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Chapter 33 OhmOhm ’’ss LawLaw

Topics Covered in Chapter 3 3-1: The Current = / 3-2: The 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 reserved. TopicsTopics CoveredCovered inin ChapterChapter 33

 3-6: The Linear Proportion between V and I  3-7: Electric  3-8: Power Dissipation in Resistance  3-9: Power Formulas  3-10: Choosing a 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 reserved. OhmOhm ’’ss LawLaw

 ' law states that, in an electrical circuit, the current passing through most materials is directly proportional to the potential difference applied across them. 33--11——33--3:3: OhmOhm ’’ss LawLaw FormulasFormulas

 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. 33--1:1: TheThe CurrentCurrent II == V/RV/R

 I = V/R  In practical units, this law may be stated as:  = /

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

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 33--4:4: PracticalPractical UnitsUnits

 The three forms of Ohm’s law can be used to define the practical units of current, voltage, and resistance:  1 = 1 / 1 ohm  1 volt = 1 ampere × 1 ohm  1 ohm = 1 volt / 1 ampere 33--4:4: PracticalPractical UnitsUnits

Applying Ohm’s Law V ? I R 20 V 20 V 4 Ω I = = 5 A 4 Ω 1 A

? 12 Ω V = 1A × 12 Ω = 12 V

3 A 6 V 6 V ? R = = 2 Ω 3 A ProblemProblem

 Solve for the resistance, R, when V and I are known a. V = 14 V, I = 2 A, R = ? . V = 25 V, I = 5 A, R = ? . V = 6 V, I = 1.5 A, R = ? . V = 24 V, I = 4 A, R = ? 33--5:5: MultipleMultiple andand SubmultipleSubmultiple UnitsUnits

 Units of Voltage  The basic unit of voltage is the volt (V).  Multiple units of voltage are:  kilovolt (kV) 1 thousand volts or 10 3 V  megavolt (MV) 1 million volts or 10 6 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 33--5:5: MultipleMultiple andand SubmultipleSubmultiple UnitsUnits

 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 33--5:5: MultipleMultiple andand SubmultipleSubmultiple UnitsUnits

 Units of Resistance  The basic unit of resistance is the Ohm ( Ω).  Multiple units of resistance are:  kilohm ( Ω) 1 thousand ohms or 10 3 Ω  Megohm ( Ω) 1 million ohms or 10 6 Ω ProblemProblem

 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 k Ω resistance whose current is 250 µA? 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II  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. 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II

 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 . 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II

 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 are examples of linear resistors. 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II

1 Ω 2 Ω 4 3 + 4 Ω 2 0 to 9 Volts 2 Ω Amperes _ 1

0 1 2 3 4 5 6 7 8 9 Volts The smaller the resistor, the steeper the slope. 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II

 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 Amperes

Volts 33--6:6: TheThe LinearLinear ProportionProportion betweenbetween VV andand II

 Another nonlinear resistance is a .  A thermistor is a resistor whose resistance value changes with its operating .  As an NTC (negative temperature coefficient) thermistor gets hot, its resistance decreases.

Thermistor Amperes Amperes

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Volts 33--7:7: ElectricElectric PowerPower

 The basic unit of power is the ().  Multiple units of power are:  kilowatt (kW): 1000 or 10 3 W  megawatt (MW): 1 million watts or 10 6 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 33--7:7: ElectricElectric PowerPower

 and are basically the same, with identical units.

 Power is different. It is the rate of doing work.  Power = work / time.  Work = power × time. 33--7:7: ElectricElectric PowerPower

 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 1 1 volt = and 1 ampere = 1 coulomb 1 33--7:7: ElectricElectric PowerPower

Power = Volts × Amps, or P = V × I

1 joule 1 coulomb 1 joule Power (1 watt) = × = 1 coulomb 1 second 1 second 33--7:7: ElectricElectric PowerPower

 Kilowatt  The kilowatt (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. 33--7:7: ElectricElectric PowerPower

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 ) 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 ProblemProblem

 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 is 7¢/kWh. 33--8:8: PowerPower DissipationDissipation inin ResistanceResistance

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

 Heat is evidence that power is used in producing current. 33--8:8: PowerPower DissipationDissipation inin ResistanceResistance

 The amount of power dissipated in a resistance may be calculated using any one of three formulas, depending on which factors are known:  P = I 2×R  P = V 2 / R  P = V×I ProblemProblem

 Solve for the power, P, dissipated by the resistance, R a. I = 1 A, R = 100 Ω , P = ? b. I = 20 mA, R = 1 k Ω , P = ? c. V = 5 V, R = 150 Ω , P = ? d. V = 22.36 V, R = 1 k Ω , P = ?

 How much power is dissipated by an 8-Ω load if the current in the load is 200 mA? 33--9:9: PowerPower FormulasFormulas

There are three basic power formulas, but each can be in three forms for nine combinations. V2 P = VI P = I 2R P = R

P P 2 I = R = V V 2 R = I P P P V = I = V= PR I R

Where: P = Power V = Voltage I = Current R=Resistance 33--9:9: PowerPower FormulasFormulas

 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 33--9:9: PowerPower FormulasFormulas

 Combining Ohm’s Law and the Power Formula  Substitute V/R for I to obtain:  P = VI = V × V/ R = V 2 / R 33--9:9: PowerPower FormulasFormulas

 Applying Power Formulas:

5 A P = VI = 20 × 5 = 100 W 20 V 4 ΩΩΩ P = I 2R = 25 × 4 = 100 W V2 400 P = = = 100 W R 4 ProblemProblem

 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? 33--10:10: ChoosingChoosing aa ResistorResistor forfor aa CircuitCircuit

 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 .  Ideally, the power dissipation in a resistor should never be more than 50% of its power rating. ProblemProblem

 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. 33--10:10: ChoosingChoosing aa ResistorResistor forfor aa CircuitCircuit

 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. 33--10:10: ChoosingChoosing aa ResistorResistor forfor aa CircuitCircuit

 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: V max = Prating × R

 Exceeding Vmax causes the resistor’s power dissipation to exceed its power rating 33--11:11: ElectricElectric ShockShock

 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 .

 Keep yourself insulated from earth ground.

 Hand-to-hand shocks can be very dangerous because current is likely to flow through the heart! 33--12:12: OpenOpen --CircuitCircuit andand ShortShort --CircuitCircuit TroublesTroubles

An open circuit has zero current flow.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 33--12:12: OpenOpen --CircuitCircuit andand ShortShort --CircuitCircuit TroublesTroubles

A short circuit has excessive current flow.

As R approaches 0, I approaches ∞.

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