Chapter 22 Current Electricity Electric Current The flow of electrons Electric Circuit A closed loop in which electrons can move The flow is due to a potential difference which is maintained by a pump,

Chapter 22 Current Electricity Forms of electrical energy Chemical Solar Hydroelectric Wind Nuclear

Chapter 22 Current Electricity Simple Electric Cell Zn Electrode (-) Carbon Electrode (+) _ + Two dissimilar metals or carbon rods in acid Zn+ ions enter acid leaving terminal negative Electrons leave carbon leaving it positive Terminals connected to external circuit ‘Battery’ referred to several cells originally Zn+ Zn+ Zn+ Zn+ Sulfuric acid

Chapter 22 Current Electricity Electric Current The flow of electrons The flow is due to a potential difference which is maintained by a pump,

Chapter 22 Current Electricity An Electric Circuit has three components 1..A source of electrical potential difference or voltage. (typically a battery or electrical outlet) 2. A conductive path which would allow for the movement of charges. (typically made of wire) 3. An electrical resistance (resistor) which is loosely defined as any object that uses electricity to do work. (a light bulb, electric motor, heating element, speaker, etc.)

Chapter 22 Current Electricity

Chapter 22 Current Electricity Generator A device using available energy to produce a potential difference. Electric potential energy can be changed to kinetic energy. Electric current is measured in amperes (coulomb/second)

Chapter 22 Current Electricity Electric Power The product of the current and the potential difference P = VI (measured in watts)

Chapter 22 Current Electricity Ohm’ Law For wires and other circuit devices, the current is proportional to the voltage applied to its ends: I  V The current also depends on the amount of resistance that the wire offers to the electrons for a given voltage V. We define a quantity called resistance R such that V = I R (Ohm’s Law) The unit of resistance is the ohm which is represented by the Greek capital omega (). Thus

Chapter 22 Current Electricity    Honey   F x d   R SIMP

Chapter 22 Current Electricity

Chapter 22 Current Electricity Find the current and the power in the circuit 6 Volts 4 Ω I = V/R I = 6V / 4Ω = 1.5 A P = VI = 6V * 1.5 A P = 9 watts

Chapter 22 Current Electricity Draw a circuit with a 9 volt battery and a 6 Ω resistor. Calculate the current and the power in the circuit. 9 Volts 6 Ω I = V/R I = 9V / 6Ω = 1.5 A P = VI = 9V * 1.5 A P = 13.5 watts

Chapter 22 Current Electricity

Chapter 22 Current Electricity Heating effects: P = VI and V = IR so P = I2R E = Pt = I2Rt Q = I2Rt measured in joules

Chapter 22 Current Electricity Kilowatt Hour The rate of energy consumption (power) multiplied by one hour 1 Kwh = (Kj/s)(3600 s) = 3.6 x 106 J

Chapter 22 Current Electricity A clock has a resistance of 12000  and is plugged into a 115 V outlet. a. How much current does it draw? b. How much power does it use? c. If it costs 9¢ per kWh, how much does it cost to operate the clock for 30 days? a. I = V/R = 115V/12000  = .0096 A b. P = VI = (115 V)(.0096 A) = 1.1 W c. Cost = (1.1 x 10-3 kW)($0.09/kWh)(30 days)(24h/day) = $0.07

Chapter 23 Series and Parallel Circuits Series Circuit: Electric current has a single pathway through the electrical circuit; therefore the current passing through the electrical devices is the same everywhere. The current is resisted by the first device, the second, third, etc. The total resistance is the sum of the individual resistors. Rt = R1+ R2+ … + Rt The current (I) is equal to the voltage divided by the total resistance I = V/RT. The total voltage across a series circuit divides among the individual resistors so that the sum of the voltage drops is equal to the total voltage. Voltage is directly proportional to the resistance.

It= V/R = 2 Amps V1= IR = (2A)(30Ω) = 60 V V2= IR = (2A)(15Ω) = 30 V V1 +V2 + V3 = 120 V V3= IR = (2A)(15Ω) = 30 V

Chapter 23 Series and Parallel Circuits Three resistors of 3 , 4 , and 5  are connected in series across a 12 V battery. What is the equivalent resistance? What is the current through each resistor? What is the voltage drop across each resistor? Find the total voltage drop across each resistor.

It= V/R = 12/12 = 1 A V1= IR = (1A)(3Ω) = 3 V V2= IR = (1A)(4Ω) = 4 V 3  4  5  RT = 3 + 4 + 5 = 12  12 V V1= IR = (1A)(3Ω) = 3 V V2= IR = (1A)(4Ω) = 4 V V1 +V2 + V3 = 12 V V3= IR = (1A)(5Ω) = 5 V

Chapter 23 Series and Parallel Circuits Every device connects to the same two points of the circuit. Therefore the voltage is the same. Current divides among the parallel branches. It follows the path of least resistance. The current (I) is equal to the sum of the current in the parallel branches. As the number of parallel branches increases, the overall resistance of the circuit decreases. 1/Rt = 1/R1+1/R2+ … + 1/Rt

RT = 10 Ω I = V/R = 90/10 = 9 Amps I1 = V/R1 = 90 V/60Ω = 1.5 A I2 = V/R2 = 90 V/ 30Ω = 3 A I3 = V/R3 = 90 V/20Ω = 4.5 A I = 1.5 A + 3 A + 4.5 A = 9 A

Find the equivalent resistance and the readings in each meter 120 60 40 Find the equivalent resistance and the readings in each meter

V = 20 V V = 10 V I = V/RT = 90/45 = 2A Rt = 10 + 5 + 6 + 24 = 45 Ω 5.0 Ω 24.0 Ω V = 20 V V = 10 V I = V/RT = 90/45 = 2A Rt = 10 + 5 + 6 + 24 = 45 Ω I = V/R = 90/45 = 2A V = 12 V V = 48 V 20 +10 + 12 + 48 = 90 V

I = V/R = 10/15 = 2/3 A I = V/R = 90/45 = 2A I = V/R = 48/60 = .8 A

Chapter 23 & 24 Series and Parallel Circuits A 30 Ω resistor is connected in parallel with a 20 Ω resistor. The parallel connection is placed in series with a 8 Ω resistor, and the entire circuit is placed across a 60 V difference of potential. Draw the circuit. What is the effective resistance in the entire circuit? What is the voltage drop across the 8 Ω resistor? What is the voltage drop across the parallel branch? What is the current through each resistor?

Chapter 23 & 24 Series and Parallel Circuits

Chapter 23 & 24 Series and Parallel Circuits

Chapter 23 & 24 Series and Parallel Circuits RT = 20 Ω I = 3 A V8Ω = IR = (3A)(8 Ω) = 24 V Vpar = IR = (3A)(12 Ω) = 36 V I8Ω = V/R = 24/8 = 3 A I30Ω = V/R = 1.2 A I20Ω = V/R = 1.8 A

Chapter 23 & 24 Series and Parallel Circuits Does the fuse melt?

Chapter 23 & 24 Series and Parallel Circuits Does the fuse melt? I = V/R = 120/6 = 20 A YES 6 Ω

Chapter 23 & 24 Series and Parallel Circuits Voltmeters: Connected in parallel, high resistance. Ammeters: Connected in series, low resistance.

Measuring in Circuits How to use a multimeter!

Taking Measurements Placement of meter depends on what you want to measure. #1 Goal: the meter should not alter the behavior of the circuit.

Measuring Current The current must flow through the meter. Circuit must be broken to place the meter in series. Ammeters must have very low resistance!

Measuring Potential The voltmeter is connected in parallel between the two points where the measurement is to be made. The voltmeter provides a parallel pathway and should take very little current. A voltmeter must have very high resistance!

Measuring Resistance The component must be removed from the circuit. Ohmmeters work by running a current through the component being tested!

The Meter Always start HIGH! Volts AC Volts DC Always start HIGH! If you are unsure of the measurement range, begin with larger values and reduce multiplier until you get a reading. Amps CURRENT Red lead Ohms Volts and Ohms Red lead Ground Black lead