Electricity & Magnetism Unit

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Electricity & Magnetism Unit. Understanding Circuits. Objectives. Recognize and sketch examples of series and parallel circuits. Describe a short circuit and why a short circuit may be a dangerous hazard. - PowerPoint PPT Presentation

Text of Electricity & Magnetism Unit

Electricity & Magnetism Unit

Electricity &Magnetism UnitUnderstanding Circuits1ObjectivesRecognize 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.2Chapter 20 Vocabulary Termsseries circuit parallel circuit short circuit network circuit circuit analysispower Kirchhoffs voltage law voltage drop direct current (DC)

alternating current (AC)kilowatt Kirchhoffs current law horsepower power factor circuit breakerwatt kilowatt-hour

20.1 Series and Parallel Circuits

Key Question:How do series and parallel circuits work?*Students read Section 20.1 AFTER Investigation 20.1420.1 Series and Parallel CircuitsIn series circuits, current can only take one path.The amount of current is the same at all points in a series circuit.

20.1 Adding resistances in seriesEach resistance in a series circuit adds to the total resistance of the circuit.Rtotal = R1 + R2 + R3...Total resistance(ohms)

Individual resistances (W)

20.1 Total resistance in a series circuitLight bulbs, resistors, motors, and heaters usually have much greater resistance than wires and batteries.

20.1 Calculate currentHow much current flows in a circuit with a 1.5-volt battery and three 1 ohm resistances (bulbs) in series?

101) You are asked to calculate current.2) You are given the voltage and resistances.3) Use Ohms law, I = VR, and add the resistance in series.4) Solve:Resistance = R1 + R2 + R3 = 1 + 1 + 1 = 3Current, I = (1.5 V) (3)=0.5 A

20.1 Voltage in a series circuitEach 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 energyOhms law is used to calculate the voltage drop across each resistor.

20.1 Series and Parallel CircuitsIn parallel circuits the current can take more than one path.Because there are multiple branches, the current is not the same at all points in a parallel circuit.

20.1 Series and Parallel CircuitsSometimes 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 Kirchhoffs current law.20.1 Voltage and current in a parallel circuitIn 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.20.1 Advantages of parallel circuitsParallel 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.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.

20.1 Calculate currentTwo 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.

201) You are asked for the current.2) You are given the voltage and resistance.3) Use Ohms law: I = V R.4) For the 3 bulb:I = (3 V) (3 ) = 1 A.For the 0.5 bulb:I = (3 V) (0.5 ) = 6 A.The battery must supply the current for both bulbs, which adds up to 7 amps.

20.1 Resistance in parallel circuitsAdding 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.

20.1 Adding resistance in parallel circuitsA circuit contains a 2 ohm resistor and a 4 ohm resistor in parallel. Calculate the total resistance of the circuit.

231) 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 + 1/4 = 2/4 +1/4 = 3/4 R= 4/3 = 1.33

20.2 Solving circuit problemsIdentify 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. 20.2 Solving circuit problemsIf you know the total resistance and current, use Ohms law as V = IR to calculate voltages or voltage drops. If you know the resistance and voltage, use Ohms law as I = V R to calculate the current.An unknown resistance can be found using Ohms law as R = V I, if you know the current and the voltage drop through the resistor.Use Kirchhoffs current and voltage laws as necessary.

20.2 Solving circuit problemsA bulb with a resistance of 1 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?

261) 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 Ohms law, R = V I, and add the resistance in series.4) Solve:Total resistance = 6V 1A = 6.SInce the bulb is 1, the additional resistor must be 5 to get a total 6 of resistance.

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 Kirchhoffs laws.20.2 Calculate using network circuitsThree bulbs, each with a resistance of 3, 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?

281) You are asked to calculate the currents.2) You are given the circuit diagram, voltages, and resistances.3) Use Ohms 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 + 1/3 = 2/3 R= 3/2 = 1.5 5) Calculate the total resistance of 4.5 by adding up the remaining series resistances.Calculate the total current using Ohms law: I = 3V 4.5 = 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.

20.3 Electric Power, AC, and DC ElectricityThe 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.

20.3 Reviewing terms

20.3 Power in electric circuitsOne 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. 20.3 PowerP = VICurrent (amps)Voltage (volts) Power (watts)

20.3 Calculate powerA light bulb with a resistance of 1.5 is connected to a 1.5-volt battery in the circuit shown at right. Calculate the power used by the light bulb.

331) You are asked to find the power used by the light bulb.2) You are given the voltage of the battery and the bulbs resistance.3) Use Ohms law, I = V/R, to calculate the current; then use the power equation, P=VI, to calculate the power.4) Solve: I = 1.5V 1.5 = 1AP = 1.5V 1A = 1.5 W; the bulb uses 1.5 watts of electric power.

20.3 Paying for electricityElectric 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.

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?

351) 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 mont