much for the similarities. Now let us have a quick look at the difference between the two circuits. Have you noticed? The voltage drops are different when the resistors are different. In fact, the voltage drops are proportional to the values of the resistors. For instance, if one resistor is twice as large as another, its voltage drop, too, will be twice as large. This, of course, can be confirmed by use of Ohm’s law. We compute the following for the circuit in Figure 10–3:

From now on, as you can see, we must use matching subscripts whenever we use Ohm’s law or the power equations. This idea will be illustrated by the solved sample problems in the next two sections.

The characteristics of series circuits can be summed up in three rules, which can be employed using Ohm’s law to solve the values of any series circuit that contains only one power source. These rules are:

1. The sum of the voltage drops across the individual circuit components is equal to the applied voltage. (Voltage drops add.)

2. The current is the same through all components in a series circuit. (Current remains the same.)

3. The total resistance is equal to the sum of all resistive components in the circuit.

(Resistance adds.)

10–4 POWER CONSUMPTION IN SERIES CIRCUITS

You should recall, from our discussion in Chapter 7, that the wattage rating of resistors is an important specification. With our knowledge of the power equations, determining the power dissipation of an individual resistor, as well as an entire circuit, becomes an easy task. The following example will illustrate.

10–5 CALCULATION OF SERIES CIRCUIT QUANTITIES

This section presents a number of solved problems to demonstrate the proper techniques of solving series circuit problems. Study them carefully.

EXAMPLE 10–2

Given: An antique radio circuit in which the heating elements of five radio tubes are connected as shown in Figure 10–5 and a voltmeter connected across R5 that reads V. (Historical note: This was a common method of radio construction until the late 1950s, when vacuum tubes began to be supplanted by solid-state electronic devices.)

Series Circuits

10–1 CHARACTERISTICS OF SERIES CIRCUITS

When any number of devices are connected so that there is only a single circuit path for electrons, the devices are in series. Each device has the same amount of current in it; see Figure 10–1.

Christmas tree lights are generally connected in series. For many years such an arrangement could be bothersome because all lamps would fail when any lamp burned out. Recent innovations have eliminated this nuisance problem by “shunting” a bypass around the defective filament. This desirable improvement was necessary because when any one lamp burns out, its filament is removed from the circuit, thereby interrupting the current. Each device in a series circuit has the same current through it as every other device in the circuit, since there is only one path for the electrons to travel.

10–2 THE VOLTAGE DROP

The eight lamps shown in Figure 10–1 share the 120 volts supplied and, assuming all lamps in the circuit to be of equal size, they will share alike. In other words, each lamp has a potential difference of 15 volts (120/8) across its terminals. These individual voltages appearing across each resistance of a series circuit are known as voltage drops.

To reinforce this idea, let us consider the circuit of Figure 10–2. Let us first examine the nomenclature used in this schematic diagram. Note that all resistors have been labeled with subscripts: R1, R2, and R3. Since all the resistors are equal in size, we may assume that they will share the supply voltage of 24 volts equally among themselves. These voltage drops of 8 volts each are labeled correspondingly: E1, E2, and E3. After 8 volts are dropped at R1, only 16 volts are left for the remainder of the circuit.

These examples should confirm that

The total voltage of a series circuit is equal to the sum of all the voltage drops.

Mathematically stated,

ET = E1 + E2 + E3 + . . . + En

This equation is sometimes explained by saying that the sum of the voltage drops equals the sum of the voltage rises. In this case, E1, E2, E3, etc., represent the voltage drops, and ET the voltage rise. A voltage rise can thus be explained as a voltage source, as shown in Figure 10–2.

10–3 RESISTANCE AND CURRENT IN SERIES CIRCUITS

Compare the circuit in Figure 10–2 with that of Figure 10–3. Can you detect the similarities and the differences? Both circuits consist of three resistors attached in series to a 24-volt supply. Both circuits have 2 amperes flowing through the resistors. By Ohm’s law, we compute the total resistance (RT) that is connected to the battery.

This may have been obvious to you right along. After all, if you add all the resistances in each circuit, you obtain the same result for RT, namely, 12 ohms. In other words,

The total resistance of a series circuit is equal to the sum of all individual resistors.

Mathematically stated,

This idea of adding individual values does not apply to the current. You may remember our earlier statement, in Section 10–1, that the current in a series circuit has the same value everywhere. The current in these two circuits is 2 amperes at any given point. Yet it is customary to label the current with different subscripts as it flows through different components. We refer to it as I1 when it flows through resistor R1 and, correspondingly, I2 when it flows through R2. The current in the supply line is named IT to correspond to the supply voltage ET. Some people prefer to use the subscript S for the word supply, and write IS and ES. Just remember, by whatever name you call it,

The current in a series circuit is the same everywhere.

Mathematically stated,

So much for the similarities. Now let us have a quick look at the difference between the two circuits. Have you noticed? The voltage drops are different when the resistors are different. In fact, the voltage drops are proportional to the values of the resistors. For instance, if one resistor is twice as large as another, its voltage drop, too, will be twice as large. This, of course, can be confirmed by use of Ohm’s law. We compute the following for the circuit in Figure 10–3:

From now on, as you can see, we must use matching subscripts whenever we use Ohm’s law or the power equations. This idea will be illustrated by the solved sample problems in the next two sections.

The characteristics of series circuits can be summed up in three rules, which can be employed using Ohm’s law to solve the values of any series circuit that contains only one power source. These rules are:

1. The sum of the voltage drops across the individual circuit components is equal to the applied voltage. (Voltage drops add.)

2. The current is the same through all components in a series circuit. (Current remains the same.)

3. The total resistance is equal to the sum of all resistive components in the circuit.

(Resistance adds.)

10–4 POWER CONSUMPTION IN SERIES CIRCUITS

You should recall, from our discussion in Chapter 7, that the wattage rating of resistors is an important specification. With our knowledge of the power equations, determining the power dissipation of an individual resistor, as well as an entire circuit, becomes an easy task. The following example will illustrate.

10–5 CALCULATION OF SERIES CIRCUIT QUANTITIES

This section presents a number of solved problems to demonstrate the proper techniques of solving series circuit problems. Study them carefully.

EXAMPLE 10–2

Given: An antique radio circuit in which the heating elements of five radio tubes are connected as shown in Figure 10–5 and a voltmeter connected across R5 that reads V. (Historical note: This was a common method of radio construction until the late 1950s, when vacuum tubes began to be supplanted by solid-state electronic devices.)

9–4 ENERGY AND COST CALCULATIONS

Electrical energy is commercially sold in units called kilowatt-hours. The consumption of energy is not only a function of the power rating of the appliance, but also of

the amount of time for which it is employed. Obviously, it makes a difference on your electrical bill whether you burn a lamp for 1 hour or for 10 hours. Take, for instance, a 100-watt lightbulb. If you burn it for 1 hour, you have expended 100 watts 3 1 hour 5 100 watt-hours, which is equivalent to 0.1 kilowatt-hour (kWh). Burning it for 10 hours will require 10 times as much energy:

EXAMPLE 9–14

Given: Five 60-watt lamps and four 100-watt lamps burning daily for 5 hours.

Find: The cost of operating the lamps for a billing period of 30 days, at the cost of $0.056 per kilowatt-hour.

Solution

The Kilowatt-Hour Meter

The electrical energy delivered to consumers is measured by an instrument known as the kilowatt-hour meter. Such a meter generally has four dials geared to each other with a ratio of 10 to 1. Thus, when the unit dial makes ten revolutions, the tens dial makes one revolution. Similarly, the tens dial makes ten revolutions for every one revolution of the hundreds scale. The dials are driven by a small motor with a turning effect proportional to the product of voltage and current (E 3 I). The reading of each dial represents a whole number, with any fractional part of that number being found on the next lower dial.

Figure 9–8 represents the four dials of such a kWh meter. In reading such a meter, some people start with the unit dial and then read from right to left; others read the dials from left to right, taking care to record the last number the pointer has passed. No matter how you do it, remember this:

If the pointer is between two digits, select the lower of the two numbers.

Put this knowledge to use and read the meter in Figure 9–8. If you do it correctly, the answer should be 4,294 kWh.

Now look at the dials in Figure 9–9. These dials represent the same meter read 1 month later. Do you agree that the new value is 4,579 kWh? By subtraction we find the difference between the two meter readings: 4,579 2 4,294 5 285 kWh. Thus, we have determined the amount of energy used during that month.

9–5 EFFICIENCY OF ENERGY CONVERSION

“Energy is never created (from nothing), and energy never vanishes” is one way of stating a principle long known as the law of conservation of energy. Each kilowatt- hour used comes from the burning of a fuel or from the release of some stored water. The energy of coal is still responsible for batteries used to light lamps; that is, coal is oxidized to release the zinc or lead used in the battery from other elements when the metal is refined. The electrical energy that we use daily is soon converted to heat by one process or another. Electrical energy can be stored briefly by charging batteries or

by pumping water into a storage tank. However, examples of energy storage are few. In general, energy is converted from one form to another. The efficiency of this energy conversion is a way of measuring how well the energy-converting device accomplishes its task.

EXAMPLE 9–15

Given: A DC motor taking 4.2 amperes on a 120-volt line and delivering 0.5 horsepower.

Find: The efficiency of the motor.

Solution

The power output is 0.5 hp, or 373 W (1 hp = 746 W).

The power input is 120 V x 4.2 A = 504 W.

The efficiency of the motor is 373/504 = 0.74 or 74%.

The efficiency of any device can be no greater than 100%. In other words, the device cannot give out more energy than it takes in. The efficiency of all electrical heating devices is 100%, because the electrical production of heat is easy. Heating devices may vary, however, in how effectively they deliver the heat from the coils in which it is produced to the place where it is to be used. In Example 9–15, we found that the motor is 74% efficient. The other 26% of the energy used appears as heat. If the motor stalls and produces no mechanical power, then 100% of the energy is converted to heat.

EXAMPLE 9–16

Given: An electrical generator with a 10-horsepower input producing 50 amperes at 100 volts.

Find: The efficiency of the generator.

Solution

Power output: 50 A x 100 V = 5,000 W

Power input: 10 hp = 7,460 W

Efficiency = 5,000/7,460 = 0.67 or 67%

You may ask yourself, “Can I drive an electrical generator with an electric motor and let the generated current run the motor?” The answer is, “This scheme will not work very well.” The reason is apparent from the previous discussion and Example 9–16. Both the motor and the generator waste some of the energy applied, with the result that one device is not going to produce enough energy to run the other device.

9–6 A PRACTICAL APPLICATION—RESISTANCE HEATING

The purpose of this section is twofold:

1. To encourage a review of theories and formulas covered up to this point, and

2. To demonstrate a practical application of many laws and formulas.

Most common heating devices, such as toasters, irons, and electric ranges, are heated by current in a coil of nickel-chromium alloy wire or ribbon. Electrical devices are widely used for localized production of small amounts of heat because of their ease of control. The appeal of nickel-chromium (nichrome, chromel, etc.) alloys lies in their small size and reasonable relationship of cost to durability.

To design a heavy-duty resistance-type soldering iron that develops heat at a rate of about 180 watts, consult Figure 9–10 for wire recommendations. Number 26-gauge wire might be used.

All the formulas necessary to complete the calculations have already been learned. The current in the wire is found from

Watts = Volts x Amperes

180 =120X

X = 1.5 A

FIGURE 9–10 Nichrome wire recommendations

The resistance of the wire is found from Ohm’s law:

as shown in the section “Calculating the Resistance of a Wire” in Chapter 7. The value of K is 600, as specified in Figure 7–2, and the cross-sectional area A for #26 wire is given as 254 circular mils in the wire gauge table, Figure A–2 of the Appendix.

Rewriting the above equation for l, we obtain:

Assuming that the construction of a heating element with 33.8 feet of #26 wire would result in an element too large or too bulky for its application, a smaller and lighter element could be constructed using #28 wire instead. The same mathematical formula can be used to calculate the required length of #28 wire. The answer is that only 21.3 feet would be required instead of 33.8 feet. Thus it would appear that the use of #28 wire would be a superior choice because the heating coil would be smaller, lighter, and cheaper. However, there are trade-offs to be considered. When the required wattage is produced in a smaller volume of wire, the surface temperature of the wire is higher. Therefore, #28 wire will oxidize faster and have a shorter life.

Some heating appliances often use a heater element consisting of a coiled resistance wire enclosed in a steel tube. The space around the wire is packed with magnesium oxide, or similar filler, to insulate the wire from the protecting tube.

SUMMARY

• Work 5 Force 3 Distance.

• Energy is the ability to do work.

• Power is the rate of using energy.

• Power x Time = Energy.

• Watts = Volts x Amperes.

• I² R is synonymous with electrical power.

• Watts and kilowatts measure power, which is a rate.

• Watt-hours and kilowatt-hours measure energy.

Achievement Review

1. The word rate implies division by a unit of time. For example, the rate of speed for a car that travels 200 miles in 4 hours is 200 ÷ 4 or 50 miles per hour. Electric power, too, is defined as a rate. What is this definition?

2. A certain motor consumes 1,492 watts. What is the power rating in:

a. Horsepower?

b. Kilowatts?

3. If it were desired to obtain more light in a room, should a bulb with a smaller or greater wattage rating be used? Why?

4. What is meant by I2R losses?

5. What is the equation for finding P, if E and I are given?

6. What is the equation for finding E, if P and R are given?

7. What is the equation for finding I, if R and P are given?

8. What is the equation for finding R, if P and E are given?

9. Calculate the wattage for a 60-volt, 10-ampere arc lamp.

10. Find the power used by a 12-ampere, 110-volt heater.

11. What is the wattage rating of a 100-ohm resistor carrying 0.5 amperes?

12. Calculate the watts for a 2-ohm resistor on a 6-volt line.

13. Calculate resistance of a 60-watt, 120-volt lamp when operating.

14. Find operating current for an 800-watt, 115-volt toaster.

15. Find the cost of operating the toaster of question 14 for 5 hours per month if en- ergy costs $0.135 cents per kilowatt-hour.

16. A 4-ohm resistor is rated at 144 watts. What is the maximum current it can carry without burning up?

17. Another resistance is rated at 25 watts. This particular resistor has 400 ohms.

How much current can be pushed through this resistor before it might burn up?

18. A 750-watt electric iron, when connected to 120 volts, will take how much current?

19. How much current will flow through a wire supplying a 1-horsepower motor at 120 volts?

20. What would be the power, in watts, consumed if a heater rated at 7.5 amperes were connected to a 220-volt circuit?

21. A small pilot light is rated at 5.4 watts. What is the voltage needed to produce a current of 0.3 amperes?

22. How much will it cost to operate a washing machine for 15 hours a month if its motor is rated at 1⁄3 horsepower and the price of electric energy is $0.12 per kilowatt-hour? Assume 60% efficiency.

23. What is the resistance of a toaster whose label reads 1,200 W/120 V?

24. A long power transmission line suffers a power loss of 3.2 megawatts when 800 amperes are flowing through it. What is the resistance of the cable?

25. Two hundred forty milliamperes are flowing through a 4.7-kilohm resistance. What should be the minimum wattage rating of the resistor, including a 10% margin for safety?

26. A 6-volt battery is charged at the rate of 5 amperes for 24 hours. What is the amount of energy charge in kilowatt-hours?

27. One 1⁄4-horsepower DC motor is 70% efficient. What is the current through the motor, in amperes, when it is operating on a 120-volt line and delivering 1⁄4 horsepower?

28. A DC motor takes 5 amperes on a 110-volt line and is 60% efficient. Find the horsepower output. Find the heating rate in watts.

29. One pound of coal releases 12,000 Btu of energy when it is burned. What is this value expressed in foot-pounds?

30. If the energy of a pound of coal is converted into electrical energy (kilowatt- hours) by means of equipment that has an overall efficiency of 30%, what is the energy in kilowatt-hours obtained from a pound of coal?

31. Assuming that 50,000 Btu per hour are required to heat a house in cold weather, how many watts of electrical heat are needed to produce heat at this rate? (Electrical heat is 100% efficient.)

32. Calculate the power in watts that must be supplied to a 40-gallon electrical water heater to raise the temperature of a tank of water from 50°F to 150°F in 2 hours. Assume that no heat is wasted (1 gallon 5 81⁄3 lb). Calculate the cost of heating the water at $0.04 per kilowatt-hour.

33. Find the amount of time required for a 1,600-watt heating element to warm 30 gallons of water from 50°F to 150°F.

34. Find the time required to heat 1,900 grams of water (about 2 quarts) from 10°C to boiling (100°C) on a 660-watt hotplate.

35. Find the time required for the heating operation in question 34 if the process is 90% efficient; that is, only 90% of the heat input is delivered to the water.

36. The heating element of a 1,800-watt/120-volt hotplate has to be replaced. The heating coil will be made of nichrome, and the size of the wire must yield a maxi- mum life span.

a. How many ohms are required?

b. What size should the wire be?

c. How long should the wire be?

9–4 ENERGY AND COST CALCULATIONS

Electrical energy is commercially sold in units called kilowatt-hours. The consumption of energy is not only a function of the power rating of the appliance, but also of

the amount of time for which it is employed. Obviously, it makes a difference on your electrical bill whether you burn a lamp for 1 hour or for 10 hours. Take, for instance, a 100-watt lightbulb. If you burn it for 1 hour, you have expended 100 watts 3 1 hour 5 100 watt-hours, which is equivalent to 0.1 kilowatt-hour (kWh). Burning it for 10 hours will require 10 times as much energy:

EXAMPLE 9–14

Given: Five 60-watt lamps and four 100-watt lamps burning daily for 5 hours.

Find: The cost of operating the lamps for a billing period of 30 days, at the cost of $0.056 per kilowatt-hour.

Solution

The Kilowatt-Hour Meter

The electrical energy delivered to consumers is measured by an instrument known as the kilowatt-hour meter. Such a meter generally has four dials geared to each other with a ratio of 10 to 1. Thus, when the unit dial makes ten revolutions, the tens dial makes one revolution. Similarly, the tens dial makes ten revolutions for every one revolution of the hundreds scale. The dials are driven by a small motor with a turning effect proportional to the product of voltage and current (E 3 I). The reading of each dial represents a whole number, with any fractional part of that number being found on the next lower dial.

Figure 9–8 represents the four dials of such a kWh meter. In reading such a meter, some people start with the unit dial and then read from right to left; others read the dials from left to right, taking care to record the last number the pointer has passed. No matter how you do it, remember this:

If the pointer is between two digits, select the lower of the two numbers.

Put this knowledge to use and read the meter in Figure 9–8. If you do it correctly, the answer should be 4,294 kWh.

Now look at the dials in Figure 9–9. These dials represent the same meter read 1 month later. Do you agree that the new value is 4,579 kWh? By subtraction we find the difference between the two meter readings: 4,579 2 4,294 5 285 kWh. Thus, we have determined the amount of energy used during that month.

9–5 EFFICIENCY OF ENERGY CONVERSION

“Energy is never created (from nothing), and energy never vanishes” is one way of stating a principle long known as the law of conservation of energy. Each kilowatt- hour used comes from the burning of a fuel or from the release of some stored water. The energy of coal is still responsible for batteries used to light lamps; that is, coal is oxidized to release the zinc or lead used in the battery from other elements when the metal is refined. The electrical energy that we use daily is soon converted to heat by one process or another. Electrical energy can be stored briefly by charging batteries or

by pumping water into a storage tank. However, examples of energy storage are few. In general, energy is converted from one form to another. The efficiency of this energy conversion is a way of measuring how well the energy-converting device accomplishes its task.

EXAMPLE 9–15

Given: A DC motor taking 4.2 amperes on a 120-volt line and delivering 0.5 horsepower.

Find: The efficiency of the motor.

Solution

The power output is 0.5 hp, or 373 W (1 hp = 746 W).

The power input is 120 V x 4.2 A = 504 W.

The efficiency of the motor is 373/504 = 0.74 or 74%.

The efficiency of any device can be no greater than 100%. In other words, the device cannot give out more energy than it takes in. The efficiency of all electrical heating devices is 100%, because the electrical production of heat is easy. Heating devices may vary, however, in how effectively they deliver the heat from the coils in which it is produced to the place where it is to be used. In Example 9–15, we found that the motor is 74% efficient. The other 26% of the energy used appears as heat. If the motor stalls and produces no mechanical power, then 100% of the energy is converted to heat.

EXAMPLE 9–16

Given: An electrical generator with a 10-horsepower input producing 50 amperes at 100 volts.

Find: The efficiency of the generator.

Solution

Power output: 50 A x 100 V = 5,000 W

Power input: 10 hp = 7,460 W

Efficiency = 5,000/7,460 = 0.67 or 67%

You may ask yourself, “Can I drive an electrical generator with an electric motor and let the generated current run the motor?” The answer is, “This scheme will not work very well.” The reason is apparent from the previous discussion and Example 9–16. Both the motor and the generator waste some of the energy applied, with the result that one device is not going to produce enough energy to run the other device.

9–6 A PRACTICAL APPLICATION—RESISTANCE HEATING

The purpose of this section is twofold:

1. To encourage a review of theories and formulas covered up to this point, and

2. To demonstrate a practical application of many laws and formulas.

Most common heating devices, such as toasters, irons, and electric ranges, are heated by current in a coil of nickel-chromium alloy wire or ribbon. Electrical devices are widely used for localized production of small amounts of heat because of their ease of control. The appeal of nickel-chromium (nichrome, chromel, etc.) alloys lies in their small size and reasonable relationship of cost to durability.

To design a heavy-duty resistance-type soldering iron that develops heat at a rate of about 180 watts, consult Figure 9–10 for wire recommendations. Number 26-gauge wire might be used.

All the formulas necessary to complete the calculations have already been learned. The current in the wire is found from

Watts = Volts x Amperes

180 =120X

X = 1.5 A

FIGURE 9–10 Nichrome wire recommendations

The resistance of the wire is found from Ohm’s law:

as shown in the section “Calculating the Resistance of a Wire” in Chapter 7. The value of K is 600, as specified in Figure 7–2, and the cross-sectional area A for #26 wire is given as 254 circular mils in the wire gauge table, Figure A–2 of the Appendix.

Rewriting the above equation for l, we obtain:

Assuming that the construction of a heating element with 33.8 feet of #26 wire would result in an element too large or too bulky for its application, a smaller and lighter element could be constructed using #28 wire instead. The same mathematical formula can be used to calculate the required length of #28 wire. The answer is that only 21.3 feet would be required instead of 33.8 feet. Thus it would appear that the use of #28 wire would be a superior choice because the heating coil would be smaller, lighter, and cheaper. However, there are trade-offs to be considered. When the required wattage is produced in a smaller volume of wire, the surface temperature of the wire is higher. Therefore, #28 wire will oxidize faster and have a shorter life.

Some heating appliances often use a heater element consisting of a coiled resistance wire enclosed in a steel tube. The space around the wire is packed with magnesium oxide, or similar filler, to insulate the wire from the protecting tube.

SUMMARY

• Work 5 Force 3 Distance.

• Energy is the ability to do work.

• Power is the rate of using energy.

• Power x Time = Energy.

• Watts = Volts x Amperes.

• I² R is synonymous with electrical power.

• Watts and kilowatts measure power, which is a rate.

• Watt-hours and kilowatt-hours measure energy.

Achievement Review

1. The word rate implies division by a unit of time. For example, the rate of speed for a car that travels 200 miles in 4 hours is 200 ÷ 4 or 50 miles per hour. Electric power, too, is defined as a rate. What is this definition?

2. A certain motor consumes 1,492 watts. What is the power rating in:

a. Horsepower?

b. Kilowatts?

3. If it were desired to obtain more light in a room, should a bulb with a smaller or greater wattage rating be used? Why?

4. What is meant by I2R losses?

5. What is the equation for finding P, if E and I are given?

6. What is the equation for finding E, if P and R are given?

7. What is the equation for finding I, if R and P are given?

8. What is the equation for finding R, if P and E are given?

9. Calculate the wattage for a 60-volt, 10-ampere arc lamp.

10. Find the power used by a 12-ampere, 110-volt heater.

11. What is the wattage rating of a 100-ohm resistor carrying 0.5 amperes?

12. Calculate the watts for a 2-ohm resistor on a 6-volt line.

13. Calculate resistance of a 60-watt, 120-volt lamp when operating.

14. Find operating current for an 800-watt, 115-volt toaster.

15. Find the cost of operating the toaster of question 14 for 5 hours per month if en- ergy costs $0.135 cents per kilowatt-hour.

16. A 4-ohm resistor is rated at 144 watts. What is the maximum current it can carry without burning up?

17. Another resistance is rated at 25 watts. This particular resistor has 400 ohms.

How much current can be pushed through this resistor before it might burn up?

18. A 750-watt electric iron, when connected to 120 volts, will take how much current?

19. How much current will flow through a wire supplying a 1-horsepower motor at 120 volts?

20. What would be the power, in watts, consumed if a heater rated at 7.5 amperes were connected to a 220-volt circuit?

21. A small pilot light is rated at 5.4 watts. What is the voltage needed to produce a current of 0.3 amperes?

22. How much will it cost to operate a washing machine for 15 hours a month if its motor is rated at 1⁄3 horsepower and the price of electric energy is $0.12 per kilowatt-hour? Assume 60% efficiency.

23. What is the resistance of a toaster whose label reads 1,200 W/120 V?

24. A long power transmission line suffers a power loss of 3.2 megawatts when 800 amperes are flowing through it. What is the resistance of the cable?

25. Two hundred forty milliamperes are flowing through a 4.7-kilohm resistance. What should be the minimum wattage rating of the resistor, including a 10% margin for safety?

26. A 6-volt battery is charged at the rate of 5 amperes for 24 hours. What is the amount of energy charge in kilowatt-hours?

27. One 1⁄4-horsepower DC motor is 70% efficient. What is the current through the motor, in amperes, when it is operating on a 120-volt line and delivering 1⁄4 horsepower?

28. A DC motor takes 5 amperes on a 110-volt line and is 60% efficient. Find the horsepower output. Find the heating rate in watts.

29. One pound of coal releases 12,000 Btu of energy when it is burned. What is this value expressed in foot-pounds?

30. If the energy of a pound of coal is converted into electrical energy (kilowatt- hours) by means of equipment that has an overall efficiency of 30%, what is the energy in kilowatt-hours obtained from a pound of coal?

31. Assuming that 50,000 Btu per hour are required to heat a house in cold weather, how many watts of electrical heat are needed to produce heat at this rate? (Electrical heat is 100% efficient.)

32. Calculate the power in watts that must be supplied to a 40-gallon electrical water heater to raise the temperature of a tank of water from 50°F to 150°F in 2 hours. Assume that no heat is wasted (1 gallon 5 81⁄3 lb). Calculate the cost of heating the water at $0.04 per kilowatt-hour.

33. Find the amount of time required for a 1,600-watt heating element to warm 30 gallons of water from 50°F to 150°F.

34. Find the time required to heat 1,900 grams of water (about 2 quarts) from 10°C to boiling (100°C) on a 660-watt hotplate.

35. Find the time required for the heating operation in question 34 if the process is 90% efficient; that is, only 90% of the heat input is delivered to the water.

36. The heating element of a 1,800-watt/120-volt hotplate has to be replaced. The heating coil will be made of nichrome, and the size of the wire must yield a maxi- mum life span.

a. How many ohms are required?

b. What size should the wire be?

c. How long should the wire be?

Electrical Power and Energy

9–1 ENERGY

In Section 2–8 we discussed the conversion of electrical energy to perform work. The type of work or change accomplished involves force and motion. The following changes all require energy: mechanical movement, the production of heat or light, the production of sound, the conversion of one chemical compound into another, and the production of radio waves. The amount of energy required for these changes can be measured, although we can- not see it. How this is accomplished is the subject of this chapter.

In common usage the words work and energy have broad meanings. The physical work or energy with which we are concerned does not include such things as the work done by someone counting the cars that pass a corner, the work done in getting people to change their minds, or the energy with which one tackles an arithmetic problem. Instead, words like work and energy have precise, scientific meanings, which should be understood before we discuss the concept of power. The following sections will help clarify these ideas.

Note: Some courses of study, under constraints of time, may bypass these discussions and proceed directly to Section 9–3. However, many programs of electrical apprenticeship demand some rudimentary knowledge of physical science related to these concepts.

Mechanical Energy

The lifting of a weight illustrates the meaning of a unit of measurement of energy. One foot-pound of energy, or 1 foot-pound of work, is required to lift a 1-pound weight a distance of 1 foot. (The words work and energy can be interchanged where measurement is concerned.)

A foot-pound is defined as the energy used when a 1-pound force moves an object a distance of 1 foot. The 1-foot movement is in the same direction as the applied force.

How much work is done in lifting a 20-pound weight 5 feet vertically, as shown in Figure 9–1?

Work = Force x Distance

A 20-pound force traveling 1 foot accomplishes 20 foot-pounds of work. If the force must travel 5 feet, then 100 foot-pounds of work is accomplished.

Foot-pounds = Feet x Pounds

Potential Energy

What happens to the 100 foot-pounds of energy? It is saved up or conserved. When the 20-pound weight is lifted 5 feet, it has 100 foot-pounds of energy that it did not

have when it was on the ground. This energy is called potential energy. When the weight is permitted to fall back to earth, it delivers 100 foot-pounds of energy to the earth; see Figure 9–2.

How much work is done when a 200-pound box is dragged horizontally along the floor a distance of 6 feet? The question cannot be answered when asked in this manner. Since 200 pounds is a vertical force, it is not in the same direction as the motion. Using a spring scale, however, we find that the horizontal force is 50 pounds. The amount of work done can now be computed.

50 lb x 6 ft = 300 ft-lb

What becomes of this 300 foot-pounds of energy? The energy is converted into heat by the process of friction against the floor. The box does not gain potential energy.

Kinetic Energy

How much work is done by a ball player whose hand moves 6 feet while it is applying an average 8-pound force to a 3⁄4-pound ball?

Work = Force x Distance = 8 lb x 6 ft x 48 ft-lb

What becomes of this energy? It exists as the energy of motion of the ball and is called kinetic energy. The ball in flight has 48 foot-pounds of energy it will deliver when it strikes its target.

Let us return to the example of potential energy shown in Figure 9–2. What happens when the 20-pound weight is allowed to fall the 5-foot distance to the floor? The potential energy of the weight is 100 foot-pounds. This energy becomes the kinetic energy of motion as the weight falls through 5 feet.

The purpose of these mechanical examples is to illustrate the meaning of such terms as work, energy, potential energy, and kinetic energy. These terms are often used in electrical energy discussions, but they are easier to visualize in mechanical energy examples.

9–2 UNITS OF ENERGY

The foot-pound is the energy unit commonly used in the British system of measurements. The metric unit of energy is called a joule. Most common electrical units are based on the joule as the unit of energy. The joule is the work done when a force of 1 newton is exerted through a distance of 1 meter. (A newton is 100,000 dynes, which is about 31⁄2 ounces; a meter is about 39 inches.)

1 J = 0.738 ft-lb

1 ft-lb = 1.35 J

There are two units of measurement that are used to express heat energy. The unit in the British system for heat is called Btu (British thermal unit). A Btu, equivalent to 1,055 joules, is the amount of heat needed to raise the temperature of a pound of water 1 degree Fahrenheit.

How much heat is needed to warm 10 pounds of water from 50°F to 65°F? The temperature increase is 65° – 50° = 15°F. Therefore, 10 lb x 15°F = 150 Btu.

The metric system unit of heat is called a calorie or a gram-calorie. The calorie is the energy needed to raise the temperature of 1 gram of water 1 degree Celsius. (The calorie used in food energy calculations is a kilogram-calorie; this amount of energy will raise 1,000 grams of water 1 degree.)

1 Btu = 778 ft-lb

1 Btu = 252 cal

9–3 POWER

The word power (P), as commonly used, means a variety of things. In technical language, power means how fast work is done or how fast energy is transferred. Two useful definitions of the term power are as follows:

Power is the rate of doing work. Power is the rate of energy conversion.

It is important to understand what is meant by the statement: power is a rate. We do not buy or sell electrical power. What we buy or sell is electrical energy. Power indicates how fast the energy is used or produced. A mechanical example will illustrate the meaning of power and energy.

EXAMPLE 9–1

Given: An elevator lifts 3,500 pounds a distance of 40 feet in 25 seconds.

Find:

a. The work done in 25 seconds (foot-pounds).

b. The rate of work (foot-pounds per second).

Solution

Naturally, power can also be expressed in foot-pounds per minute. The next example will make this clear.

EXAMPLE 9–2

Given: A pump takes 20 minutes to lift 5,000 pounds of water 66 feet.

Find: The rate of doing work.

Solution

The Horsepower

When James Watt started to sell steam engines, he needed to express the capacity of his engines in terms of the horses they were to replace. He found that an average horse, working at a steady rate, could do 550 foot-pounds of work per second. This rate is the definition of 1 horsepower (hp), Figure 9–3.

Can you see that the elevator in Example 9–1 is doing work at the rate of approximately 10 horsepower? Remember, its rate of work was found to be 5,600 foot pounds per second. To convert this to horsepower, we compute

Now try to compute the horsepower rating of the pump described in Example 9–2. Your answer should be approximately 0.5 horsepower. Do you agree?

Horsepower can also be expressed in units of electrical power called watts (W).

1 hp = 746 W

Watt—The Unit of Electrical Power

Recall that the metric unit of energy is called a joule. In the metric system, power is measured in joules per second. This unit of measurement corresponds to foot-pounds per second in the British system. Because the unit joules per second is used so often, it is replaced with the single term watt. (Figure 9–4 gives a comparison of various power units with the watt.)

One watt is a rate of 1 joule per second.

Another way of explaining the term watt is to say that 1 watt of power is dissipated (in the form of heat) when 1 volt of electrical pressure forces 1 ampere of current through the resistance of 1 ohm, Figure 9–5. This relationship can be expressed by the power equation

Recall from Chapter 8 the Ohm’s law circle used as a memory device. The power equation lends itself to another such memory circle, often referred to as a PIE diagram, Figure 9–6. Cover the unknown quantity to be found and read off the equation. Thus,

EXAMPLE 9–5

Given: The label of a television that states 720 W/120 V.

Find: The current in the supply line to the TV.

Solution

A Second Equation for Power

The power equation P 5 E 3 I may be combined with Ohm’s law to yield the hybrid equation P 5 I2R. Let us see how this is derived.

Assume that we need to calculate the electrical power of a circuit for which the volt- age is unknown. From Ohm’s law we know that

P = I²R is an important formula to remember. The synonymous word power and the term I²R have found their way into the vocabulary of people in the electrical trades. Thus, we speak of I2R losses (power losses in the form of heat), I2R heating, and I²R ratings.

Mathematically, the quantities of this equation may be transposed to yield two other equations, namely:

The following examples will explain the application of these formulas in finding an unknown circuit quantity.

EXAMPLE 9–6

Given: A current of 2 amperes flowing through a 250-ohm resistance.

Find: The electrical power converted into heat by the resistor. (Remember, all resistors develop heat when a current flows through them.)

Solution

A Third Equation for Power

It is possible to develop yet a third formula by combining the power equation with Ohm’s law. Beginning with P 5 E 3 I, let us assume that the quantity I is unknown. From Ohm’s law, we substitute the equality

EXAMPLE 9–11

Given: A lamp rated at 300 watts, 120 volts.

Find: The resistance of the lamp while it is operating.

The PIRE Wheel

Up to this point we have become acquainted with 12 mathematical expressions that relate to Ohm’s law and electrical power. For your convenience, these 12 equations are summarized in the circular chart in Figure 9–7.

The four letters P, I, R, and E, shown in the inner circle, represent the unknown quantities that may need to be found. Radiating outward from each of these letters are three choices of equalities that can be used for calculating the unknown.

Electrical Power and Energy

9–1 ENERGY

In Section 2–8 we discussed the conversion of electrical energy to perform work. The type of work or change accomplished involves force and motion. The following changes all require energy: mechanical movement, the production of heat or light, the production of sound, the conversion of one chemical compound into another, and the production of radio waves. The amount of energy required for these changes can be measured, although we can- not see it. How this is accomplished is the subject of this chapter.

In common usage the words work and energy have broad meanings. The physical work or energy with which we are concerned does not include such things as the work done by someone counting the cars that pass a corner, the work done in getting people to change their minds, or the energy with which one tackles an arithmetic problem. Instead, words like work and energy have precise, scientific meanings, which should be understood before we discuss the concept of power. The following sections will help clarify these ideas.

Note: Some courses of study, under constraints of time, may bypass these discussions and proceed directly to Section 9–3. However, many programs of electrical apprenticeship demand some rudimentary knowledge of physical science related to these concepts.

Mechanical Energy

The lifting of a weight illustrates the meaning of a unit of measurement of energy. One foot-pound of energy, or 1 foot-pound of work, is required to lift a 1-pound weight a distance of 1 foot. (The words work and energy can be interchanged where measurement is concerned.)

A foot-pound is defined as the energy used when a 1-pound force moves an object a distance of 1 foot. The 1-foot movement is in the same direction as the applied force.

How much work is done in lifting a 20-pound weight 5 feet vertically, as shown in Figure 9–1?

Work = Force x Distance

A 20-pound force traveling 1 foot accomplishes 20 foot-pounds of work. If the force must travel 5 feet, then 100 foot-pounds of work is accomplished.

Foot-pounds = Feet x Pounds

Potential Energy

What happens to the 100 foot-pounds of energy? It is saved up or conserved. When the 20-pound weight is lifted 5 feet, it has 100 foot-pounds of energy that it did not

have when it was on the ground. This energy is called potential energy. When the weight is permitted to fall back to earth, it delivers 100 foot-pounds of energy to the earth; see Figure 9–2.

How much work is done when a 200-pound box is dragged horizontally along the floor a distance of 6 feet? The question cannot be answered when asked in this manner. Since 200 pounds is a vertical force, it is not in the same direction as the motion. Using a spring scale, however, we find that the horizontal force is 50 pounds. The amount of work done can now be computed.

50 lb x 6 ft = 300 ft-lb

What becomes of this 300 foot-pounds of energy? The energy is converted into heat by the process of friction against the floor. The box does not gain potential energy.

Kinetic Energy

How much work is done by a ball player whose hand moves 6 feet while it is applying an average 8-pound force to a 3⁄4-pound ball?

Work = Force x Distance = 8 lb x 6 ft x 48 ft-lb

What becomes of this energy? It exists as the energy of motion of the ball and is called kinetic energy. The ball in flight has 48 foot-pounds of energy it will deliver when it strikes its target.

Let us return to the example of potential energy shown in Figure 9–2. What happens when the 20-pound weight is allowed to fall the 5-foot distance to the floor? The potential energy of the weight is 100 foot-pounds. This energy becomes the kinetic energy of motion as the weight falls through 5 feet.

The purpose of these mechanical examples is to illustrate the meaning of such terms as work, energy, potential energy, and kinetic energy. These terms are often used in electrical energy discussions, but they are easier to visualize in mechanical energy examples.

9–2 UNITS OF ENERGY

The foot-pound is the energy unit commonly used in the British system of measurements. The metric unit of energy is called a joule. Most common electrical units are based on the joule as the unit of energy. The joule is the work done when a force of 1 newton is exerted through a distance of 1 meter. (A newton is 100,000 dynes, which is about 31⁄2 ounces; a meter is about 39 inches.)

1 J = 0.738 ft-lb

1 ft-lb = 1.35 J

There are two units of measurement that are used to express heat energy. The unit in the British system for heat is called Btu (British thermal unit). A Btu, equivalent to 1,055 joules, is the amount of heat needed to raise the temperature of a pound of water 1 degree Fahrenheit.

How much heat is needed to warm 10 pounds of water from 50°F to 65°F? The temperature increase is 65° – 50° = 15°F. Therefore, 10 lb x 15°F = 150 Btu.

The metric system unit of heat is called a calorie or a gram-calorie. The calorie is the energy needed to raise the temperature of 1 gram of water 1 degree Celsius. (The calorie used in food energy calculations is a kilogram-calorie; this amount of energy will raise 1,000 grams of water 1 degree.)

1 Btu = 778 ft-lb

1 Btu = 252 cal

9–3 POWER

The word power (P), as commonly used, means a variety of things. In technical language, power means how fast work is done or how fast energy is transferred. Two useful definitions of the term power are as follows:

Power is the rate of doing work. Power is the rate of energy conversion.

It is important to understand what is meant by the statement: power is a rate. We do not buy or sell electrical power. What we buy or sell is electrical energy. Power indicates how fast the energy is used or produced. A mechanical example will illustrate the meaning of power and energy.

EXAMPLE 9–1

Given: An elevator lifts 3,500 pounds a distance of 40 feet in 25 seconds.

Find:

a. The work done in 25 seconds (foot-pounds).

b. The rate of work (foot-pounds per second).

Solution

Naturally, power can also be expressed in foot-pounds per minute. The next example will make this clear.

EXAMPLE 9–2

Given: A pump takes 20 minutes to lift 5,000 pounds of water 66 feet.

Find: The rate of doing work.

Solution

The Horsepower

When James Watt started to sell steam engines, he needed to express the capacity of his engines in terms of the horses they were to replace. He found that an average horse, working at a steady rate, could do 550 foot-pounds of work per second. This rate is the definition of 1 horsepower (hp), Figure 9–3.

Can you see that the elevator in Example 9–1 is doing work at the rate of approximately 10 horsepower? Remember, its rate of work was found to be 5,600 foot pounds per second. To convert this to horsepower, we compute

Now try to compute the horsepower rating of the pump described in Example 9–2. Your answer should be approximately 0.5 horsepower. Do you agree?

Horsepower can also be expressed in units of electrical power called watts (W).

1 hp = 746 W

Watt—The Unit of Electrical Power

Recall that the metric unit of energy is called a joule. In the metric system, power is measured in joules per second. This unit of measurement corresponds to foot-pounds per second in the British system. Because the unit joules per second is used so often, it is replaced with the single term watt. (Figure 9–4 gives a comparison of various power units with the watt.)

One watt is a rate of 1 joule per second.

Another way of explaining the term watt is to say that 1 watt of power is dissipated (in the form of heat) when 1 volt of electrical pressure forces 1 ampere of current through the resistance of 1 ohm, Figure 9–5. This relationship can be expressed by the power equation

Recall from Chapter 8 the Ohm’s law circle used as a memory device. The power equation lends itself to another such memory circle, often referred to as a PIE diagram, Figure 9–6. Cover the unknown quantity to be found and read off the equation. Thus,

EXAMPLE 9–5

Given: The label of a television that states 720 W/120 V.

Find: The current in the supply line to the TV.

Solution

A Second Equation for Power

The power equation P 5 E 3 I may be combined with Ohm’s law to yield the hybrid equation P 5 I2R. Let us see how this is derived.

Assume that we need to calculate the electrical power of a circuit for which the volt- age is unknown. From Ohm’s law we know that

P = I²R is an important formula to remember. The synonymous word power and the term I²R have found their way into the vocabulary of people in the electrical trades. Thus, we speak of I2R losses (power losses in the form of heat), I2R heating, and I²R ratings.

Mathematically, the quantities of this equation may be transposed to yield two other equations, namely:

The following examples will explain the application of these formulas in finding an unknown circuit quantity.

EXAMPLE 9–6

Given: A current of 2 amperes flowing through a 250-ohm resistance.

Find: The electrical power converted into heat by the resistor. (Remember, all resistors develop heat when a current flows through them.)

Solution

A Third Equation for Power

It is possible to develop yet a third formula by combining the power equation with Ohm’s law. Beginning with P 5 E 3 I, let us assume that the quantity I is unknown. From Ohm’s law, we substitute the equality

EXAMPLE 9–11

Given: A lamp rated at 300 watts, 120 volts.

Find: The resistance of the lamp while it is operating.

The PIRE Wheel

Up to this point we have become acquainted with 12 mathematical expressions that relate to Ohm’s law and electrical power. For your convenience, these 12 equations are summarized in the circular chart in Figure 9–7.

The four letters P, I, R, and E, shown in the inner circle, represent the unknown quantities that may need to be found. Radiating outward from each of these letters are three choices of equalities that can be used for calculating the unknown.

7–11 POWER RATINGS

Resistors also have a power rating in watts. Exceeding this rating will damage the resistor. The amount of heat that must be dissipated by the resistor can be determined with one of the following formulas:

EXAMPLE 7–10

The resistor shown in Figure 7–29 has a value of 100 ohms and a power rating of 1/2 watt. If the resistor is connected to a 10-volt power supply, will it be damaged?

Since the resistor has a power rating of 1/2 watt and the amount of heat that will be dissipated is 1 watt, the resistor will be damaged.

7–12 VARIABLE RESISTORS

A variable resistor is a resistor whose values can be changed or varied over a range. Variable resistors can be obtained in different case styles and power ratings. Figure 7–30 illustrates how a variable resistor is constructed. In this example, a resistive wire is wound in a circular pattern, and a sliding tap makes contact with the wire. The value of resistance can be adjusted between one end of the resistive wire and the sliding tap. If the resistive wire has a total value of 100 ohms, the resistor can be set between the values of 0 and 100 ohms.

A variable resistor with three terminals is shown in Figure 7–31. This type of resis- tor has a wiper arm inside the case that makes contact with the resistive element. The full resistance value is between the two outside terminals, and the wiper arm is connected to the center terminal. The resistance between the center terminal and either of the two outside terminals can be adjusted by turning the shaft and changing the position of the wiper arm. There are wire wound variable resistors of this type also; see Figure 7–32. The advantage of the wire wound type is a higher power rating.

The resistor shown in Figure 7–31 can be adjusted from its minimum to maximum value by turning the control approximately three-quarters of a turn. In some types of

electrical equipment this range of adjustment may be too coarse to allow for sensitive adjustments. When this becomes a problem, a multiturn resistor can be used. Multiturn variable resistors operate by moving the wiper arm with a screw that has some number of turns, generally from three to ten. For example, a ten-turn variable resistor requires ten turns of the control knob to move the wiper from one end of the resistor to the other instead of three-quarters of a turn.

Variable Resistor Terminology

Variable resistors are known by several common names. The most popular is pot, which is shortened from the word potentiometer. A potentiometer describes how a variable resistor is used rather than some specific type of resistor. The word potentiometer comes from the word potential, or voltage. Thus a potentiometer is used to provide a variable voltage, as shown in Figure 7–33. In this example, one end of the variable resistor is connected to 112 volts, and the other end is connected to ground. The middle terminal, or wiper, is connected to the 1 terminal of a voltmeter and the 2 lead is connected to ground. If the wiper is moved to the upper end of the resistor, the voltmeter will indicate a potential of 12 volts. If the wiper is moved to the bottom, the voltmeter will indicate a value of 0 volts. The wiper can be adjusted to provide any value of voltage between 12 and 0 volts.

Another common name is rheostat. A rheostat is actually a variable resistor with only two terminals, instead of three, but three-terminal variable resistors are often referred to as rheostats also. A rheostat is used to increase or decrease resistance in a circuit; see Figure 7–34. If resistance is decreased, the lamp will burn brighter. If resistance is increased, the lamp will become dimmer.

7–13 SCHEMATIC SYMBOLS

Figure 7–35 illustrates several symbols used to represent both fixed and variable resistors in schematics. Unfortunately, the symbol used is not standard.

SUMMARY

• Every conductor, no matter how good, has some resistance.

• The amount of resistance a wire has depends on its material composition, length, cross-sectional area, and temperature.

• The statement above is mathematically expressed by the equation

• Cross-sectional area (CSA) of a round conductor is figured in circular mils.

• Cross-sectional area of a rectangular bus bar is figured in square mils.

• 1 CM 5 0.7854 sq mil.

• Resistance is more for a long wire than a short one, a thin wire than a thick one, a hot wire than a cold one, and iron than copper.

• A change in resistance due to temperature change is equal to the original resistance times temperature coefficient times degrees change.

• Resistors must be specified by resistance value and power rating.

• Fixed resistors have only one ohmic value that cannot be adjusted.

• The resistance value and tolerance of many resistors can be determined by bands of color.

• Variable resistors are known as potentiometers or rheostats, depending on their circuit function.

• The wire table gives the resistance values in ohms per 1,000 feet for copper and aluminum.

Achievement Review

1. Write definitions for the words (a) resistance, (b) mil, (c) circular mil, (d) mil-foot.

2. There is no perfect conductor. Every wire has at least some resistance. Four factors determine the amount of resistance. Explain.

3. What is the name of the special resistance wire used in heating appliances?

4. If you want to obtain a lot of heat from such heating wire, should it have a lot of resistance or just a little bit? Explain. (Hint: Heating power is equivalent to P 5 E 3 I or P 5 I2R.)

5. Which has more resistance, a wire 20 feet long or a wire 20 inches long, if both are taken from the same stock? Explain.

6. Which has more resistance, a given length of #12 wire or #14 wire? Explain.

7. What happens to the resistance of a lamp as it heats up? Explain.

8. Does a hot lamp draw as much current as a cold lamp? Explain.

9. Two pieces of wire are cut off the same coil. One is 17 feet long, the other is 204 inches long. Which of the two has more resistance? Explain.

10. The diameter of #14 copper wire is 64 mils. Find the cross-sectional area of the wire in circular mils (CM).

11. What is the cross-sectional area of:

a. A wire 0.012 inch in diameter

b. A wire 0.0155 inch in diameter

12. Find the diameter of a wire that has a cross-sectional area of 81 CM.

13. Find the resistance of the wires in parts a–d, using

a. 100 feet of #14 aluminum

b. 25 feet of #20 nichrome

c. 1 mile of #8 iron

d. 6 inches of #18 copper

14. A power line requires 12,000 feet of wire. The total resistance of this wire is not to exceed 5 ohms.

a. What size copper wire meets these requirements? What will 12,000 feet of the wire weigh?

b. What is the smallest size aluminum wire that meets the requirements? What will this wire weigh?

15. The specific resistances of copper and aluminum are 10.4 ohms and 17 ohms, respectively, at 68°F. (See Figure 7–2.) Calculate the specific resistances of the wires at:

a. 0°F (–18°C)

b. 104°F (40°C)

16. A resistance of 0.0005 ohm is required in the connecting wire of an ammeter.

What length of #10 copper wire must be used?

17. The field magnet winding of a 230-volt DC generator has a resistance of 54.5 ohms at 20°C. Find the resistance of the copper winding when the temperature rises to 50°C.

18. A coil of copper wire has 150 ohms resistance at 20°C. After several hours of operation, the resistance of the coil is 172 ohms. Find the temperature of the coil. Of what use is a calculation of this type?

19. Find the diameter (in inches) of a wire whose cross-sectional area is 4,096 CM.

20. A flexible copper cable has 74 strands each 8.75 mils in diameter. Compute the cross-sectional area. (Hint: Compute first the CSA of one strand.)

21. A cable with a cross-sectional area of 1,200 MCM is made up of 19 strands. Find the diameter of each strand.

22. The cross-sectional area of a #10 wire is 10,382 CM. Find its diameter.

23. A 300 MCM cable is composed of 37 strands of copper wire of equal size. Find the diameter of each strand.

24. Find the length of a copper wire that has 4,000 CM and has 2.5 ohms.

25. Find the resistance of a 0.1-inch-diameter aluminum wire that is 100 feet long.

26. Assume that 25 feet of #24 manganin wire is used as a resistance element on a 120-volt circuit. Find the current flowing in this circuit. (Note: Manganin has a resistivity of 260.)

27. Use your wire table to find the resistance of 200 feet of #23 copper wire at 68°F.

28. How heavy (in pounds) is a #2 copper wire, 600 feet long?

COLOR CODE EXERCISE

Part A

For each of the following resistors, state the value and tolerance.

Part B

For each of the following resistors, state the color code.

7–11 POWER RATINGS

Resistors also have a power rating in watts. Exceeding this rating will damage the resistor. The amount of heat that must be dissipated by the resistor can be determined with one of the following formulas:

EXAMPLE 7–10

The resistor shown in Figure 7–29 has a value of 100 ohms and a power rating of 1/2 watt. If the resistor is connected to a 10-volt power supply, will it be damaged?

Since the resistor has a power rating of 1/2 watt and the amount of heat that will be dissipated is 1 watt, the resistor will be damaged.

7–12 VARIABLE RESISTORS

A variable resistor is a resistor whose values can be changed or varied over a range. Variable resistors can be obtained in different case styles and power ratings. Figure 7–30 illustrates how a variable resistor is constructed. In this example, a resistive wire is wound in a circular pattern, and a sliding tap makes contact with the wire. The value of resistance can be adjusted between one end of the resistive wire and the sliding tap. If the resistive wire has a total value of 100 ohms, the resistor can be set between the values of 0 and 100 ohms.

A variable resistor with three terminals is shown in Figure 7–31. This type of resis- tor has a wiper arm inside the case that makes contact with the resistive element. The full resistance value is between the two outside terminals, and the wiper arm is connected to the center terminal. The resistance between the center terminal and either of the two outside terminals can be adjusted by turning the shaft and changing the position of the wiper arm. There are wire wound variable resistors of this type also; see Figure 7–32. The advantage of the wire wound type is a higher power rating.

The resistor shown in Figure 7–31 can be adjusted from its minimum to maximum value by turning the control approximately three-quarters of a turn. In some types of

electrical equipment this range of adjustment may be too coarse to allow for sensitive adjustments. When this becomes a problem, a multiturn resistor can be used. Multiturn variable resistors operate by moving the wiper arm with a screw that has some number of turns, generally from three to ten. For example, a ten-turn variable resistor requires ten turns of the control knob to move the wiper from one end of the resistor to the other instead of three-quarters of a turn.

Variable Resistor Terminology

Variable resistors are known by several common names. The most popular is pot, which is shortened from the word potentiometer. A potentiometer describes how a variable resistor is used rather than some specific type of resistor. The word potentiometer comes from the word potential, or voltage. Thus a potentiometer is used to provide a variable voltage, as shown in Figure 7–33. In this example, one end of the variable resistor is connected to 112 volts, and the other end is connected to ground. The middle terminal, or wiper, is connected to the 1 terminal of a voltmeter and the 2 lead is connected to ground. If the wiper is moved to the upper end of the resistor, the voltmeter will indicate a potential of 12 volts. If the wiper is moved to the bottom, the voltmeter will indicate a value of 0 volts. The wiper can be adjusted to provide any value of voltage between 12 and 0 volts.

Another common name is rheostat. A rheostat is actually a variable resistor with only two terminals, instead of three, but three-terminal variable resistors are often referred to as rheostats also. A rheostat is used to increase or decrease resistance in a circuit; see Figure 7–34. If resistance is decreased, the lamp will burn brighter. If resistance is increased, the lamp will become dimmer.

7–13 SCHEMATIC SYMBOLS

Figure 7–35 illustrates several symbols used to represent both fixed and variable resistors in schematics. Unfortunately, the symbol used is not standard.

SUMMARY

• Every conductor, no matter how good, has some resistance.

• The amount of resistance a wire has depends on its material composition, length, cross-sectional area, and temperature.

• The statement above is mathematically expressed by the equation

• Cross-sectional area (CSA) of a round conductor is figured in circular mils.

• Cross-sectional area of a rectangular bus bar is figured in square mils.

• 1 CM 5 0.7854 sq mil.

• Resistance is more for a long wire than a short one, a thin wire than a thick one, a hot wire than a cold one, and iron than copper.

• A change in resistance due to temperature change is equal to the original resistance times temperature coefficient times degrees change.

• Resistors must be specified by resistance value and power rating.

• Fixed resistors have only one ohmic value that cannot be adjusted.

• The resistance value and tolerance of many resistors can be determined by bands of color.

• Variable resistors are known as potentiometers or rheostats, depending on their circuit function.

• The wire table gives the resistance values in ohms per 1,000 feet for copper and aluminum.

Achievement Review

1. Write definitions for the words (a) resistance, (b) mil, (c) circular mil, (d) mil-foot.

2. There is no perfect conductor. Every wire has at least some resistance. Four factors determine the amount of resistance. Explain.

3. What is the name of the special resistance wire used in heating appliances?

4. If you want to obtain a lot of heat from such heating wire, should it have a lot of resistance or just a little bit? Explain. (Hint: Heating power is equivalent to P 5 E 3 I or P 5 I2R.)

5. Which has more resistance, a wire 20 feet long or a wire 20 inches long, if both are taken from the same stock? Explain.

6. Which has more resistance, a given length of #12 wire or #14 wire? Explain.

7. What happens to the resistance of a lamp as it heats up? Explain.

8. Does a hot lamp draw as much current as a cold lamp? Explain.

9. Two pieces of wire are cut off the same coil. One is 17 feet long, the other is 204 inches long. Which of the two has more resistance? Explain.

10. The diameter of #14 copper wire is 64 mils. Find the cross-sectional area of the wire in circular mils (CM).

11. What is the cross-sectional area of:

a. A wire 0.012 inch in diameter

b. A wire 0.0155 inch in diameter

12. Find the diameter of a wire that has a cross-sectional area of 81 CM.

13. Find the resistance of the wires in parts a–d, using

a. 100 feet of #14 aluminum

b. 25 feet of #20 nichrome

c. 1 mile of #8 iron

d. 6 inches of #18 copper

14. A power line requires 12,000 feet of wire. The total resistance of this wire is not to exceed 5 ohms.

a. What size copper wire meets these requirements? What will 12,000 feet of the wire weigh?

b. What is the smallest size aluminum wire that meets the requirements? What will this wire weigh?

15. The specific resistances of copper and aluminum are 10.4 ohms and 17 ohms, respectively, at 68°F. (See Figure 7–2.) Calculate the specific resistances of the wires at:

a. 0°F (–18°C)

b. 104°F (40°C)

16. A resistance of 0.0005 ohm is required in the connecting wire of an ammeter.

What length of #10 copper wire must be used?

17. The field magnet winding of a 230-volt DC generator has a resistance of 54.5 ohms at 20°C. Find the resistance of the copper winding when the temperature rises to 50°C.

18. A coil of copper wire has 150 ohms resistance at 20°C. After several hours of operation, the resistance of the coil is 172 ohms. Find the temperature of the coil. Of what use is a calculation of this type?

19. Find the diameter (in inches) of a wire whose cross-sectional area is 4,096 CM.

20. A flexible copper cable has 74 strands each 8.75 mils in diameter. Compute the cross-sectional area. (Hint: Compute first the CSA of one strand.)

21. A cable with a cross-sectional area of 1,200 MCM is made up of 19 strands. Find the diameter of each strand.

22. The cross-sectional area of a #10 wire is 10,382 CM. Find its diameter.

23. A 300 MCM cable is composed of 37 strands of copper wire of equal size. Find the diameter of each strand.

24. Find the length of a copper wire that has 4,000 CM and has 2.5 ohms.

25. Find the resistance of a 0.1-inch-diameter aluminum wire that is 100 feet long.

26. Assume that 25 feet of #24 manganin wire is used as a resistance element on a 120-volt circuit. Find the current flowing in this circuit. (Note: Manganin has a resistivity of 260.)

27. Use your wire table to find the resistance of 200 feet of #23 copper wire at 68°F.

28. How heavy (in pounds) is a #2 copper wire, 600 feet long?

COLOR CODE EXERCISE

Part A

For each of the following resistors, state the value and tolerance.

Part B

For each of the following resistors, state the color code.

Ohm’s Law

8–1 VOLTAGE, CURRENT, AND RESISTANCE

The three quantities E, I, and R relate to one another in a very significant way. This relationship, first formulated by the German scientist Georg S. Ohm, states that the current (I) is directly proportional to the voltage (E) but inversely proportional to the resistance (R). In other words,

This equation is known as Ohm’s law and can be algebraically transposed and rewritten into other forms, namely,

Ohm’s law is one of the most important formulas in electrical theory. It enables us to compute unknown circuit quantities, thereby helping us to predict or confirm proper

circuit operation. Memorize Ohm’s law, because you will encounter it time and again in your future studies of electricity and electronics.

A popular memory device, for some students, has Ohm’s law arranged in a circle, as shown in Figure 8–1. To use this circle, place a finger on the letter representing the unknown quantity. The other two letters will show themselves in the proper arrangement to indicate either multiplication or division.

EXAMPLE 8–1

Given: A resistance (R) of 4 ohms connected to a 12-volt battery (E).

Find: The current (I).

Solution

Covering the letter I in the memory circle suggests the correct formula:

EXAMPLE 8–2

Given: A current (I) of 6 amperes flowing through a 4-ohm resistance (R). Find: The voltage (E) needed to force the current through the resistance.

Solution

Wanting to find the voltage (E), cover the letter E within the memory circle. This would suggest that

EXAMPLE 8–3

Given: A source of 6 volts (E) forcing a current of 1.5 amperes (I) through a resistor (R). Find: The value of the unknown resistor.

Solution

Placing your finger on the unknown value (R) of your memory circle will suggest that

8–2 OHM’S LAW WITH METRIC PREFIXES

Ohm’s law is designed for use with basic units only. That means:

• Current must be expressed in amperes.

• Voltage must be expressed in volts.

• Resistance must be expressed in ohms.

Special precaution must be observed when given values are stated with metric pre- fixes instead of basic units. In such cases it is recommended to convert the metric prefixes to powers of 10.

EXAMPLE 8–4

Given: An electromotive force of 100 volts and a current of 20 milliamperes.

Find: The resistance.

Solution

EXAMPLE 8–5

Given: A 3-kilohm resistor is connected to a 12-volt power source.

Find: The current.

Solution

EXAMPLE 8–6

Given: A current of 2 milliamperes is flowing through a 5-kilohm resistance.

Find: The voltage.

SUMMARY

• Ohm’s law states that:

1. Current and voltage are directly proportional.

2. Current and resistance are inversely proportional.

• Mathematically, Ohm’s law may be expressed by the equations

• When solving Ohm’s law problems, numerical values must be stated in basic units (volts, amperes, and ohms).

Achievement Review

1. Given voltage = 120 volts and current = 4 amperes, find resistance.

2. Given resistance = 6 ohms and current = 3 amperes, find voltage.

3. Given voltage = 24 volts and resistance = 8 ohms, find current.

4. Given voltage = 150 volts and resistance = 450 ohms, find current.

5. Given current = 5.5 amperes and voltage = 440 volts, find resistance.

6. Given voltage = 6 volts and resistance = 0.05 ohms, find current.

7. Given current = 0.022 ampere and voltage = 660 volts, find resistance.

8. Given voltage = 60 volts and current = 0.01 ampere, find resistance.

9. Given resistance = 48 ohms and current = 2.5 amperes, find voltage.

10. Given voltage = 12 volts and current = 0.002 ampere, find resistance.

11. Given resistance = 50,000 ohms and voltage = 250 volts, find current.

12. Given current = 0.01 ampere and resistance = 3,000 ohms, find voltage.

13. Given resistance = 7.5 ohms and current = 4.5 amperes, find voltage.

14. Given resistance = 31.4 ohms and current = 3.99 amperes, find voltage.

15. Given resistance = 480 ohms and current = 220 milliamps, find voltage.

16. Given current = 50 milliamps and resistance = 2 kilohms, find voltage.

17. Given resistance = 30,000 ohms and current = 3 milliamps, find voltage.

18. Given voltage = 100 volts and resistance = 10 kilohms, find current.

19. Given voltage = 240 volts and current = 8 milliamps, find resistance.

Given current = 0.002 ampere and voltage = 6.62 volts, find resistanc

Ohm’s Law

8–1 VOLTAGE, CURRENT, AND RESISTANCE

The three quantities E, I, and R relate to one another in a very significant way. This relationship, first formulated by the German scientist Georg S. Ohm, states that the current (I) is directly proportional to the voltage (E) but inversely proportional to the resistance (R). In other words,

This equation is known as Ohm’s law and can be algebraically transposed and rewritten into other forms, namely,

Ohm’s law is one of the most important formulas in electrical theory. It enables us to compute unknown circuit quantities, thereby helping us to predict or confirm proper

circuit operation. Memorize Ohm’s law, because you will encounter it time and again in your future studies of electricity and electronics.

A popular memory device, for some students, has Ohm’s law arranged in a circle, as shown in Figure 8–1. To use this circle, place a finger on the letter representing the unknown quantity. The other two letters will show themselves in the proper arrangement to indicate either multiplication or division.

EXAMPLE 8–1

Given: A resistance (R) of 4 ohms connected to a 12-volt battery (E).

Find: The current (I).

Solution

Covering the letter I in the memory circle suggests the correct formula:

EXAMPLE 8–2

Given: A current (I) of 6 amperes flowing through a 4-ohm resistance (R). Find: The voltage (E) needed to force the current through the resistance.

Solution

Wanting to find the voltage (E), cover the letter E within the memory circle. This would suggest that

EXAMPLE 8–3

Given: A source of 6 volts (E) forcing a current of 1.5 amperes (I) through a resistor (R). Find: The value of the unknown resistor.

Solution

Placing your finger on the unknown value (R) of your memory circle will suggest that

8–2 OHM’S LAW WITH METRIC PREFIXES

Ohm’s law is designed for use with basic units only. That means:

• Current must be expressed in amperes.

• Voltage must be expressed in volts.

• Resistance must be expressed in ohms.

Special precaution must be observed when given values are stated with metric pre- fixes instead of basic units. In such cases it is recommended to convert the metric prefixes to powers of 10.

EXAMPLE 8–4

Given: An electromotive force of 100 volts and a current of 20 milliamperes.

Find: The resistance.

Solution

EXAMPLE 8–5

Given: A 3-kilohm resistor is connected to a 12-volt power source.

Find: The current.

Solution

EXAMPLE 8–6

Given: A current of 2 milliamperes is flowing through a 5-kilohm resistance.

Find: The voltage.

SUMMARY

• Ohm’s law states that:

1. Current and voltage are directly proportional.

2. Current and resistance are inversely proportional.

• Mathematically, Ohm’s law may be expressed by the equations

• When solving Ohm’s law problems, numerical values must be stated in basic units (volts, amperes, and ohms).

Achievement Review

1. Given voltage = 120 volts and current = 4 amperes, find resistance.

2. Given resistance = 6 ohms and current = 3 amperes, find voltage.

3. Given voltage = 24 volts and resistance = 8 ohms, find current.

4. Given voltage = 150 volts and resistance = 450 ohms, find current.

5. Given current = 5.5 amperes and voltage = 440 volts, find resistance.

6. Given voltage = 6 volts and resistance = 0.05 ohms, find current.

7. Given current = 0.022 ampere and voltage = 660 volts, find resistance.

8. Given voltage = 60 volts and current = 0.01 ampere, find resistance.

9. Given resistance = 48 ohms and current = 2.5 amperes, find voltage.

10. Given voltage = 12 volts and current = 0.002 ampere, find resistance.

11. Given resistance = 50,000 ohms and voltage = 250 volts, find current.

12. Given current = 0.01 ampere and resistance = 3,000 ohms, find voltage.

13. Given resistance = 7.5 ohms and current = 4.5 amperes, find voltage.

14. Given resistance = 31.4 ohms and current = 3.99 amperes, find voltage.

15. Given resistance = 480 ohms and current = 220 milliamps, find voltage.

16. Given current = 50 milliamps and resistance = 2 kilohms, find voltage.

17. Given resistance = 30,000 ohms and current = 3 milliamps, find voltage.

18. Given voltage = 100 volts and resistance = 10 kilohms, find current.

19. Given voltage = 240 volts and current = 8 milliamps, find resistance.

Given current = 0.002 ampere and voltage = 6.62 volts, find resistanc