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Published byEdmund McBride Modified over 4 years ago

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Ch. 7 - Electricity SPS10. Students will investigate the properties of electricity and magnetism. a. Investigate static electricity in terms of friction, induction, conduction b. Explain the flow of electrons in terms of alternating and direct current. the relationship among voltage, resistance and current. simple series and parallel circuits. 1 1

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**Ch. 7 - Electricity I. Electric Charge Static Electricity Conductors**

Insulators Electroscope 1 1

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**A. Static Electricity Static Electricity**

the net accumulation of electric charges on an object Electric Field force exerted by an e- on anything that has an electric charge opposite charges attract like charges repel

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**A. Static Electricity Static Discharge**

the movement of electrons to relieve a separation in charge

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**A. Static Electricity Three ways to achieve static electricity**

Friction Rubbing together two objects Induction Inducing a positive or negative charge Conduction Transferring a positive or negative charge Demo on Static Electricity

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**B. Conductors Conductor**

material that allows electrons to move through it easily e- are loosely held ex: metals like copper and silver

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**C. Insulators Insulator**

material that doesn’t allow electrons to move through it easily e- are tightly held ex: plastic, wood, rubber, glass

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**D. Electroscope Electroscope**

instrument that detects the presence of electrical charges leaves separate when they gain either a + or - charge

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**Ch. 7 - Electricity II. Electric Current Circuit Potential Difference**

Resistance Ohm’s Law

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A. Circuit Circuit closed path through which electrons can flow

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**A. Potential Difference**

Potential Difference (voltage) difference in electrical potential between two places large separation of charge creates high voltage the “push” that causes e- to move from - to + measured in volts (V)

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**B. Current Current flow of electrons through a conductor**

depends on # of e- passing a point in a given time measured in amperes (A)

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**C. Resistance Resistance opposition the flow of electrons**

electrical energy is converted to thermal energy & light measured in ohms () Tungsten - high resistance Copper - low resistance

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**C. Resistance Resistance depends on… the conductor wire thickness**

less resistance in thicker wires wire length less resistance in shorter wires temp - less resistance at low temps

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**V = I × R E. Ohm’s Law V: potential difference (V) I: current (A)**

R: resistance () V = I × R Voltage increases when current increases. Voltage decreases when resistance increases.

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**E. Ohm’s Law R = 160 V= I * R V = 120 V I = V ÷ R I = ?**

A light bulb with a resistance of 160 is plugged into a 120-V outlet. What is the current flowing through the bulb? GIVEN: R = 160 V = 120 V I = ? WORK: V= I * R Divide both sides by R I = V ÷ R I = (120 V) ÷ (160 ) I = 0.75 A

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**E. Ohm’s Law R = 50 V= I * R V = 110 V I = V ÷ R I = ?**

A cell phone with a resistance of 50 is plugged into a 110-V outlet. What is the current flowing through the bulb? GIVEN: R = 50 V = 110 V I = ? WORK: V= I * R Divide both sides by R I = V ÷ R I = (110 V) ÷ (50 ) I = 2.2 A

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**E. Ohm’s Law R = ? V= I * R V = 120 V R = V ÷ I I = 4.25**

A Television is plugged into a 120-V outlet. How much resistance does it have if the current is measured to be 4.25 A? GIVEN: R = ? V = 120 V I = 4.25 WORK: V= I * R Divide both sides by R R = V ÷ I R = (120 V) ÷ (4.25 A) R = 28.2

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**III. Electrical Circuits**

Ch. 7 - Electricity III. Electrical Circuits Circuit components Series circuits Parallel circuits Household circuits

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**A. Circuit Components A - battery C - light bulb**

B - switch D - resistor

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**B. Series Circuits Series Circuit current travels in a single path**

one break stops the flow of current current is the same throughout circuit lights are equal brightness each device receives a fraction of the total voltage get dimmer as lights are added

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**C. Parallel Circuits Parallel Circuits**

current travels in multiple paths one break doesn’t stop flow current varies in different branches takes path of least resistance “bigger” light would be dimmer each device receives the total voltage no change when lights are added

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**D. Household Circuits Combination of parallel circuits**

too many devices can cause wires to overheat Safety Features: fuse - metal melts, breaking circuit circuit breaker - bimetallic strip bends when hot, breaking circuit

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**IV. Measuring Electricity Electrical Power Electrical Energy**

Ch. 7 - Electricity IV. Measuring Electricity Electrical Power Electrical Energy

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**P = I × V A. Electrical Power P: power (W) I: current (A) V: potential**

rate at which electrical energy is converted to another form of energy P: power (W) I: current (A) V: potential difference (V) P = I × V

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**A. Electrical Power I = 0.01 A P = I · V V = 9 V P = (0.01 A) (9 V)**

A calculator has a 0.01-A current flowing through it. It operates with a potential difference of 9 V. How much power does it use? GIVEN: I = 0.01 A V = 9 V P = ? WORK: P = I · V P = (0.01 A) (9 V) P = 0.09 W

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**A. Electrical Power I = 0.05 A P = I · V V = 8 V P = (0.05 A) (8 V)**

A flashlight has a 0.05-A current flowing through it. It operates with a potential difference of 8 V. How much power does it use? GIVEN: I = 0.05 A V = 8 V P = ? WORK: P = I · V P = (0.05 A) (8 V) P = 0.4 W

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**E = P × t B. Electrical Energy E: energy (kWh) P: power (kW)**

energy use of an appliance depends on power required and time used E: energy (kWh) P: power (kW) t: time (h) E = P × t

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**B. Electrical Energy P = 700 W E = P · t = 0.7 kW E = (0.7 kW) (10 h)**

A refrigerator is a major user of electrical power. If it uses 700 W and runs 10 hours each day, how much energy (in kWh) is used in one day? GIVEN: P = 700 W = 0.7 kW t = 10 h E = ? WORK: E = P · t E = (0.7 kW) (10 h) E = 7 kWh

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**B. Electrical Energy P = 900 W E = P · t = 0.9 kW E = (0.9 kW) (14 h)**

A refrigerator is a major user of electrical power. If it uses 700 W and runs 10 hours each day, how much energy (in kWh) is used in one day? GIVEN: P = 900 W = 0.9 kW t = 14 h E = ? WORK: E = P · t E = (0.9 kW) (14 h) E = 12.6 kWh

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