Chapter 7: Electricity Recall: Electrons have a negative charge. Electrons can be transferred from one atom to another. Objects that have extra electrons have a negative charge. Objects that have lost electrons have a positive charge. When you scuff your feet across a carpet, electrons are transferred from the carpet fibers to your body and you build up a STATIC CHARGE. When you touch something (or someone) that has a different amount of charge, the excess electrons are transferred resulting a static shock.

Charges exert a force one another. Opposite charges attract. + - Like charges repel as the quantity of charge INCREASES as the distance between the charges DECREASES The magnitude of the force of repulsion/attraction between the charges INCREASES

Electric Fields When an object has a static charge the charges distribute themselves over the surface of the object. The distributed charge exerts a force field on any other charges in the area. The direction of the force is in the same direction a POSITIVE charge would move

Conductors: materials in which electrons can move easily. (Ex: metals, salt water) Insulators: materials in which electrons CANNOT move easily. (Ex: plastics, wood, rubber, glass) Charging by contact: materials acquire a charge by touching each other. Charging by induction: materials acquire a charge as a result of being in an electric field. There is no direct contact involved. Grounding: A conductor is used to allow excess charge to transfer to the Earth. (Ex: Lightning rods provide a route for the large static charge from lightning to get to Earth)

(A)Negative charge builds up on lower part of cloud. Electric field from negative charge in cloud causes positive charge to accumulate on ground. (B)Moisture in air helps to conduct negative toward ground. (C)Built up positive charge arcs up to cloud causing the bright lightning flash. (A)(B) (C)

Electric Current Movement of electric charges is electric current. The units of current are amperes (symbol is A) Charge flow High voltage Low voltage The amount of current that flows depends upon the voltage. Electric charge flows from regions of HIGH voltage to low voltage.

A closed path that electric current follows is a circuit. If the path is broken, the current stops flowing. A battery provides the voltage difference that causes the current to flow. circuit Q: Which end of the battery has the higher voltage? A: The negative terminal. Charge flows from High to low voltage.

Electrodes collect charge. The carbon rod is the positive electrode. The zinc container is the negative electrode. The moist paste is an electrolyte and allows the charges to move to their specific electrode. It is called a cell rather than a battery because there is only one pair of electrodes. It is called a dry cell because it doesn’t contain a liquid electrolyte.

A car battery consists of 6 wet cells. Each wet cell consists of solid lead positive electrode plates and lead dioxide negative electrode plates. Each wet cell generates a voltage of 2 V, so the battery has a combined voltage of 12 V. A sulfuric acid solution is the electrolyte.

Resistance: the tendency for a material to oppose the flow of electrons causing some of the electrical energy to be converted into thermal energy and light Measured in units of ohms (  ). Increases with temperature for most materials. Increases with length for most materials. Increases as the conductor’s width decreases. 47,000 

Resistor Color Code 4 7 x ,000 

Resistor Color Code 1 5 x 1 = 15  6 5 x 10,000 = 650,000  3 0 x 100 = 3000 

Ohm’s Law V = voltage, units are Volts (V) I = current, units are Amperes or Amps (A) R = resistance, units are ohms (  ) V IR Example: 1. What voltage is needed to produce a 0.3 A current through a 100  resistor? V = IR = (0.3 A)(100  ) = 30 V

2. What current will flow through a 100 ohm resistor connected to a 9 V dry cell? 3. What resistance is needed to produce a 0.1 A current in a circuit with a 12 volt battery? V = IRI = V/R = 9V/100  = 0.09 A V = IRR = V/I= 12 V/0.1 A = 120 

Simple Circuits: One resistor R = 10  V = 10 V I = V/R = (10 V)/(10  ) = 1 A Circuits with more than one resistor in SERIES 10  20  V = 9 V The current has to go through both resistors so the two resistances add together R tot = 10  + 20  = 30  I = V/R tot = (9 V)/(30  ) = 0.3 A

10  20  V = 9 V How much voltage is lost at each resistor? Already know that the current through the circuit is 0.3 A so... V = IR = (0.3 A)(10  ) = 3 V V = IR = (0.3 A)(20  ) = 6 V The total voltage dropped at all of the resistors in the circuit have to add up to the voltage of the original power source.

10  25  V = 5 V 15  1. What is the total resistance of the circuit? 2. What is the current through the circuit? 3. What is the voltage drop at each of the resistors? R tot = 10  + 25  + 15  = 50  I = V/Rtot = (5 V)/(50  ) = 0.1 A V = IR = (0.1 A)(10  ) = 1 V V = IR = (0.1 A)(25  ) = 2.5 V V = IR = (0.1 A)(15  ) = 1.5 V V tot = 5 V

What does the current do if the resistors are in PARALLEL? 10  5  V = 10 V The voltage source ‘sees’ the two resistors as being just ONE. The voltage drop across both of them is the same. I = V/R = (10 V)/(10  ) = 1 A I = V/R = (10 V)/(5  ) = 2 A The current through the 10  branch of the circuit is: The current through the 5  branch of the circuit is: The total current through the circuit is 3 A.

10  5  V = 10 V How do the resistors combine in a Parallel circuit? R tot = 3.3  Just to check: I = V/R = (10 V)/(3.3  ) = 3 A

Power Like with work, electrical power is the amount of electrical work done per second. The units of power are Watts (W) It can be calculated from current and voltage using: Or it can be calculated from voltage and resistance using: P = IV P = V 2 R

1. How much power is dissipated in an appliance using 9 V and 0.1 A of current? 2. How much power is dissipated in a 100  resistor with a 12 V voltage drop across it? W of power is dissipated by a resistor with a 15 V voltage drop across it. What is the current through the resistor? P = IV = (0.1 A)(9 V) = 0.9 W P = (V 2 )/R = (12 V) 2 /(100  ) = 144/100 = 1.44 W I = P/V = (150 W)/(15 V) = 10 A