Contributed by:

The Current I = V/R, The Voltage V = IR, The Resistance R = V/I, Practical Units, Multiple and Submultiple Units, The Linear Proportion between V and I, Electric Power, Power Dissipation in Resistance, Power Formulas, Choosing a Resistor for a Circuit, Electric Shock, Open-Circuit and Short-Circuit Troubles

1.
Chapter

3

Ohm’s Law

Topics Covered in Chapter 3

3-1: The Current I = V/R

3-2: The Voltage V = IR

3-3: The Resistance R = V/I

3-4: Practical Units

3-5: Multiple and Submultiple Units

© 2007 The McGraw-Hill Companies, Inc. All rights rese

3

Ohm’s Law

Topics Covered in Chapter 3

3-1: The Current I = V/R

3-2: The Voltage V = IR

3-3: The Resistance R = V/I

3-4: Practical Units

3-5: Multiple and Submultiple Units

© 2007 The McGraw-Hill Companies, Inc. All rights rese

2.
Topics Covered in Chapter 3

3-6: The Linear Proportion between V and I

3-7: Electric Power

3-8: Power Dissipation in Resistance

3-9: Power Formulas

3-10: Choosing a Resistor for a Circuit

3-11: Electric Shock

3-12: Open-Circuit and Short-Circuit Troubles

McGraw-Hill © 2007 The McGraw-Hill Companies, Inc. All rights rese

3-6: The Linear Proportion between V and I

3-7: Electric Power

3-8: Power Dissipation in Resistance

3-9: Power Formulas

3-10: Choosing a Resistor for a Circuit

3-11: Electric Shock

3-12: Open-Circuit and Short-Circuit Troubles

McGraw-Hill © 2007 The McGraw-Hill Companies, Inc. All rights rese

3.
Ohm’s Law

Ohm's law states that, in an

electrical circuit, the current

passing through most

materials is directly

proportional to the potential

difference applied across

them.

Ohm's law states that, in an

electrical circuit, the current

passing through most

materials is directly

proportional to the potential

difference applied across

them.

4.
3-1—3-3: Ohm’s Law Formulas

There are three forms of

Ohm’s Law:

I = V/R

V = IR

R = V/I

where:

I = Current

V = Voltage

R = Resistance

Fig. 3-4: A circle diagram to help in memorizing the Ohm’s Law formulas V = IR, I = V/R,

and R= V/I. The V is always at the top.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

There are three forms of

Ohm’s Law:

I = V/R

V = IR

R = V/I

where:

I = Current

V = Voltage

R = Resistance

Fig. 3-4: A circle diagram to help in memorizing the Ohm’s Law formulas V = IR, I = V/R,

and R= V/I. The V is always at the top.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

5.
3-1: The Current I = V/R

I = V/R

In practical units, this law

may be stated as:

amperes = volts / ohms

Fig. 3-1: Increasing the applied voltage V produces more current I to light the bulb with

more intensity.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

I = V/R

In practical units, this law

may be stated as:

amperes = volts / ohms

Fig. 3-1: Increasing the applied voltage V produces more current I to light the bulb with

more intensity.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6.
3-4: Practical Units

The three forms of Ohm’s law can be used to define the

practical units of current, voltage, and resistance:

1 ampere = 1 volt / 1 ohm

1 volt = 1 ampere × 1 ohm

1 ohm = 1 volt / 1 ampere

The three forms of Ohm’s law can be used to define the

practical units of current, voltage, and resistance:

1 ampere = 1 volt / 1 ohm

1 volt = 1 ampere × 1 ohm

1 ohm = 1 volt / 1 ampere

7.
3-4: Practical Units

Applying Ohm’s Law V

? I R

20 V

20 V 4 I = =5A

4

1A

? 12 V = 1A × 12 = 12 V

3A

6V

6V ? R = =2

3A

Applying Ohm’s Law V

? I R

20 V

20 V 4 I = =5A

4

1A

? 12 V = 1A × 12 = 12 V

3A

6V

6V ? R = =2

3A

8.
Problem

Solve for the resistance, R, when V and I are known

a. V = 14 V, I = 2 A, R = ?

b. V = 25 V, I = 5 A, R = ?

c. V = 6 V, I = 1.5 A, R = ?

d. V = 24 V, I = 4 A, R = ?

Solve for the resistance, R, when V and I are known

a. V = 14 V, I = 2 A, R = ?

b. V = 25 V, I = 5 A, R = ?

c. V = 6 V, I = 1.5 A, R = ?

d. V = 24 V, I = 4 A, R = ?

9.
3-5: Multiple and Submultiple

Units

Units of Voltage

The basic unit of voltage is the volt (V).

Multiple units of voltage are:

kilovolt (kV)

1 thousand volts or 103 V

megavolt (MV)

1 million volts or 106 V

Submultiple units of voltage are:

millivolt (mV)

1-thousandth of a volt or 10-3 V

microvolt (μV)

1-millionth of a volt or 10-6 V

Units

Units of Voltage

The basic unit of voltage is the volt (V).

Multiple units of voltage are:

kilovolt (kV)

1 thousand volts or 103 V

megavolt (MV)

1 million volts or 106 V

Submultiple units of voltage are:

millivolt (mV)

1-thousandth of a volt or 10-3 V

microvolt (μV)

1-millionth of a volt or 10-6 V

10.
3-5: Multiple and Submultiple

Units

Units of Current

The basic unit of current is the ampere (A).

Submultiple units of current are:

milliampere (mA)

1-thousandth of an ampere or 10-3 A

microampere (μA)

1-millionth of an ampere or 10-6 A

Units

Units of Current

The basic unit of current is the ampere (A).

Submultiple units of current are:

milliampere (mA)

1-thousandth of an ampere or 10-3 A

microampere (μA)

1-millionth of an ampere or 10-6 A

11.
3-5: Multiple and Submultiple

Units

Units of Resistance

The basic unit of resistance is the Ohm (Ω).

Multiple units of resistance are:

kilohm (kΩ)

1 thousand ohms or 103 Ω

Megohm (MΩ)

1 million ohms or 106 Ω

Units

Units of Resistance

The basic unit of resistance is the Ohm (Ω).

Multiple units of resistance are:

kilohm (kΩ)

1 thousand ohms or 103 Ω

Megohm (MΩ)

1 million ohms or 106 Ω

12.
Problem

How much is the current, I, in a 470-kΩ resistor if its

voltage is 23.5 V?

How much voltage will be dropped across a 40 kΩ

resistance whose current is 250 µA?

How much is the current, I, in a 470-kΩ resistor if its

voltage is 23.5 V?

How much voltage will be dropped across a 40 kΩ

resistance whose current is 250 µA?

13.
3-6: The Linear Proportion

between V and I

The Ohm’s Law formula I = V/R states that V and I are

directly proportional for any one value of R.

Fig. 3.5: Experiment to show that I increases in direct proportion to V with the same R. (a)

Circuit with variable V but constant R. (b) Table of increasing I for higher V. (c) Graph of V

and I values. This is a linear volt-ampere characteristic. It shows a direct proportion

between V and I.

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between V and I

The Ohm’s Law formula I = V/R states that V and I are

directly proportional for any one value of R.

Fig. 3.5: Experiment to show that I increases in direct proportion to V with the same R. (a)

Circuit with variable V but constant R. (b) Table of increasing I for higher V. (c) Graph of V

and I values. This is a linear volt-ampere characteristic. It shows a direct proportion

between V and I.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

14.
3-6: The Linear Proportion

between V and I

When V is constant:

I decreases as R increases.

I increases as R decreases.

Examples:

If R doubles, I is reduced by half.

If R is reduced to ¼, I increases by 4.

This is known as an inverse relationship.

between V and I

When V is constant:

I decreases as R increases.

I increases as R decreases.

Examples:

If R doubles, I is reduced by half.

If R is reduced to ¼, I increases by 4.

This is known as an inverse relationship.

15.
3-6: The Linear Proportion

between V and I

Linear Resistance

A linear resistance has a constant value of ohms. Its R

does not change with the applied voltage, so V and I

are directly proportional.

Carbon-film and metal-film resistors are examples of

linear resistors.

between V and I

Linear Resistance

A linear resistance has a constant value of ohms. Its R

does not change with the applied voltage, so V and I

are directly proportional.

Carbon-film and metal-film resistors are examples of

linear resistors.

16.
3-6: The Linear Proportion

between V and I

1 2

4

3

Amperes

+ 4

0 to 9 Volts 2

2

_ 1

0 1 2 3 4 5 6 7 8 9

Volts

The smaller the resistor, the steeper the slope.

between V and I

1 2

4

3

Amperes

+ 4

0 to 9 Volts 2

2

_ 1

0 1 2 3 4 5 6 7 8 9

Volts

The smaller the resistor, the steeper the slope.

17.
3-6: The Linear Proportion

between V and I

Nonlinear Resistance

In a nonlinear resistance, increasing the applied V

produces more current, but I does not increase in the

same proportion as the increase in V.

Example of a Nonlinear Volt–Ampere Relationship:

As the tungsten filament in a light bulb gets hot, its

resistance increases.

Amperes

Volts

between V and I

Nonlinear Resistance

In a nonlinear resistance, increasing the applied V

produces more current, but I does not increase in the

same proportion as the increase in V.

Example of a Nonlinear Volt–Ampere Relationship:

As the tungsten filament in a light bulb gets hot, its

resistance increases.

Amperes

Volts

18.
3-6: The Linear Proportion

between V and I

Another nonlinear resistance is a thermistor.

A thermistor is a resistor whose resistance value

changes with its operating temperature.

As an NTC (negative temperature coefficient)

thermistor gets hot, its resistance decreases.

Amperes

Thermistor

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Volts

between V and I

Another nonlinear resistance is a thermistor.

A thermistor is a resistor whose resistance value

changes with its operating temperature.

As an NTC (negative temperature coefficient)

thermistor gets hot, its resistance decreases.

Amperes

Thermistor

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Volts

19.
3-7: Electric Power

The basic unit of power is the watt (W).

Multiple units of power are:

kilowatt (kW):

1000 watts or 103 W

megawatt (MW):

1 million watts or 106 W

Submultiple units of power are:

milliwatt (mW):

1-thousandth of a watt or 10-3 W

microwatt (μW):

1-millionth of a watt or 10-6 W

The basic unit of power is the watt (W).

Multiple units of power are:

kilowatt (kW):

1000 watts or 103 W

megawatt (MW):

1 million watts or 106 W

Submultiple units of power are:

milliwatt (mW):

1-thousandth of a watt or 10-3 W

microwatt (μW):

1-millionth of a watt or 10-6 W

20.
3-7: Electric Power

Work and energy are basically the same, with identical

units.

Power is different. It is the time rate of doing work.

Power = work / time.

Work = power × time.

Work and energy are basically the same, with identical

units.

Power is different. It is the time rate of doing work.

Power = work / time.

Work = power × time.

21.
3-7: Electric Power

Practical Units of Power and Work:

The rate at which work is done (power) equals the

product of voltage and current. This is derived as

follows:

First, recall that:

1 joule 1 coulomb

1 volt = and 1 ampere =

1 coulomb 1 second

Practical Units of Power and Work:

The rate at which work is done (power) equals the

product of voltage and current. This is derived as

follows:

First, recall that:

1 joule 1 coulomb

1 volt = and 1 ampere =

1 coulomb 1 second

22.
3-7: Electric Power

Power = Volts × Amps, or

P=V×I

1 joule 1 coulomb 1 joule

Power (1 watt) = × =

1 coulomb 1 second 1 second

Power = Volts × Amps, or

P=V×I

1 joule 1 coulomb 1 joule

Power (1 watt) = × =

1 coulomb 1 second 1 second

23.
3-7: Electric Power

Kilowatt Hours

The kilowatt hour (kWh) is a unit commonly used for

large amounts of electrical work or energy.

For example, electric bills are calculated in kilowatt

hours. The kilowatt hour is the billing unit.

The amount of work (energy) can be found by

multiplying power (in kilowatts) × time in hours.

Kilowatt Hours

The kilowatt hour (kWh) is a unit commonly used for

large amounts of electrical work or energy.

For example, electric bills are calculated in kilowatt

hours. The kilowatt hour is the billing unit.

The amount of work (energy) can be found by

multiplying power (in kilowatts) × time in hours.

24.
3-7: 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

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

25.
Problem

How much is the output voltage of a power supply if it

supplies 75 W of power while delivering a current of 5

A?

How much does it cost to light a 300-W light bulb for

30 days if the cost of the electricity is 7¢/kWh.

How much is the output voltage of a power supply if it

supplies 75 W of power while delivering a current of 5

A?

How much does it cost to light a 300-W light bulb for

30 days if the cost of the electricity is 7¢/kWh.

26.
3-8: Power Dissipation in

Resistance

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.

Resistance

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.

27.
3-8: Power Dissipation in

Resistance

The amount of power dissipated in a resistance may be

calculated using any one of three formulas, depending

on which factors are known:

P = I2×R

P = V2 / R

P = V×I

Resistance

The amount of power dissipated in a resistance may be

calculated using any one of three formulas, depending

on which factors are known:

P = I2×R

P = V2 / R

P = V×I

28.
Problem

Solve for the power, P, dissipated by the resistance,

R

a. I = 1 A, R = 100Ω , P = ?

b. I = 20 mA, R = 1 kΩ , P = ?

c. V = 5 V, R = 150Ω , P = ?

d. V = 22.36 V, R = 1 kΩ , P = ?

How much power is dissipated by an 8-Ω load if the

current in the load is 200 mA?

Solve for the power, P, dissipated by the resistance,

R

a. I = 1 A, R = 100Ω , P = ?

b. I = 20 mA, R = 1 kΩ , P = ?

c. V = 5 V, R = 150Ω , P = ?

d. V = 22.36 V, R = 1 kΩ , P = ?

How much power is dissipated by an 8-Ω load if the

current in the load is 200 mA?

29.
3-9: Power Formulas

There are three basic power formulas, but each can be

in three forms for nine combinations.

Where:

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

There are three basic power formulas, but each can be

in three forms for nine combinations.

Where:

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

30.
3-9: Power Formulas

Combining Ohm’s Law and the Power Formula

All nine power formulas are based on Ohm’s Law.

V = IR P = VI

I= V

R

Substitute IR for V to obtain:

P = VI

= (IR)I

= I 2R

Combining Ohm’s Law and the Power Formula

All nine power formulas are based on Ohm’s Law.

V = IR P = VI

I= V

R

Substitute IR for V to obtain:

P = VI

= (IR)I

= I 2R

31.
3-9: Power Formulas

Combining Ohm’s Law and the Power Formula

Substitute V/R for I to obtain:

P = VI

= V × V/ R

= V2 / R

Combining Ohm’s Law and the Power Formula

Substitute V/R for I to obtain:

P = VI

= V × V/ R

= V2 / R

32.
3-9: Power Formulas

Applying Power Formulas:

5A P = VI = 20 × 5 = 100 W

20 V 4 2

P = I R = 25 × 4 = 100 W

2

V 400

P= = = 100 W

R 4

Applying Power Formulas:

5A P = VI = 20 × 5 = 100 W

20 V 4 2

P = I R = 25 × 4 = 100 W

2

V 400

P= = = 100 W

R 4

33.
Problem

What is the resistance of a device that dissipates 1.2

kW of power when its current is 10 A?

How much current does a 960 W coffeemaker draw

from the 120 V power line?

What is the resistance of a 20 W, 12 V halogen lamp?

What is the resistance of a device that dissipates 1.2

kW of power when its current is 10 A?

How much current does a 960 W coffeemaker draw

from the 120 V power line?

What is the resistance of a 20 W, 12 V halogen lamp?

34.
3-10: Choosing a Resistor

for a Circuit

Follow these steps when choosing a resistor for a

circuit:

Determine the required resistance value as R = V / I.

Calculate the power dissipated by the resistor using any

of the power formulas.

Select a wattage rating for the resistor that will provide

an adequate cushion between the actual power

dissipation and the resistor’s power rating.

Ideally, the power dissipation in a resistor should never

be more than 50% of its power rating.

for a Circuit

Follow these steps when choosing a resistor for a

circuit:

Determine the required resistance value as R = V / I.

Calculate the power dissipated by the resistor using any

of the power formulas.

Select a wattage rating for the resistor that will provide

an adequate cushion between the actual power

dissipation and the resistor’s power rating.

Ideally, the power dissipation in a resistor should never

be more than 50% of its power rating.

35.
Problem

Determine the required resistance and appropriate

wattage rating of a carbon-film resistor to meet the

following requirements. The resistor has a 54-V IR

drop when its current is 20 mA. The resistors available

have the following wattage ratings:

1/8 W, 1/4 W, 1/2 W, 1 W, and 2 W.

Determine the required resistance and appropriate

wattage rating of a carbon-film resistor to meet the

following requirements. The resistor has a 54-V IR

drop when its current is 20 mA. The resistors available

have the following wattage ratings:

1/8 W, 1/4 W, 1/2 W, 1 W, and 2 W.

36.
3-10: Choosing a Resistor

for a Circuit

Maximum Working Voltage Rating

A resistor’s maximum working voltage rating is the

maximum voltage a resistor can withstand without

internal arcing.

The higher the wattage rating of the resistor, the higher

the maximum working voltage rating.

for a Circuit

Maximum Working Voltage Rating

A resistor’s maximum working voltage rating is the

maximum voltage a resistor can withstand without

internal arcing.

The higher the wattage rating of the resistor, the higher

the maximum working voltage rating.

37.
3-10: Choosing a Resistor

for a Circuit

Maximum Working Voltage Rating

With very large resistance values, the maximum

working voltage rating may be exceeded before the

power rating is exceeded.

For any resistor, the maximum voltage which produces

the rated power dissipation is:

Vmax = Prating × R

Exceeding Vmax causes the resistor’s power dissipation

to exceed its power rating

for a Circuit

Maximum Working Voltage Rating

With very large resistance values, the maximum

working voltage rating may be exceeded before the

power rating is exceeded.

For any resistor, the maximum voltage which produces

the rated power dissipation is:

Vmax = Prating × R

Exceeding Vmax causes the resistor’s power dissipation

to exceed its power rating

38.
3-11: Electric Shock

When possible, work only on circuits that have the

power shut off.

If the power must be on, use only one hand when

making voltage measurements.

Keep yourself insulated from earth ground.

Hand-to-hand shocks can be very dangerous because

current is likely to flow through the heart!

When possible, work only on circuits that have the

power shut off.

If the power must be on, use only one hand when

making voltage measurements.

Keep yourself insulated from earth ground.

Hand-to-hand shocks can be very dangerous because

current is likely to flow through the heart!

39.
3-12: Open-Circuit and

Short-Circuit Troubles

An open circuit has zero current flow.

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Short-Circuit Troubles

An open circuit has zero current flow.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

40.
3-12: Open-Circuit and

Short-Circuit Troubles

A short circuit has excessive current flow.

As R approaches 0, I approaches .

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Short-Circuit Troubles

A short circuit has excessive current flow.

As R approaches 0, I approaches .

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.