# Chapter Twenty One: Electrical Systems

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Chapter Twenty One: Electrical Systems
21.1 Series Circuits 21.2 Parallel Circuits 21.3 Electrical Power

21.1 Electrical Systems In a series circuit, current can only take one path, so the current is the same at all points in the circuit.

21.1 Electrical Systems Inexpensive strings of holiday lights are wired with the bulbs in series. If you remove one of the bulbs from its socket, the whole string of mini bulbs will go out.

21.1 Current and resistance in series circuits
If you know the resistance of each device, you can find the total resistance of the circuit by adding up the resistance of each device.

21.1 Current and resistance in series circuits

21.1 Current and resistance in series circuits
Everything has some resistance, even wires.

Q u i z T i m e! A series circuit contains a 12-V battery and three bulbs with resistances of1Ω, 2 Ω, and 3 Ω. What is the current in the circuit?

Q u i z a n s w e r ! Total = 6 Ω Looking for: Given Relationships:
…current (amps) Given …Voltage = 12V; resistances = 1Ω, 2 Ω, 3 Ω. Relationships: Rtot = R1+R2+R3 Ohm’s Law I = V ÷ R Solution Rtot = 6 Ω I = 12 V ÷ 6 Ω = 2 amps Total = 6 Ω

21.1 Voltage drop As each device in series uses power, the power carried by the current is reduced. As a result, the voltage is lower after each device that uses power. This is known as the voltage drop.

21.1 Voltage drop The law of conservation of energy also applies to a circuit. In this circuit, each bulb has a resistance of 1 ohm, so each has a voltage drop of 1 volt when 1 amp flows through the circuit.

21.1 Kirchhoff’s Voltage Law
Kirchhoff’s voltage law states that the total of all the voltage drops must add up to the battery’s voltage.

L a s t Q u i z T i m e t o d a y ! The circuit shown contains a 9-volt battery, a 1-ohm bulb, and a 2- ohm bulb. Calculate the circuit’s total resistance and current. Then find each bulb’s voltage drop.

Q u i z a n s w e r ! = 3 Ω Looking for: Given Relationships:
…total resistance; voltage drop each bulb Given …Voltage = 9V; resistances = 1Ω, 2 Ω Relationships: Rtot = R1+R2+R3 Ohm’s Law I = V ÷ R Solution- part 1 Rtot = 3 Ω I = 9 V ÷ 3 Ω = 3 amps = 3 Ω

Voltage total = (3v + 6v ) = 9 V
Q u i z a n s w e r ! Solution- part 2 Use resistance to find current I = 9 V ÷ 3 ohms = 3 amps Solution- part 3 Rearrange Ohm’s law to solve for voltage Use current to find each voltage drop V = I x R V1 = (3 A) x (1 ohms) = 3 volts V2 = (3 A) x (2 ohms) = 6 volts Voltage total = (3v + 6v ) = 9 V

Parallel Circuits In parallel circuits the current can take more than one path.

21.2 Kirchhoff’s Current Law
All of the current entering a branch point must exit again. This is known as Kirchhoff’s current law.

21.2 Voltage and parallel circuits
If the voltage is the same along a wire, then the same voltage appears across each branch of a parallel circuit.

21.2 Voltage and parallel circuits
Parallel circuits have two advantages over series circuits. Each device in the circuit has a voltage drop equal to the full battery voltage. Each device in the circuit may be turned off independently without stopping the current in the other devices in the circuit.

21.2 Current and parallel circuits
Each branch works independently so the total current in a parallel circuit is the sum of the currents in each branch.

21.2 Calculating in circuits
In a series circuit, adding an extra resistor increases the total resistance of the circuit. In a parallel circuit, more current flows so the total resistance decreases.

RESISTORS IN PARALLEL If no other resistance is present, the potential difference across each resistor equals the potential difference across the terminals of the battery. The equivalent resistance (R) of a parallel combination is always less than the smallest of the individual resistors. The formula for the equivalent resistance is as follows: 1/R = 1/RI + 1/R2 + 1/R3 The potential difference across each resistor in the arrangement is the same, i. e. V = VI = V2 = V3

21.2 Parallel vs. Series Remember: series/same/current; parallel/same/voltage. Use Ohm’s law for both.

Q u i z T i m e! All of the electrical outlets in Jonah’s living room are on one parallel circuit. The circuit breaker cuts off the current if it exceeds 15 amps. Will the breaker trip if he uses a light (240 Ω), stereo (150 Ω), and an air conditioner (10 Ω)?

Q u i z a n s w e r ! Looking for: Given: Relationships: Solution:
whether current exceeds 15 amps Given: ……resistances = 240 W150 W; 10 W Relationships: Assume voltage for each branch = 120 V Ohm’s Law I = V ÷ W Kirchhoff’s Current Law Itotal = I1 +I2 +I3 Solution: Ilight = 120 V ÷ 240 W = 0.5 amps Istereo = 120 V ÷ 150 W = 0.8 amps Ia/c = 120 V ÷ 10 W = 12 amps 0.5 0.8 +12.0 13.3 Breaker will not trip

Solving Problems Given: ……resistances = 240 W 150 W; 10 W
1/R = 1/ 240 W + 1/ 150 W + 1/ 10 W 1/ R = 240/ / /36000 1/R = 3990/36000 R = 36000/ R = 9.02 W Ohm’s Law I = V ÷ R I = 120 v / 9.02 W I = 13.3 A Breaker will not trip

21.2 Short circuits A short circuit is a parallel path in a circuit with very low resistance. A short circuit can be created accidentally by making a parallel branch with a wire.

21.2 Short circuits Each circuit has its own fuse or circuit breaker that stops the current if it exceeds the safe amount, usually 15 or 20 amps If you turn on too many appliances in one circuit at the same time, the circuit breaker or fuse cuts off the current. To restore the current, you must FIRST disconnect some or all of the appliances.

21.2 Fuses In newer homes, flip the tripped circuit breaker.
In older homes you must replace the blown fuse (in older homes). Fuses are also used in car electrical systems and in electrical devices such as televisions or in electrical meters used to test circuits.

Chapter Twenty One: Electrical Systems
21.1 Series Circuits 21.2 Parallel Circuits 21.3 Electrical Power

21.3 Electrical Power Electrical power is measured in watts, just like mechanical power. Power is the rate at which energy is changed into other forms of energy such as heat, sound, or light. Anything that “uses” electricity is actually converting electrical energy into some other type of energy.

21.3 Power Power is a “rate” and is measured using current and voltage .

21.3 Different forms of the Power Equation

21.3 Important review

Watts Used to measure power of light bulbs and small appliances
An electric bill is measured in kW/hrs. 1 kilowatt = 1000 W

Horsepower (hp) = about 746 watts
Traditionally associated with engines. (car,motorcycle,lawn-mower) The term horsepower was developed to quantify power. A strong horse could move a 746 N object one meter in one second.

What is the SI unit of power?
Q u i z T i m e! What is the SI unit of power? Watt

21.3 Electrical Power The watt is an abbreviation for one joule per second. A 100-watt light bulb uses 100 joules of energy every second.

21.3 Kilowatt Most electrical appliances have a label that lists the power in watts (W) or kilowatts (kW). The kilowatt is used for large amounts of power.

A 12-volt battery is connected in series to two identical light bulbs.
Solving Problems A 12-volt battery is connected in series to two identical light bulbs. The current in the circuit is 3 amps. Calculate the power output of the battery.

Looking for: Given: Relationships: Solution: Solving Problems
…power of battery Given: …voltage = 12 V; current = 3 amps Relationships: Power: P = I x V Solution: P = 3 A x 12 V = 36 watts

21.3 Buying Electricity Utility companies charge customers for the number of kilowatt-hours (kWh) used each month. A kilowatt-hour is a unit of energy. The number of kilowatt-hours used equals the number of kilowatts multiplied by the number of hours the appliance was turned on.

21.3 Buying Electricity There are many simple things you can do to use less electricity. When added up, these simple things can mean many dollars of savings each month.

How much does it cost to run a 3,000 W electric stove for 2 hours?
Solving Problems How much does it cost to run a 3,000 W electric stove for 2 hours? Use an electricity cost of \$0.15 per kilowatt-hour. Looking for: …cost to run stove for 2h Given: … P = 3,000W; T = 2h; price \$0.15/kW

Relationships: Solution: = \$ 0.90 Solving Problems 1000 watts = 1 kW
Charge in kWh Solution: 3000 W x 1 kW = 3 kW 1000 W Charge = 3 kW x 2 h = 6 kWh Cost = 6 kWh x \$ 0.15 1 kWh = \$ 0.90

L a s t Q u i z T i m e t o d a y ! How much power does a 100 watt light bulb use if it is turned on for 30 seconds? One more duh! 100 watts!!!!!!!!!!!!!!!!!!!