Ohm ’s Law

1 OhmOhm s Laws LawTopics Covered in Chapter 3 3-1: The Current I = V/R3-2: The Voltage V= IR3-3: The ResistanceR= V/I3-4: Practical Units3-5: Multiple and Submultiple UnitsChapterChapter33 2007 The McGraw-Hill Companies, Inc. All rights Covered in Chapter 3 Topics Covered in Chapter 3 3-6: The Linear Proportion between Vand 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 TroublesMcGraw-Hill 2007 The McGraw-Hill Companies, Inc. All rights s Laws 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 33--3: Ohm3: Ohm s Law Formulass Law Formulas There are three forms of Ohm s Law: I = V/R V = IR R = V/I where: I = Current V= Voltage R= ResistanceFig.

2 3-4: A circle diagram to help in memorizing the Ohm s Law formulas V = IR, I = V/R, and R= V/I. The Vis always at the top. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or : The Current 1: The Current I I = = V/RV/R I = V/R In practical units, this law may be stated as: amperes = volts / ohmsFig. 3-1: Increasing the applied voltage Vproduces more current Ito light the bulb with more intensity. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or : Practical Units4: 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 ampere33--4: Practical Units4: Practical Units?

3 20 V4 I = 20 V4 =5 A1 A?12 V= 1A 12 = 12 V3 A6 V?R=6 V3 A=2 Applying Ohm s LawVIRP roblemProblem Solve for the resistance, R, when V and I are = 14 V, I = 2 A, R = ? = 25 V, I = 5 A, R = ? = 6 V, I = A, R = ? = 24 V, I = 4 A, R = ?33--5: Multiple and Submultiple Units5: 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 103V megavolt (MV)1 million volts or 106V Submultiple units of voltage are: millivolt (mV)1-thousandth of a volt or 10-3V microvolt ( V)1-millionth of a volt or 10-6V33--5: Multiple and Submultiple Units5: Multiple and Submultiple Units Units of Current The basic unit of current is the ampere (A).

4 Submultiple units of current are: milliampere (mA)1-thousandth of an ampere or 10-3A microampere ( A)1-millionth of an ampere or 10-6A33--5: Multiple and Submultiple Units5: Multiple and Submultiple Units Units of Resistance The basic unit of resistance is the Ohm ( ). Multiple units of resistance are: kilohm (k )1 thousand ohms or 103 Megohm (M )1 million ohms or 106 ProblemProblem How much is the current, I, in a 470-k resistor if its voltage is V? How much voltage will be dropped across a 40 k resistance whose current is 250 A?33--6: The Linear Proportion between 6: The Linear Proportion between VVand and II The Ohm s Law formula I = V/Rstates that VandI are directly proportional for any one value of : Experiment to show that Iincreases in direct proportion to Vwith the same R.

5 (a) Circuit with variable Vbut constant R. (b) Table of increasing Ifor higher V.(c) Graph of Vand Ivalues. This is a linear volt-ampere characteristic. It shows a direct proportion between Vand The McGraw-Hill Companies, Inc. Permission required for reproduction or : The Linear Proportion between 6: The Linear Proportion between VVand and II When V is constant: Idecreases as Rincreases. I increases as Rdecreases. Examples: If Rdoubles,Iis reduced by half. If Ris reduced to , Iincreases by 4. This is known as an inverse : The Linear Proportion between 6: The Linear Proportion between VVand and II Linear Resistance A linear resistance has a constant value of ohms.

6 Its Rdoes not change with the applied voltage, so Vand I are directly proportional. Carbon-film and metal-film resistors are examples of linear : The Linear Proportion between 6: The Linear Proportion between VVand and II0 1 2 3 4 5 6 7 8 91234 VoltsAmperes2 +_0 to 9 Volts2 1 4 The smaller the resistor , the steeper the : The Linear Proportion between 6: The Linear Proportion between VVand and II Nonlinear Resistance In a nonlinear resistance, increasing the applied Vproduces 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 : The Linear Proportion between 6: The Linear Proportion between VVand and II Another nonlinear resistance is a thermistor.

7 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. VoltsAmperesThermistorVoltsAmperesThermi storCopyright The McGraw-Hill Companies, Inc. Permission required for reproduction or : Electric Power7: Electric Power The basic unit of power is the watt (W). Multiple units of power are: kilowatt (kW):1000 watts or 103W megawatt (MW):1 million watts or 106W Submultiple units of power are: milliwatt (mW):1-thousandth of a watt or 10-3W microwatt ( W):1-millionth of a watt or 10-6W33--7: Electric Power7: Electric Power Work and energy are basically the same, with identical units.

8 Power is different. It is the time rateof doing work. Power = work / time. Work = power : Electric Power7: 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 volt =1 coulomb1 joule1 coulomb1 second1 ampere =and33--7: Electric Power7: Electric PowerPower = Volts Amps, orP = V IPower (1 watt) =1 joule1 coulomb 1 coulomb1 second1 joule1 second=33--7: Electric Power7: Electric Power Kilowatt Hours The kilowatt hour (kWh) is a unit commonly used for large amounts of electrical work or energy.

9 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 : Electric Power7: Electric PowerTo 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 = kilowatts Multiply by hours: (Assume it runs half the day) energy = kW 12 hours = kWh Multiply by rate: (Assume a rate of $ kWh) cost = $ = $ per dayProblemProblem How much is the output voltage of a power supply if it supplies 75 W of power while delivering a current of 5 A?

10 How much does it cost to light a 300-W light bulb for 30 days if the cost of the electricity is 7 : Power Dissipation in Resistance8: 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 : Power Dissipation in Resistance8: 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 IProblemProblem Solve for the power, P, dissipated by the resistance, = 1 A, R = 100 , P = ?