Presentation on theme: "Foundations of Physics"— Presentation transcript:
1 Foundations of Physics CPO ScienceFoundations of PhysicsUnit 7, Chapter 20
2 Unit 7: Electricity and Magnetism Chapter 20 Electric Circuits and Power20.1 Series and Parallel Circuits20.2 Analysis of Circuits20.3 Electric Power, AC, and DC Electricity
3 Chapter 20 ObjectivesRecognize and sketch examples of series and parallel circuits.Describe a short circuit and why a short circuit may be a dangerous hazard.Calculate the current in a series or parallel circuit containing up to three resistances.Calculate the total resistance of a circuit by combining series or parallel resistances.Describe the differences between AC and DC electricity.Calculate the power used in an AC or DC circuit from the current and voltage.
4 Chapter 20 Vocabulary Terms series circuitparallel circuitshort circuitnetwork circuitcircuit analysispowerKirchhoff’s voltage lawvoltage drop directcurrent (DC)alternating current (AC)kilowattKirchhoff’s current lawhorsepowerpower factorcircuit breakerwattkilowatt-hour
5 20.1 Series and Parallel Circuits Key Question:How do series and parallel circuits work?*Students read Section AFTER Investigation 20.1
6 20.1 Series and Parallel Circuits In series circuits, current can only take one path.The amount of current is the same at all points in a series circuit.
10 20.1 Total resistance in a series circuit Light bulbs, resistors, motors, and heaters usually have much greater resistance than wires and batteries.
11 20.1 Calculate current1) You are asked to calculate current.2) You are given the voltage and resistances.3) Use Ohm’s law, I = V÷R, and add the resistance in series.4) Solve:Resistance = R1 + R2 + R3 = 1Ω + 1Ω + 1Ω = 3ΩCurrent, I = (1.5 V) ÷ (3Ω)=0.5 AHow much current flows in a circuit with a 1.5-volt battery and three 1 ohm resistances (bulbs) in series?
13 20.1 Voltage in a series circuit Each separate resistance creates a voltage drop as the current passes through.As current flows along a series circuit, each type of resistor transforms some of the electrical energy into another form of energyOhm’s law is used to calculate the voltage drop across each resistor.
17 20.1 Series and Parallel Circuits Sometimes these paths are called branches.The current through a branch is also called the branch current.When analyzing a parallel circuit, remember that the current always has to go somewhere.The total current in the circuit is the sum of the currents in all the branches.At every branch point the current flowing out must equal the current flowing in.This rule is known as Kirchhoff’s current law.
19 20.1 Voltage and current in a parallel circuit In a parallel circuit the voltage is the same across each branch because each branch has a low resistance path back to the battery.The amount of current in each branch in a parallel circuit is not necessarily the same.The resistance in each branch determines the current in that branch.
20 20.1 Advantages of parallel circuits Parallel circuits have two big advantages over series circuits:1. Each device in the circuit sees the full battery voltage.2. Each device in the circuit may be turned off independently without stopping the current flowing to other devices in the circuit.
21 20.1 Short circuitA short circuit is a parallel path in a circuit with zero or very low resistance.Short circuits can be made accidentally by connecting a wire between two other wires at different voltages.Short circuits are dangerous because they can draw huge amounts of current.
22 20.1 Calculate current1) You are asked for the current.2) You are given the voltage and resistance.3) Use Ohm’s law: I = V ÷ R.4) For the 3Ω bulb:I = (3 V) ÷ (3 Ω) = 1 A.For the 0.5 Ω bulb:I = (3 V) ÷ (0.5 Ω) = 6 A.The battery must supply the current for both bulbs, which adds up to 7 amps.Two bulbs with different resistances are connected in parallel to batteries with a total voltage of 3 volts.Calculate the total current supplied by the battery.
23 20.1 Resistance in parallel circuits Adding resistance in parallel provides another path for current, and more current flows.When more current flows for the same voltage, the total resistance of the circuit decreases.This happens because every new path in a parallel circuit allows more current to flow for the same voltage.
25 20.1 Adding resistance in parallel circuits 1) You are asked for the resistance.2) You are given the circuit diagram and resistances.3) Use the rule for parallel resistances.4) Solve:1/R total = 1/2 Ω + 1/4 Ω = 2/4 Ω +1/4 Ω = 3/4 ΩR= 4/3 Ω = ΩA circuit contains a 2 ohm resistor and a 4 ohm resistor in parallel.Calculate the total resistance of the circuit.
26 20.2 Analysis of Circuits Key Question: How do we analyze network circuits?*Students read Section AFTER Investigation 20.2
27 20.2 Analysis of CircuitsAll circuits work by manipulating currents and voltages.The process of circuit analysis means figuring out what the currents and voltages in a circuit are, and also how they are affected by each other.Three basic laws are the foundation of circuit analysis.
29 20.2 Voltage divider circuit The total resistance of this circuit is 10 ohms (9Ω + 1Ω).The total current is 1 amp from Ohm’s law (10V ÷ 10Ω).What is the voltage at point (A)?We can use Kirchhoff’s voltage law to find the answer.Assume the current flows according to the light blue loop.The battery starts at +10V.Resistor R1 drops the voltage by 9V.This voltage drop is calculated from Ohm’s law and the resistance, V = 1A × 9Ω.Resistor R2 drops the voltage by 1V (1A × 1Ω).Around the whole loop the sum of the voltage gains and drops is zero ( = 0).
30 20.2 Voltage divider V0 = R1 Vi R1 + R2 A circuit divides any supplied voltage by a ratio of the resistors.V0 = R ViR1 + R2Inputvoltage(volts)Outputvoltage(volts)resistor ratio(W)
31 20.2 Solving circuit problems Identify what the problem is asking you to find. Assign variables to the unknown quantities.Make a large clear diagram of the circuit. Label all of the known resistances, currents, and voltages. Use the variables you defined to label the unknowns.You may need to combine resistances to find the total circuit resistance. Use multiple steps to combine series and parallel resistors.
32 20.2 Solving circuit problems If you know the total resistance and current, use Ohm’s law as V = IR to calculate voltages or voltage drops. If you know the resistance and voltage, use Ohm’s law as I = V ÷ R to calculate the current.An unknown resistance can be found using Ohm’s law as R = V ÷ I, if you know the current and the voltage drop through the resistor.Use Kirchhoff’s current and voltage laws as necessary.
33 20.2 Solving circuit problems A bulb with a resistance of 1Ω is to be used in a circuit with a 6-volt battery.The bulb requires 1 amp of current.If the bulb were connected directly to the battery, it would draw 6 amps and burn out instantly.To limit the current, a resistor is added in series with the bulb.What size resistor is needed to make the current 1 amp?1) You are asked to calculate the resistance.2) You are told it is a series circuit and given the voltage, total current, and one resistance.3) Use Ohm’s law, R = V ÷ I, and add the resistance in series.4) Solve:Total resistance = 6V ÷ 1A = 6Ω.SInce the bulb is 1Ω, the additional resistor must be 5Ω to get a total 6Ω of resistance.
34 20.2 Network circuitsIn many circuits, resistors are connected both in series and in parallel.Such a circuit is called a network circuit.There is no single formula for adding resistors in a network circuit.For very complex circuits, electrical engineers use computer programs that can rapidly solve equations for the circuit using Kirchhoff’s laws.
35 20.2 Calculate using network circuits Three bulbs, each with a resistance of 3Ω, are combined in the circuit in the diagramThree volts are applied to the circuit.Calculate the current in each of the bulbs.From your calculations, do you think all three bulbs will be equally bright?1) You are asked to calculate the currents.2) You are given the circuit diagram, voltages, and resistances.3) Use Ohm’s law, R = V ÷ I, and the series and parallel resistance formulas.4) First, reduce the circuit by combining the two parallel resistances.1/R total = 1/3 Ω + 1/3 Ω = 2/3 ΩR= 3/2 Ω = 1.5 Ω5) Calculate the total resistance of 4.5Ω by adding up the remaining series resistances.Calculate the total current using Ohm’s law: I = 3V ÷ 4.5Ω = 0.67A.The two bulbs in parallel have the same resistance, so they divide the current equally;each one gets 0.33 amps.The single bulb in series gets the full current of 0.67 amps, but the other two bulbsget only 0.33 amps each. That means the bulbs in parallel will be much dimmer sincethey only get half the current.
36 20.3 Electric Power, AC, and DC Electricity Key Question:How much does electricity cost and what do you pay for?*Students read Section AFTER Investigation 20.3
37 20.3 Electric Power, AC, and DC Electricity The watt (W) is a unit of power.Power is the rate at which energy moves or is used.Since energy is measured in joules, power is measured in joules per second.One joule per second is equal to one watt.
39 20.3 Power in electric circuits One watt is a pretty small amount of power.In everyday use, larger units are more convenient to use.A kilowatt (kW) is equal to 1,000 watts.The other common unit of power often seen on electric motors is the horsepower.One horsepower is 746 watts.
41 20.3 Calculate powerA light bulb with a resistance of 1.5Ω is connected to a 1.5-volt battery in the circuit shown at right.Calculate the power used by the light bulb.1) You are asked to find the power used by the light bulb.2) You are given the voltage of the battery and the bulb’s resistance.3) Use Ohm’s law, I = V/R, to calculate the current; then use the power equation, P=VI, to calculate the power.4) Solve: I = 1.5V ÷ 1.5Ω = 1AP = 1.5V × 1A = 1.5 W; the bulb uses 1.5 watts of electric power.
42 20.3 Paying for electricity Electric companies charge for the number of kilowatt-hours used during a set period of time, often a month.One kilowatt-hour (kWh) means that a kilowatt of power has been used for one hour.Since power multiplied by time is energy, a kilowatt-hour is a unit of energy.One kilowatt-hour is 3.6 x 106 joules.
43 20.3 Calculate powerYour electric company charges 14 cents per kilowatt-hour. Your coffee maker has a power rating of 1,050 watts.How much does it cost to use the coffee maker one hour per day for a month?1) You are asked to find the cost of using the coffee maker.2) You are given the power in watts and the time.3) Use the power formula P = VI and the fact that 1 kWh = 1kW x 1h.4) Solve: Find the number of kilowatts of power that the coffee maker uses.1,050 W × 1 kW/1,000 W = 1.05 kWFind the kilowatt-hours used by the coffee maker each month.1.05 kW × 1 hr/day x 30 days/month = 31.5 kWh per month.Find the cost of using the coffee maker.31.5 kWh/month × $0.14/kWh = $4.41 per month.
44 20.3 Alternating and direct current The current from a battery is always in the same direction.One end of the battery is positive and the other end is negative.The direction of current flows from positive to negative.This is called direct current, or DC.
45 20.3 Alternating and direct current If voltage alternates, so does current.When the voltage is positive, the current in the circuit is clockwise.When the voltage is negative the current is the opposite direction.This type of current is called alternating current, or AC.
46 20.3 Alternating and direct current AC current is used for almost all high-power applications because it is easier to generate and to transmit over long distances.The 120 volt AC (VAC) electricity used in homes and businesses alternates between peak values of +170 V and -170 V at a frequency of 60 Hz.AC electricity is usually identified by the average voltage, (120 VAC) not the peak voltage.
48 20.3 Power in AC circuitsFor a circuit containing a motor, the power calculation is a little different from that for a simple resistance like a light bulb.Because motors store energy and act like generators, the current and voltage are not in phase with each other.The current is always a little behind the voltage.
49 20.3 Power for AC circuits P = VI x pf Electrical engineers use a power factor (pf) to calculate power for AC circuits with motorsAvg. voltage(volts)Avg. current (amps)P = VI x pfPower (watts)power factor0-100%