chapter 22 current electricity zelectric current the flow of electrons zelectric circuit a closed...

Chapter 22Current 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,


Forms of electrical energyChemicalSolarHydroelectricWindNuclear


Simple Electric Cell

Sulfuric acid

Zn+ Zn+

Zn+ Zn+

+++

___

Carbon Electrode

(+)

Zn 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


Electric Current

The flow of electrons

The flow is due to a potential difference which is maintained by a pump,


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.)


Generator A device using available energy to produce apotential difference.

Electric potential energy can be changed to kinetic energy.

Electric current is measured in amperes (coulomb/second)


•Electric Power The product of the current and the potential difference• P = VI (measured in watts)


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

A

V1


F x d

e

Honey

R SIMP


6 Volts

4 Ω

Find the current and the power in the circuit

I = V/R

I = 6V / 4Ω = 1.5 A

P = VI = 6V * 1.5 A

P = 9 watts


9 Volts

6 Ω

Draw a circuit with a 9 volt battery and a 6 Ωresistor. Calculate the current and the power in the circuit.

I = V/R

I = 9V / 6Ω = 1.5 A

P = VI = 9V * 1.5 A

P = 13.5 watts


Heating effects:

P = VI and V = IR so P = I2R

E = Pt = I2Rt

Q = I2Rt measured in joules


Kilowatt Hour The rate of energy consumption (power) multipliedby one hour

1 Kwh = (Kj/s)(3600 s) = 3.6 x 106 J


A clock has a resistance of 12000 and is plugged into a 115 Voutlet.

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 23Series and Parallel Circuits

Series Circuit:1. Electric current has a single pathway through the electrical

circuit; therefore the current passing through the electrical devices is the same everywhere.

2. 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

3. The current (I) is equal to the voltage divided by the total resistance I = V/RT.

4. 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.

5. Voltage is directly proportional to the resistance.

It= V/R = 2 Amps

V1= IR = (2A)(30Ω) = 60 V

V2= IR = (2A)(15Ω) = 30 VV3= IR = (2A)(15Ω) = 30 VV1 +V2 + V3 = 120 V


Three resistors of 3 , 4 , and 5 are

connected in series across a 12 V battery.

a. What is the equivalent resistance?

b. What is the current through each resistor?

c. What is the voltage drop across each resistor?

d. 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 VV3= IR = (1A)(5Ω) = 5 VV1 +V2 + V3 = 12 V

12 V

3 4 5

RT = 3 + 4 + 5 = 12


Parallel Circuit:

1. Every device connects to the same two points of the circuit. Therefore the voltage is the same.

2. Current divides among the parallel branches. It follows the path of least resistance.

3. The current (I) is equal to the sum of the current in the parallel branches.

4. As the number of parallel branches increases, the overall resistance of the circuit decreases. 1/Rt = 1/R1+1/R2+ … + 1/Rt

1RT

160

130

120

660

110

RT = 10 ΩI = 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 AI = 1.5 A + 3 A + 4.5 A = 9 A

12 V120 60 40

Find the equivalent resistanceand the readings in each meter

40

1

60

1

120

11

tR

20tR AV

R

VI 6.

20

12

AV

R

VI 1.

120

121

AV

R

VI 2.

60

122

AV

R

VI 3.

400

123

I = V/RT = 90/45 = 2A

5.0 Ω

24.0 Ω

V = 20 V V = 10 V

V = 12 VV = 48 V20 +10 + 12 + 48 = 90 V

Rt = 10 + 5 + 6 + 24 = 45 ΩI = V/R = 90/45 = 2A

I = V/R = 10/15 = 2/3 A

I = V/R = 48/60 = .8 AI = V/R = 90/45 = 2A

Chapter 23 & 24Series 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.

a. Draw the circuit.b. What is the effective resistance in the entire circuit?c. What is the voltage drop across the 8 Ω resistor?d. What is the voltage drop across the parallel branch?e. What is the current through each resistor?


b. RT = 20 Ω I = 3 A

c. V8Ω = IR = (3A)(8 Ω) = 24 V

d. Vpar = IR = (3A)(12 Ω) = 36 V

e. I8Ω = V/R = 24/8 = 3 A I30Ω = V/R = 1.2 A I20Ω = V/R = 1.8 A


Does the fuse melt?


Does the fuse melt?

6 Ω

I = V/R = 120/6 = 20 A

YES


Voltmeters: Connected in parallel, high resistance.

Ammeters: Connected in series, low resistance.

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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

GroundBlack lead

CURRENTRed lead

Volts and OhmsRed lead

Always start HIGH!

If you are unsure of the measurement range, begin with larger values and reduce multiplier until you get a reading.

Ohms

VoltsDC

VoltsAC

Amps

chapter 22 current electricity zelectric current the flow of electrons zelectric circuit a closed...

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