Slide 1 99

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Electric Current amp

DC Circuits

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Slide 2 99

This material is made freely available at wwwnjctlorg and is intended for the non-commercial use of students and teachers These materials may not be used for any commercial purpose without the written permission of the owners NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers participate in a virtual professional learning community andor provide access to course materials to parents students and others

Click to go to websitewwwnjctlorg

New Jersey Center for Teaching and Learning

Progressive Science Initiative

Slide 3 99

How to Use this File

middot Each topic is composed of brief direct instruction

middot There are formative assessment questions after every topic denoted by black text and a number in the upper left

gt Students work in groups to solve these problems but use student responders to enter their own answers

gt Designed for SMART Response PE student response systems

gt Use only as many questions as necessary for a sufficient number of students to learn a topic

middot Full information on how to teach with NJCTL courses can be found at njctlorgcoursesteaching methods

Slide 4 99

Click on the topic to go to that section

middot Circuits

middot Conductors

middot Resistivity and Resistance

middot Circuit Diagrams

Electric Current amp DC Circuits

middot Measurement

httpnjctln3

middot EMF amp Terminal Voltage

middot Kirchhoffs Rules

Slide 5 99

Return toTable ofContents

Circuits

httpnjctln3

Slide 6 99

Electric Current

Electric Current is the rate of flow of electric charges (charge carriers) through space More specifically it is defined as the amount of charge that flows past a location in a material per unit time The letter I is the symbol for current

ΔQ is the amount of charge and Δt is the time it flowed past the location

The current depends on the type of material and the Electric Potential difference (voltage) across it

ΔQ ΔtI =

Slide 7 99

Electric Current

A good analogy to help understand Electric Current is to consider water flow The flow of water molecules is similar to the flow of electrons (the charge carriers) in a wire

Water flow depends on the pressure exerted on the molecules either by a pump or by a height difference such as water falling off a cliff

Electric current depends on the pressure exerted by the Electric Potential difference - the greater the Electric Potential difference the greater the Electric Current

Slide 8 99

The current has the units Coulombs per second

The units can be rewritten as Amperes (A)

1 A = 1 Cs

Amperes are often called amps

ΔQ ΔtI =

Electric Current

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Slide 7 99

Electric Current

A good analogy to help understand Electric Current is to consider water flow The flow of water molecules is similar to the flow of electrons (the charge carriers) in a wire

Water flow depends on the pressure exerted on the molecules either by a pump or by a height difference such as water falling off a cliff

Electric current depends on the pressure exerted by the Electric Potential difference - the greater the Electric Potential difference the greater the Electric Current

Slide 8 99

The current has the units Coulombs per second

The units can be rewritten as Amperes (A)

1 A = 1 Cs

Amperes are often called amps

ΔQ ΔtI =

Electric Current

httpnjctln3

Slide 8 99

The current has the units Coulombs per second

The units can be rewritten as Amperes (A)

1 A = 1 Cs

Amperes are often called amps

ΔQ ΔtI =

Electric Current

httpnjctln3

Slide 9 99

Electric Current

We know that if an Electric Potential difference is applied to a wire charges will flow from high to low potential - a current

However due to a convention set by Benjamin Franklin current in a wire is defined as the movement of positive charges (not the electrons which are really moving) and is called conventional current

Ben didnt do this to confuse future generations of electrical engineers and students It was already known that electrical phenomena came in two flavors - attractive and repulsive - Ben was the person who explained them as distinct positive and negative charges

Slide 10 99

Electric Current

He arbitrarily assigned a positive charge to a glass rod that had been rubbed with silk He could just as easily called it negative - 5050 chance

The glass rod was later found to have a shortage of electrons (they were transferred to the silk) So if the glass rod is grounded the electrons will flow from the ground to the rod

The problem comes in how Electric Potential is defined - charge carriers will be driven from high to low potential - from positive to negative For this to occur in the glass rod - ground system the conventional current will flow from the rod to the ground - opposite the direction of the movement of electrons

Slide 11 99

Electric Current

To summarize - conventional Electric Current is defined as the movement of positive charge In wires it is opposite to the direction of the electron movement

However - in the case of a particle accelerator where electrons are stripped off of an atom resulting in a positively charged ion which is then accelerated to strike a target - the direction of the conventional current is the same as the direction of the positive ions

Slide 12 99

Circuits

An electric circuit is an external path that charges can follow between two terminals using a conducting material

For charge to flow the path must be complete and unbroken

An example of a conductor used to form a circuit is copper wire Continuing the water analogy one can think of a wire as a pipe for charge to move through

httpnjctln3

Slide 10 99

Electric Current

He arbitrarily assigned a positive charge to a glass rod that had been rubbed with silk He could just as easily called it negative - 5050 chance

The glass rod was later found to have a shortage of electrons (they were transferred to the silk) So if the glass rod is grounded the electrons will flow from the ground to the rod

The problem comes in how Electric Potential is defined - charge carriers will be driven from high to low potential - from positive to negative For this to occur in the glass rod - ground system the conventional current will flow from the rod to the ground - opposite the direction of the movement of electrons

Slide 11 99

Electric Current

To summarize - conventional Electric Current is defined as the movement of positive charge In wires it is opposite to the direction of the electron movement

However - in the case of a particle accelerator where electrons are stripped off of an atom resulting in a positively charged ion which is then accelerated to strike a target - the direction of the conventional current is the same as the direction of the positive ions

Slide 12 99

Circuits

An electric circuit is an external path that charges can follow between two terminals using a conducting material

For charge to flow the path must be complete and unbroken

An example of a conductor used to form a circuit is copper wire Continuing the water analogy one can think of a wire as a pipe for charge to move through

httpnjctln3

Slide 11 99

Electric Current

To summarize - conventional Electric Current is defined as the movement of positive charge In wires it is opposite to the direction of the electron movement

However - in the case of a particle accelerator where electrons are stripped off of an atom resulting in a positively charged ion which is then accelerated to strike a target - the direction of the conventional current is the same as the direction of the positive ions

Slide 12 99

Circuits

An electric circuit is an external path that charges can follow between two terminals using a conducting material

For charge to flow the path must be complete and unbroken

An example of a conductor used to form a circuit is copper wire Continuing the water analogy one can think of a wire as a pipe for charge to move through

httpnjctln3

Slide 12 99

Circuits

An electric circuit is an external path that charges can follow between two terminals using a conducting material

For charge to flow the path must be complete and unbroken

An example of a conductor used to form a circuit is copper wire Continuing the water analogy one can think of a wire as a pipe for charge to move through

httpnjctln3

Slide 13 99

1 12 C of charge passes a location in a circuit in 10 seconds What is the current flowing past the point

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Slide 13 (Answer) 99

1 12 C of charge passes a location in a circuit in 10 seconds What is the current flowing past the point

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Slide 14 99

2 A circuit has 3 A of current How long does it take 45 C of charge to travel through the circuit

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Slide 14 (Answer) 99

2 A circuit has 3 A of current How long does it take 45 C of charge to travel through the circuit

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Slide 15 99

3 A circuit has 25 A of current How much charge travels through the circuit after 4s

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Slide 15 (Answer) 99

3 A circuit has 25 A of current How much charge travels through the circuit after 4s

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Slide 16 99

Batteries

Positive Terminal

Negative Terminal

Each battery has two terminals which are conductors The terminals are used to connect an external circuit allowing the movement of charge

Batteries convert chemical energy to electrical energy which maintains the potential difference

The chemical reaction acts like an escalator carrying charge up to a higher voltage

Click here for a Battery VoltageSimulation from PhET

httpnjctln9

Slide 17 99

Reviewing Basic CircuitsThe circuit cannot have gaps

The bulb had to be between the wire and the terminal

A voltage difference is needed to make the bulb light

The bulb still lights regardless of which side of the battery you place it on

As you watch the videoobservations and

the answers to the questions below

What is going on in the circuit

What is the role of the battery

How are the circuits similar different

Click here for video using thecircuit simulator

from PhET

httpnjctln9

Slide 18 99

The battery pushes current through the circuit A battery acts like a pump pushing charge through the circuit It is the circuits energy source

Charges do not experience an electrical force unless there is a difference in electrical potential (voltage)Therefore batteries have a potential difference between their terminals The positive terminal is at a higher voltage than the negative terminal

Batteries and Current

How will voltage affect currentclick here for a video from

Veritasiums Derek on current

httpnjctln9

Slide 19 99

Return toTable ofContents

Conductors

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Slide 20 99

Conductors

Some conductors conduct better or worse than others Reminder conducting means a material allows for the free flow of electrons

The flow of electrons is just another name for current Another way to look at it is that some conductors resist current to a greater or lesser extent

We call this resistance R Resistance is measured in ohms which is noted by the Greek symbol omega (Ω)

Click here to run another PhET simulation

How will resistance affect current

httpnjctlna

Slide 21 99

Current vs Resistance amp VoltageRaising resistance reduces currentRaising voltage increases current

We can combine these relationships in what we call Ohms Law

Another way to write this is that

OR V = IR

V

RI =

V

IR =

You can see that one = V

A click here for a Veritasiummusic video on electricity

httpnjctlna

Slide 22 99

4 A flashlight has a resistance of 25 and is connected by a wire to a 120 V source of voltage What is the current in the flashlight

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Slide 22 (Answer) 99

4 A flashlight has a resistance of 25 and is connected by a wire to a 120 V source of voltage What is the current in the flashlight

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Slide 23 99

5 How much voltage is needed in order to produce a 070 A current through a 490 resistor

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Slide 23 (Answer) 99

Slide 24 99

6 What is the resistance of a rheostat coil if 005 A of current flows through it when 6 V is applied across it

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Slide 24 (Answer) 99

6 What is the resistance of a rheostat coil if 005 A of current flows through it when 6 V is applied across it

httpnjctlng

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Slide 25 99

Electrical Power

Power is defined as work per unit time

if W = QV then substitute

if then substitute

P = W

t

P = QV

t

I = Q

t P = IV

What happens if the current is increased

What happens if the voltage is decreased

httpnjctlni

Slide 26 99

Electrical Power

Lets think about this another way

The water at the top has GPE amp KE

As the water falls it loses GPE and the wheel gets turned doing workWhen the water falls to the bottom it is now slower having done work

httpnjctlni

Slide 27 99

Electrical Power

Electric circuits are similar

A charge falls from high voltage to low voltage

In the process of falling energy may be used (light bulb run a motor etc)

What is the unit of Power

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Slide 28 99

Electrical Power

How can we re-write electrical power by using Ohms Law

P = IV(electrical power)

I = V R

(Ohms Law)

P = VV R

P = V2

R

httpnjctlni

Slide 29 99

Is there yet another way to rewrite this

P = IV(electrical power)

V = I R(Ohms Law)

P = I(IR)

P = I2R

We can substitute this into Power

I = V can be rewritten as V = IR R

Electrical Power

httpnjctlni

Slide 30 99

D C AAA AA 9 V

15 V

D C AA amp AAA have the same voltage however they differ in the amount of power they deliver

For instance D batteries can deliver more current and therefore more power

Batteries

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Slide 31 99

7 A toy cars electric motor has a resistance of 17 find the power delivered to it by a 6-V battery

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Slide 31 (Answer) 99

7 A toy cars electric motor has a resistance of 17 find the power delivered to it by a 6-V battery

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Slide 32 99

8 What is the power consumption of a flash light bulb that draws a current of 028 A when connected to a 6 V battery

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Slide 32 (Answer) 99

8 What is the power consumption of a flash light bulb that draws a current of 028 A when connected to a 6 V battery

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Slide 33 99

9 A 30Ω toaster consumes 560 W of power how much current is flowing through the toaster

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Slide 33 (Answer) 99

Slide 34 99

10 When 30 V is applied across a resistor it generates 600 W of heat what is the magnitude of its resistance

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Slide 34 (Answer) 99

10 When 30 V is applied across a resistor it generates 600 W of heat what is the magnitude of its resistance

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Slide 35 99

Return toTable ofContents

Resistivity and Resistance

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Slide 36 99

Pipe size

How could the wire in the circuit affect the current

If wire is like a pipe and current is like water that flows through the pipe

if there were pipes with water in them what could we do to the pipes to change the speed of the water (the current)

httpnjctlnr

Slide 36 (Answer) 99

Pipe size

How could the wire in the circuit affect the current

If wire is like a pipe and current is like water that flows through the pipe

if there were pipes with water in them what could we do to the pipes to change the speed of the water (the current)

httpnjctlnr

Ans

wer

change the cross-sectional area of the pipe

making it bigger will allow more water to flow

change the length of the pipeincreasing the length will increase the time it takes for the water to get to the end of its trip

Slide 37 99

Resistivity amp ResisitanceEvery conductor conducts electric charge to a greater or lesser extent

The last example also applies to conductors like copper wire Decreasing the length (L) or increasing the cross-sectional area (A) would increase conductivity

Also the measure of a conductors resistance to conduct is called its resistivity Each material has a different resistivity

Resistivity is abbreviated using the Greek letter rho ()

Combining what we know about A L and ρ we can find a conductors total resistance

R = L A

httpnjctlnr

Slide 38 99

Resistivity amp Resisitance

Resistance R is measured in Ohms (Ω) Ω is the Greek letter Omega

Cross-sectional area A is measured in m2

Length L is measured in m

Resistivity ρ is measured in Ωm

R = L A

How can we define A for a wire

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Slide 39 99

Resisitance

What is the resistance of a good conductor

Low low resistance means that electric charges are free to move in a conductor

= RA L

Click here for a PhETsimulation about Resistance

httpnjctlnr

Slide 40 99

Resistivities of Common Conductors

Resistivity (10-8 Ωm)MaterialSilver

Copper

Gold

Aluminum

Tungsten

Iron

Platinum

Mercury

Nichrome

159

168

244

265

560

971

106

98

100

Slide 41 99

11 Rank the following materials in order of best conductor to worst conductor

A Iron Copper Platinum

B Platinum Iron Copper

C Copper Iron Platinum

Resistivity (10-8 Ωm)MaterialSilver

Copper

Gold

Aluminum

Tungsten

Iron

Platinum

Mercury

Nichrome

159

168

244

265

560

971

106

98

100

httpnjctlnr

Slide 41 99

11 Rank the following materials in order of best conductor to worst conductor

A Iron Copper Platinum

B Platinum Iron Copper

C Copper Iron Platinum

Resistivity (10-8 Ωm)MaterialSilver

Copper

Gold

Aluminum

Tungsten

Iron

Platinum

Mercury

Nichrome

159

168

244

265

560

971

106

98

100

httpnjctlnr

Slide 41 (Answer) 99

11 Rank the following materials in order of best conductor to worst conductor

A Iron Copper Platinum

B Platinum Iron Copper

C Copper Iron Platinum

Resistivity (10-8 Ωm)MaterialSilver

Copper

Gold

Aluminum

Tungsten

Iron

Platinum

Mercury

Nichrome

159

168

244

265

560

971

106

98

100

httpnjctlnr

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C

Slide 42 99

12 What is the resistance of a 2 m long copper wire whose cross-sectional area of 02 mm2

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Slide 42 (Answer) 99

Slide 43 99

13 An aluminum wire with a length of 900 m and cross-sectional area of 10 mm 2 has a resistance of 25 What is the resistivity of the wire

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Slide 43 (Answer) 99

13 An aluminum wire with a length of 900 m and cross-sectional area of 10 mm 2 has a resistance of 25 What is the resistivity of the wire

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Slide 44 99

14 What diameter of 100 m long copper wire would have a resistance of 010

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Slide 44 (Answer) 99

14 What diameter of 100 m long copper wire would have a resistance of 010

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Slide 45 99

15 What is the length of a 10 Ω copper wire whose diameter is 32 mm

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Slide 45 (Answer) 99

15 What is the length of a 10 Ω copper wire whose diameter is 32 mm

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Slide 46 99

16 The length of a copper wire is cut to half By what factor does the resistance change

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A 14B 12C 2D 4

Slide 46 (Answer) 99

16 The length of a copper wire is cut to half By what factor does the resistance change

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A 14B 12C 2D 4

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B

Slide 47 99

17 The radius of a copper wire is doubled By what factor does the resistivity change

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A 14B 12C 1D 2

Slide 47 (Answer) 99

17 The radius of a copper wire is doubled By what factor does the resistivity change

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A 14B 12C 1D 2

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C

Slide 48 99

Return toTable ofContents

Circuit Diagrams

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Slide 49 99

Circuit Diagrams

Drawing realistic pictures of circuits can be very difficult For this reason we have common symbols to represent each piece

Resistor Battery Wire

Note Circuit diagrams do not show where each part is physically located

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Slide 50 99

Circuit Diagrams

Draw a simple circuit that has a 9 V battery with a 3 Ω resistor across its terminals What is the magnitude and direction of the current

Conventional current flows from the positive terminal to the negative terminal

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Slide 50 (Answer) 99

Circuit Diagrams

Draw a simple circuit that has a 9 V battery with a 3 Ω resistor across its terminals What is the magnitude and direction of the current

Conventional current flows from the positive terminal to the negative terminal

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R = 3

V = 9 V

II = 3A

Slide 51 99

There are two ways to add a second resistor to the circuit

R1 R2

V

R1

R2

V

Series Parallel

All charges must move through both resistors to get to the negative

terminal

Charges pass through either R1 or R2 but not both

Circuit Diagrams

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Slide 52 99

Are the following sets of resistors in series or parallelR1

R2

V

R1

R2V

Circuit Diagrams

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Slide 52 (Answer) 99

Are the following sets of resistors in series or parallelR1

R2

V

R1

R2V

Circuit Diagrams

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SeriesParallelThe test is to trace the shortest route around the circuit The resistors found on the same route are in series those not found on the same route are in parallel to those that were

Slide 53 99

Equivalent Resistance

Resistors and voltage from batteries determine the current

Circuits can be redrawn as if there were only a single resistor and batteryBy reducing the circuit this way the circuit becomes easier to study

The process of reducing the resistors in a circuit is called finding the equivalent resistance (Req)

R1 R2

V

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Slide 54 99

Series Circuits Equivalent Resistance

What happens to the current in the circuit to the right

R1 R2

V

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Slide 54 (Answer) 99

Series Circuits Equivalent Resistance

What happens to the current in the circuit to the right

R1 R2

V

pI4oFdjExZE+++TI-Tmp-njctl_a

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wer The current passing through

all parts of a series circuit is the same For example I = I1 = I2

Slide 55 99

Series Circuits Equivalent Resistance

What happens to the voltage as it moves around the circuit

R1 R2

V

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Slide 55 (Answer) 99

Series Circuits Equivalent Resistance

What happens to the voltage as it moves around the circuit

R1 R2

V

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The sum of the voltage drops across each of the resistors in a series circuit equals the voltage of the battery

For example V = V1 + V2

Slide 56 99

If V = V1 + V2 + V3 +

IR = I1R1 + I2R2 + I3R3

IR = IR1 + IR2 + IR3

Req = R1 + R2 + R3 +

To find the equivalent resistance (Req) of a series circuit add the resistance of all the resistorsIf you add more resistors to a series circuit what happens to the resistance

Series Circuits Equivalent Resistance

substitute Ohms Law solved for V is V = IR

but since current (I) is the same everywhere in a series circuit

I = I1 = I2 = I3

Now divide by I

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Slide 57 99

18 What is the equivalent resistance in this circuit

R1 = 5 R2 = 3

V = 9 V

httpnjctlnw

Slide 57 (Answer) 99

18 What is the equivalent resistance in this circuit

R1 = 5 R2 = 3

V = 9 V

httpnjctlnw

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Slide 58 99

19 What is the total current at any spot in the circuit

R1 = 5 R2 = 3

V = 9 V

httpnjctlnx

Slide 58 (Answer) 99

19 What is the total current at any spot in the circuit

R1 = 5 R2 = 3

V = 9 V

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Resistors in series

Slide 59 99

20 What is the voltage drop across R1

R1 = 5 R2 = 3

V = 9 V

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Slide 59 (Answer) 99

20 What is the voltage drop across R1

R1 = 5 R2 = 3

V = 9 V

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Resistors in series

Net Currentequal everywhere

Voltage Drop across R1

Slide 60 99

hint A good way to check your work is to see if the voltage drop across all resistors equals the total voltage in the circuit

21 What is the voltage drop across R2

R1 = 5 R2 = 3

V = 9 V

httpnjctlnz

Slide 60 (Answer) 99

hint A good way to check your work is to see if the voltage drop across all resistors equals the total voltage in the circuit

21 What is the voltage drop across R2

R1 = 5 R2 = 3

V = 9 V

httpnjctlnz

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Net Currentequal everywhereResistors in series

Voltage Drop across R2

Slide 61 99

22 How much power is used by R1

R1 = 5 R2 = 3

V = 9 V

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Slide 61 (Answer) 99

Slide 62 99

Parallel Circuits Equivalent Resistance

What happens to the current in the circuit to the right

R1

R2

V

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Slide 62 (Answer) 99

Parallel Circuits Equivalent Resistance

What happens to the current in the circuit to the right

R1

R2

V

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The sum of the currents through each of the resistors in a parallel circuit equals the current of the battery

For example I = I1 + I2

Slide 63 99

Parallel Circuits Equivalent Resistance

What happens to the voltage as it moves around the circuit

R1

R2

V

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Slide 63 (Answer) 99

Parallel Circuits Equivalent Resistance

What happens to the voltage as it moves around the circuit

R1

R2

V

httpnjctlo5

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wer The voltage across all the

resistors in a parallel circuit is the same

For example V = V1 = V2

Slide 64 99

If I = I1 + I2 + I3

V1

R1

VR

V3

R3

V2

R2++=

VR1

VR

VR3

VR2

++=

1R1

VR

1R3

1R2

++= V( (

1R1

1Req

1R3

1R2

++=

If you add more resistors in parallel what will happen to the resistance of the circuit

Rewrite Ohms Law for I and substitute for each resistor

Also since V = V1 = V2 = V3 so we can substitute V for any other voltage

Voltage is a common factor so factor it out

Divide by V to eliminate voltage from the equation

Parallel Circuits Equivalent Resistance

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Slide 65 99

23 What is the equivalent resistance in the circuit

R1 = 3

R2 = 6

V = 18V

httpnjctlo6

Slide 65 (Answer) 99

23 What is the equivalent resistance in the circuit

R1 = 3

R2 = 6

V = 18V

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Slide 66 99

24 What is the voltage at any spot in the circuit

R1 = 3

R2 = 6

V = 18V

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Slide 66 (Answer) 99

24 What is the voltage at any spot in the circuit

R1 = 3

R2 = 6

V = 18V

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18 V

Slide 67 99

25 What is the current through R1

R1 = 3

R2 = 6

V = 18V

httpnjctlo8

Slide 67 (Answer) 99

25 What is the current through R1

R1 = 3

R2 = 6

V = 18V

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Slide 68 99

26 What is the current through R2

R1 = 3

R2 = 6

V = 18V

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Slide 68 (Answer) 99

26 What is the current through R2

R1 = 3

R2 = 6

V = 18V

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Slide 69 99

27 What is the power used by R1

R1 = 3

R2 = 6

V = 18V

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Slide 69 (Answer) 99

27 What is the power used by R1

R1 = 3

R2 = 6

V = 18V

httpnjctloa

[This object is a pull tab]

Ans

wer

Power used by R1

Slide 70 99

28 What is the power used by R2

R1 = 3

R2 = 6

V = 18V

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Slide 70 (Answer) 99

28 What is the power used by R2

R1 = 3

R2 = 6

V = 18V

httpnjctlob

[This object is a pull tab]

Ans

wer

Power used by R2

Slide 71 99

29 What is the total current in this circuit

R1 = 3

R2 = 6

V = 18V

R3 = 4

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Slide 71 (Answer) 99

29 What is the total current in this circuit

R1 = 3

R2 = 6

V = 18V

R3 = 4

httpnjctlob

[This object is a pull tab]

Ans

wer

Slide 72 99

30 What is the voltage drop across the third resistor

R1 = 3

R2 = 6

V = 18V

R3 = 4

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Slide 72 (Answer) 99

30 What is the voltage drop across the third resistor

R1 = 3

R2 = 6

V = 18V

R3 = 4

httpnjctlob

[This object is a pull tab]

Ans

wer

Slide 73 99

31 What is the current though the first resistor

R1 = 3

R2 = 6

V = 18V

R3 = 4

httpnjctlob

Slide 73 (Answer) 99

31 What is the current though the first resistor

R1 = 3

R2 = 6

V = 18V

R3 = 4

httpnjctlob

[This object is a pull tab]

Ans

wer

Slide 74 99

Return toTable ofContents

Measurement

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Slide 75 99

Voltmeter

Voltage is measured with a voltmeter Voltmeters are connected in parallel and measure the difference in potential between two points

Since circuits in parallel have the same voltage and a voltmeter has very high resistance very little current passes through it

This means that it has little effect on the circuit

httpnjctloh

Slide 76 99

Ammeter

Current is measured using an ammeter

Ammeters are placed in series with a circuit In order to not interfere with the current the ammeter has a very low resistance

httpnjctloh

Slide 77 99

Multimeter

Although there are separate items to measure current and voltage there are devices that can measure both (one at a time)

These devices are called multimetersMultimeters can also measure resistance

Click here for a PhETsimulation on circuits

httpnjctloh

Slide 78 99

L

32 A group of students prepare an experiment with electric circuits Which of the following diagrams can be used to measure both current and voltage

A B

C D

E

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Slide 78 (Answer) 99

L

32 A group of students prepare an experiment with electric circuits Which of the following diagrams can be used to measure both current and voltage

A B

C D

E

httpnjctloi

[This object is a pull tab]

Ans

wer

E

Slide 79 99

Return toTable ofContents

EMF amp Terminal Voltage

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Slide 80 99

Electromotive Force

Req

Er

_+

A battery is a source of voltage AND a resistor

Each battery has a source of electromotive force and internal resistance

Electromotive force (EMF) is the process that carries charge from low to high voltage

Another way to think about it is that EMF is the voltage you measure when no resistance is connected to the circuit

httpnjctloj

Slide 81 99

Req

Er

_+

Terminal voltage (VT) is the voltage measured when a voltmeter is across its terminals

If there is no circuit attached no current flows and the measurement will equal the EMF

Electromotive Force

If however a circuit is attached the internal resistance will result in a voltage drop and a smaller terminal voltage (E - Ir)

httpnjctloj

Slide 82 99

Req

Er

_+

We say that the terminal voltage is

VT = E - Ir

Maximum current will occur when there is zero external current

When solving for equivalent resistance in a circuit the internal resistance of the battery is considered a series resistor

REQ = Rint + Rext

Terminal Voltage

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Slide 83 99

33 When the switch in the circuit below is open the voltmeter reading is referred to as

A EMFB CurrentC PowerD Terminal VoltageE Restivity

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Slide 83 (Answer) 99

33 When the switch in the circuit below is open the voltmeter reading is referred to as

A EMFB CurrentC PowerD Terminal VoltageE Restivity

httpnjctlok

[This object is a pull tab]

Ans

wer

A

Slide 84 99

34 When the switch in the circuit below is closed the voltmeter reading is referred to as

A Terminal VoltageB EMFC CurrentD ResistanceE Power

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Slide 84 (Answer) 99

34 When the switch in the circuit below is closed the voltmeter reading is referred to as

A Terminal VoltageB EMFC CurrentD ResistanceE Power

httpnjctlol

[This object is a pull tab]

Ans

wer

A

Slide 85 99

35 A 6V battery whose internal resistance 15 Ω is connected in series to a light bulb with a resistance of 68 Ω What is the current in the circuit

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Slide 85 (Answer) 99

Slide 86 99

36 A 6V battery whose internal resistance 15Ω is connected in series to a light bulb with a resistance of 68Ω What is the terminal voltage of the battery

httpnjctlon

Slide 86 (Answer) 99

Slide 87 99

37 A 25 Ω resistor is connected across the terminals of a battery whose internal resistance is 06 Ω What is the EMF of the battery if the current in the circuit is 075 A

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Slide 87 (Answer) 99

Slide 88 99

Return toTable ofContents

Kirchhoffs Rules

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Slide 89 99

Up until this point we have been analyzing simple circuits by combining resistors in series and parallel and using Ohms law

This works for simple circuits but in order analyze more complex circuits we need to use Kirchhoffs Rules which are based on the laws of conservation of charge and energy

Kirchhoffs Rules

Slide 90 99

Kirchhoffs First rule or junction rule is based on the law of conservation of charge It states

At any junction point the sum of all currents entering the junction point must equal the sum of all the currents exiting the junction

For example

I1 + I2 = I3

Kirchhoffs Rules

I1

I2I3

Slide 91 99

Kirchhoffs Second rule or loop rule is based on the law of conservation of energy It states

The sum of all changes in potential around any closed path must equal zero

For example

The sum of the voltage drops is equal to the voltage across the battery

V - V1 - V2 =0 OR V = V1 + V2

Kirchhoffs Rules

V

V1 V2

Slide 92 99

1 Label + and - for each battery

2 Label the current in each branch with a symbol and an arrow (Dont worry about the direction of the arrow It its incorrect the solution will be negative)

3 Apply the junction rule to each junction You need as many equations as there are unknowns (You can also use Ohms Law to reduce the number of unknowns)

4 Apply the loop rule for each loop (Pay attention to signs For a resistor the potential difference is negative For a battery the potential difference is positive)

5 Solve the equations algebraically

Problem Solving with Kirchhoffs Rules

Slide 93 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2V3 = 187 V

I2 = 26 A

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 90 99

Kirchhoffs First rule or junction rule is based on the law of conservation of charge It states

At any junction point the sum of all currents entering the junction point must equal the sum of all the currents exiting the junction

For example

I1 + I2 = I3

Kirchhoffs Rules

I1

I2I3

Slide 91 99

Kirchhoffs Second rule or loop rule is based on the law of conservation of energy It states

The sum of all changes in potential around any closed path must equal zero

For example

The sum of the voltage drops is equal to the voltage across the battery

V - V1 - V2 =0 OR V = V1 + V2

Kirchhoffs Rules

V

V1 V2

Slide 92 99

1 Label + and - for each battery

2 Label the current in each branch with a symbol and an arrow (Dont worry about the direction of the arrow It its incorrect the solution will be negative)

3 Apply the junction rule to each junction You need as many equations as there are unknowns (You can also use Ohms Law to reduce the number of unknowns)

4 Apply the loop rule for each loop (Pay attention to signs For a resistor the potential difference is negative For a battery the potential difference is positive)

5 Solve the equations algebraically

Problem Solving with Kirchhoffs Rules

Slide 93 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2V3 = 187 V

I2 = 26 A

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 91 99

Kirchhoffs Second rule or loop rule is based on the law of conservation of energy It states

The sum of all changes in potential around any closed path must equal zero

For example

The sum of the voltage drops is equal to the voltage across the battery

V - V1 - V2 =0 OR V = V1 + V2

Kirchhoffs Rules

V

V1 V2

Slide 92 99

1 Label + and - for each battery

2 Label the current in each branch with a symbol and an arrow (Dont worry about the direction of the arrow It its incorrect the solution will be negative)

3 Apply the junction rule to each junction You need as many equations as there are unknowns (You can also use Ohms Law to reduce the number of unknowns)

4 Apply the loop rule for each loop (Pay attention to signs For a resistor the potential difference is negative For a battery the potential difference is positive)

5 Solve the equations algebraically

Problem Solving with Kirchhoffs Rules

Slide 93 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2V3 = 187 V

I2 = 26 A

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 92 99

1 Label + and - for each battery

2 Label the current in each branch with a symbol and an arrow (Dont worry about the direction of the arrow It its incorrect the solution will be negative)

3 Apply the junction rule to each junction You need as many equations as there are unknowns (You can also use Ohms Law to reduce the number of unknowns)

4 Apply the loop rule for each loop (Pay attention to signs For a resistor the potential difference is negative For a battery the potential difference is positive)

5 Solve the equations algebraically

Problem Solving with Kirchhoffs Rules

Slide 93 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2V3 = 187 V

I2 = 26 A

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 93 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2V3 = 187 V

I2 = 26 A

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 94 99

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

First label + and - for each battery

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 95 99

Next label the current in each branch with a symbol and an arrow

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 96 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 96 (Answer) 99

Next apply the junction rule to each junction

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I

I1

[This object is a pull tab]

Ans

wer I = I3 + I4

= I2

I1 = I3

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 97 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 97 (Answer) 99

Next apply the loop rule to each loop

Problem Solving with Kirchhoffs Rules

R1 = 5 R2 = 3

V = 12 V

R3 = R4 = 2

I2 = 26 A

V3 = 187 V

+ _

I3 I4

I1

I

I = I3 + I4

I = I2

I1 = I3

[This object is a pull tab]

Ans

wer

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 98 99

Problem Solving with Kirchhoffs Rules

V = V3 + V1 + V2

V = V4 + V2

V1 + V3 = V4

I = I3 + I4

I = I2

I1 = I3

Current (Amps) Voltage (Volts) Resistance (Ohms)

R1 5

R2 26 3

R3 187

R4 2

Total 12

List the givens and use ohms law to solve for the unknowns

Ans

wer

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V

Slide 99 99

Find the unknowns in the following circuit

Problem Solving with Kirchhoffs Rules

V = 120 V

R1 = 10

I1 = 7 A

V1 = 70 V

R3 = 2

I3 = 10 A

V3 = 20 V

R4 = 3

I4 = 10 A

V4 = 30 V

R2 = 23

I2 = 3 A

V2 = 70 V