Topics Covered in Chapter 3 Ohm’s Law Formulas Practical Units of Resistance Multiple Units of Resistance Linear Proportion Between V and I Electric Power
Topics Covered in Chapter 3 (continued) Power Formulas Choosing Resistors Electric Shock Open- and Short-Circuits
Increasing V causes more I to flow in the bulb. I = V R R I V flow = pressure opposition amperes = volts ohms
There are three forms of Ohm’s Law: I = V/R V = IR R = V/I Where: I = Current V = Voltage R = Resistance V I R
20 V 4 ? I = 20 V 4 = 5 A ? 12 1 A V = 1A x 12 = 12 V 6 V? 3 A R = 6 V 3 A = 2 Applying Ohm’s Law V IR
Units of Voltage Submultiple units of voltage are: millivolt (mV) 1-thousandth of a Volt or 10 -3 V microvolt ( V) 1-millionth of a Volt or 10 -6 V Multiple units of voltage are: kilovolt (kV) 1-thousand Volts or 10 3 V megavolt (MV) 1-million Volts or 10 6 V The basic unit of voltage is the Volt (V).
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 microvolt ( ) 1-millionth of an Ampere or 10 -6 A
Units of Resistance Multiple units of resistance are: kilohm (k ) 1-thousand Ohms or 10 3 Megohm (M ) 1-million Ohms or 10 6 The basic unit of resistance is the Ohm ( ).
When V is constant: I decreases as R increases I increases as R decreases Examples: 16 V 4 4 A 16 V 8 2 A If R doubles, I is reduced to half. 16 A 16 V 1 If R is reduced to ¼, I increases by 4. This is known as an inverse relationship.
0 1 2 3 4 5 6 7 8 9 1 2 3 4 Volts Amperes 2 + _ 0 to 9 Volts Fixed resistors have linear volt-ampere relationships. 2 1 4 The smaller the resistor, the steeper the slope.
Volts Amperes Example of a Non-Linear Volt-Ampere Relationship As the tungsten filament gets hot, its resistance increases.
Volts Amperes 2 nd Example of a Non-Linear Volt-Ampere Relationship As the thermistor gets hot, its resistance decreases. Thermistor
Power is the time rate of doing work. The basic unit of power is the Watt (W). Submultiple units of power are: milliwatt (mW) 1-thousandth of a Watt or 10 -3 W microwatt ( W) 1-millionth of a Watt or 10 -6 W Multiple units of power are: kilowatt (kW) 1-thousand Watts or 10 3 W megawatt (MW) 1-million Watts or 10 6 W
The rate of work can be found by multiplying potential difference times flow. 1 Volt = 1 Joule 1 coulomb 1 Ampere = 1 coulomb 1 second Power = Volts x Amps 1 Joule 1 coulomb 1 second x 1 Joule 1 second = Power (1 Watt) = and First, recall that: The base unit of power is the Watt or Joule/second.
The amount of work (energy) can be found by multiplying power times time. The amount of work (energy) is used for calculating electric bills. The kilowatt-hour is the billing unit.
To calculate electric cost, start with the power: Convert to kilowatts: Multiply by hours: (Assume it runs half the day) Multiply by rate: (Assume a rate of $0.08) An air conditioner operates at 240 volts and 20 amperes. The power is P = V x I = 240 x 20 = 4800 watts. 4800 watts = 4.8 kilowatts energy = 4.8 kW x 12 hours = 57.6 kWh cost = 57.6 x $0.08 = $4.61 per day
Combining Ohm’s Law and the power formula V = IR P = VI P = (IR)I = I 2 R Substitute IR for V to obtain: Substitute V/R for I to obtain: P = V x V R V2V2 R = V R I =
Power Formulas Where: I = Current R = Resistance P = Power V = Voltage
Applying Power Formulas 20 V 4 5 A P = VI = 20 x 5 = 100 W P = I 2 R = 25 x 4 = 100 W P = V2V2 R = 400 4 = 100 W
Electrical Shock Hazard When possible, work only on circuits that have the power shut off. If the power must be on, use only one hand. Hand-to-hand shocks can be very dangerous!
Open and Short Circuits An open circuit has zero current flow. 16 V 0 A 16 V 0 excessive I A short circuit has excessive current flow. I = 16 V 0 = undefined As R approaches 0, I approaches .