Norah Ali Al- Moneef1 Sources of emf The source that maintains the current in a closed circuit is called a source of emf – Any devices that increase

Norah Ali Al- Moneef 1

Sources of emfSources of emf• The source that maintains the current in a closed

circuit is called a source of emf– Any devices that increase the potential energy of charges

circulating in circuits are sources of emf– Examples include batteries and generators

• SI units are Volts– The emf is the work done per unit charge


28.1 Electromotive Force28.1 Electromotive Force

All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery


Direct CurrentDirect Current• When the current in a circuit has a constant

direction, the current is called direct current– Most of the circuits analyzed will be assumed to

be in steady state, with constant magnitude and direction

• Because the potential difference between the terminals of a battery is constant, the battery produces direct current

• The battery is known as a source of emf

emf and Internal Resistanceemf and Internal Resistance

• A real battery has some internal resistance• Therefore, the terminal voltage is not equal to the

emf


•All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery

r+ - IrVV

• The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit

• The terminal voltage is ΔV = Vb-Va ΔV = ε – Ir

• For the entire circuit, ε = IR + Ir


• The schematic shows the internal resistance, r• The internal resistance of the source of emf is

always considered to be in a series with the external resistance present in the electric circuit

• The terminal voltage is ΔV = Vb-Va ΔV = ε – Ir

• For the entire circuit, ε = IR + Ir

the relationship between Emf, resistance, current, and terminal voltage

Circuit model looks like this:I

r

R Terminal voltage = V

V = IR= I r + I R = I (r + R)

I = /(r + R) - Ir = V = IR

The terminal voltage decrease = - Ir as the internal resistance r increases or when I increases.

If r =0 ( no internal resistance ) ( ideal battery ) = V = I R

I = V / R I = / R Norah Ali Al- Moneef 6

– Power dissipated by a resistor, is amount of energy per time that goes into heat, light, etc.

R I r I 22 IPower P = I P = I ΔΔ V V

the total power output IIεε of the batteryis delivered to the external load resistance in the amount I I 22R R and to the internal resistance in the amount I I 22rr.

rIRII 22 Total power

of emfPower dissipated as joule heat in:

Internal resistorLoad resistor


Potential gain in the batteryPotential drop at all resistancesIn an old, “used-up” battery emf is nearly the same, but internal resistance increases enormously

ε = ΔV + Ir = IR + Ir

• ε is equal to the terminal voltage when the current is zero – open-circuit voltage

I = ε / (R + r)

• R is called the load resistance• The current depends on both the resistance

external to the battery and the internal resistance

• When R >> r, r can be ignored• Power relationship: I ε = I2 R + I2 r

• When R >> r, most of the power delivered by the battery is transferred to the load resistor



• At V=10V someone measures a current of 1A through the below circuit. When she raises the voltage to 25V, the current becomes 2 A. Is the resistor Ohmic?

a) YES b) NO

What are voltmeter and ammeter readings?

Example

V 8 ) 2 2 ( -12 V

Ir - V

solutionanother

8V 4 2 V

IR V

readingVoltmeter

2 6

12

2 4

12

r R

readingAmmeter

AI


• the voltage across the internal resistor (lost voltage): v = Ir = 1.14 amps × 0.5 ohms = 0.57 volts

AI 14.1 .5 01

12

0.5 01

12

r R

From the figure find the current and the voltage across the internal resistor

example

out the terminal voltage: Terminal voltage = emf - lost voltage out the terminal voltage: Terminal voltage = emf - lost voltage = 12 - 0.57 = = 12 - 0.57 = 11.43 volts11.43 volts


ExampleExampleA battery has an emf of 12V and an internal resistance of 0.05Ω. Its terminals

are connected to a load resistance of 3Ω. Find:

a) The current in the circuit and the terminal voltage

b) The power dissipated in the load, the internal resistance, and the total power delivered by the battery

47.16W 12 3.93 I p

battery. by the deliveredpower

W0.772 0.05 3.93

.resistance linternal in the dissipatedpower

W46.3347 3 3.93

load in the dissipatedpower ) b

2r

2R

2

2

P

P

rI

RI

V 8.11 ) 0.05 3.93 ( -12 V

Ir - V

solutionanother

V 11.8 3 3.93 V

IR V

voltageterminal

93.3 .05 3

12

0.05 3

12

r R

AI


ExampleA car has a 12-V battery with internal resistance 0.020 . When

the starter motor is cranking, it draws 125 A.What’s the voltage

across the battery terminals while starting?

Voltage across battery terminals =

Battery terminals

V 9.5 ) 0.02 ( )A (125 -V 12 V

076.0 0.025-A 125

12 r-

I R

R I Ir

V

V 9.5 ) 0.02 ( )A (125 -V 12 V


ε=12 V R

I = 2 A

What does the energy balance look like in this circuit?

When current leaves the battery the battery supplies power equal to:

If current were forced to enter the battery, (as in charging it) then it absorbs the same power

P = Iε = 24W

2126 so Joule heating=I 24

2V

R R WA

Resistor: Electrical energy heat

example


Resistors in SeriesResistors in Series• When two or more resistors are connected end-to-

end, they are said to be in series• The current is the same in all resistors because any

charge that flows through one resistor flows through the other

• The sum of the potential differences across the resistors is equal to the total potential difference across the combination


28.2 Resistors in Series and Parallel28.2 Resistors in Series and Parallel


• Potentials add– ΔV = IR1 + IR2 = I (R1+R2)– Consequence of Conservation of Energy

• The equivalent resistance has the effect on the circuit as the original combination of resistors

Equivalent Resistance – Series• Req = R1 + R2 + R3 + …• The equivalent resistance of a series combination

of resistors is the algebraic sum of the individual resistances and is always greater than any of the always greater than any of the individual resistorsindividual resistors


•Current has one pathway - same in every part of the circuit•Total resistance is sum of individual resistances along path•Current in circuit equal to voltage supplied divided by total resistance•Sum of voltages across each lamp equal to total voltage•One bulb burns out - circuit broken - other lamps will not light

Resistors in Parallel

• The potential difference across each resistor is the same because each is connected directly across the battery terminals

• A junction is a point where the current can split• The current, I, that enters a point must be equal to the total

current leaving that point– I = I 1 + I 2

– The currents are generally not the sameThe currents are generally not the same– Consequence of Conservation of ChargeConsequence of Conservation of Charge


Equivalent Resistance – Parallel

• Equivalent Resistance

• The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance– The equivalent is always less than the The equivalent is always less than the

smallest resistor in the groupsmallest resistor in the group


1 2 3

1 1 1 1

eqR R R R

Equivalent Resistance – Parallel, Example

• Equivalent resistance replaces the two original resistances• Household circuits are wired so the electrical devices are

connected in parallel– Circuit breakers may be used in series with other circuit elements

for safety purposesNorah Ali Al- Moneef 20


Resistors in Parallel, Final

• In parallel, each device operates independently of the others so that if one is switched off, the others remain on

• In parallel, all of the devices operate on the same voltage• The current takes all the paths

– The lower resistance will have higher currents– Even very high resistances will have some currents

• Household circuits are wired so that electrical devices are connected in parallel

Equivalent Resistance – Series: An Example

• Four resistors are replaced with their equivalent resistance


Problem-Solving Strategy, 1

• Combine all resistors in series– They carry the same current– The potential differences across them are not the

same– The resistors add directly to give the equivalent

resistance of the series combination: Req = R1 + R2 + …



• Combine all resistors in parallel– The potential differences across them are the

same– The currents through them are not the same– The equivalent resistance of a parallel

combination is found through reciprocal addition:


321eq R

1

R

1

R

1

R

1


• A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor– Replace any resistors in series or in parallel using steps 1

or 2. – Sketch the new circuit after these changes have been

made– Continue to replace any series or parallel combinations – Continue until one equivalent resistance is found



• If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit found in step 3 and gradually work back through the circuits– Use ΔV = I R and the procedures in steps 1 and 2

– The greater the resistance in a circuit, the lower the current.


With the switch in this circuit (figure a) open, there is no current in R2. There is current in R1 and this current is measured with the ammeter at the right side of the circuit. If the switch is closed (figure b), there is current in R2. When the switch is closed, the reading on the ammeter (a) increases,

(b) decreases, or (c) remains the same.


(a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As

a result, the total current increases.

Example

You have a large supply of lightbulbs and a battery. You start with one light bulb connected to the battery and notice its

brightness. You then add one light bulb at a time, each new bulb being added in parallel to the previous bulbs. As the light bulbs are added, what happens (a) to the brightness of the bulbs? (b)

to the current in the bulbs? (c) to the power delivered by the battery? (d) to the lifetime of the battery? (e) to the terminal

voltage of the battery? Hint: Do not ignore the internal resistance of the battery.


(a) The brightness of the bulbs decreases (b) The current in the bulbs decreases (c) The power delivered by the battery increases (d) The lifetime of the battery decreases (e) The terminal voltage of the battery decreases

Example

• what is the equivalent resistance of all resistors as placed in the below circuit? If V=12V, what is the current I?


I

R1 R2

V

R3R1=3 OhmR2=3 OhmR3=3 OhmV=12V

R2 & R3 are in parallel1/R23=1/R2+1/R3=1/3+1/3=2/3R23=3/2 OhmR1 is in series with R23

R123=R1+R23=3+3/2=9/2 OhmI=V/R=12/(9/2)=24/9=8/3 A

Example

• A tree is decorated with a string of many equal lights placed in parallel. If one burns out (no current flow through it), what happens to the others? a) They all stop shiningb) the others get a bit dimmerc) the others get a bit brighterd) the brightness of the others remains the same


R

R

VI

Before the one light fails:1/Req=1/R1+1/R2+…+1/Rn

if there are 3 lights of 1 Ohm: Req=1/3I=V/Req Ij=V/Rj (if 3 lights: I=3V Ij=V/1After one fails:1/Req=1/R1+1/R2+….+1/Rn-1

if there are 2 lights left: Req=1/2I=V/Req Ij=V/Rj (if 2 lights: I=2V Ij=V/1) The total resistance increases, so the current drops. The two effects cancel each other

Example

• A tree is decorated with a string of many equal lights placed in parallel. If one burns out (no current flow through it), what happens to the others? – a) They all stop shining– b) the others get a bit dimmer– c) the others get a bit brighter– d) the brightness of the others remains the same


R

R

VI

Before the one light fails:1/Req=1/R1+1/R2+…+1/Rn

if there are 3 lights of 1 Ohm: Req=1/3I=V/Req Ij=V/Rj (if 3 lights: I=3V Ij=V/1After one fails:1/Req=1/R1+1/R2+….+1/Rn-1

if there are 2 lights left: Req=1/2I=V/Req Ij=V/Rj (if 2 lights: I=2V Ij=V/1) The total resistance increases, so the current drops. The two effects cancel each other

Example


• a person designs a new string of lights which are placed in series. One fails, what happens to the others?– a) They all stop shining– b) the others get a bit dimmer– c) the others get a bit brighter– d) the brightness of the others remains the same

Assume: If one fails, the wire inside it is broken and Current cannot flow through it any more.

Example

ExampleFind the current through the 2- resistor in the circuit.

Equivalent of parallel 2.0- & 4.0- resistors:

1 1 1

2.0 4.0R

1.33R

Total current is

Equivalent of series 1.0-, 1.33- & 3.0- resistors: 5.33

3

4.0

1.0 1.33 3.0TR

5.33T

IR E 12

5.33

V

2.25 A

Voltage across of parallel 2.0- & 4.0- resistors: 1.33 2.25 1.33V A 2.99V

Current through the 2- resistor: 2

2.99

2.0

VI

1.5A


• Equivalent Resistance


321eq R

1

R

1

R

1

R

1

11

18

18

11

18

236

9

1

6

1

3

11

eq

eq

R

R

Three resistors are connected in parallel as shown in the Figure a potential difference of 18.0 V is maintained between points a and b. (A) Find the current in each resistor ,total current ( I )

11 2 3 6 I

I I I I

A 11

1118

V 18

R

V

A 2 9

V 18

R

V

A 3 6

V 18

R

V

A 6 3

V 18

R

V

321

eq

33

22

11

II

I

I

total

I

Example


(B) Calculate the power delivered to each resistor and thetotal power delivered to the combination of resistors.

We apply the relationship P = P = I I 22R R to each resistorand obtain

This shows that the smallest resistor receives the mostpower. Summing the three quantities gives a total power of198 W.

Equivalent Resistance – Complex Circuit


Example

inthe circuit (a) below to get the equivalent resistance. b ) calculate the power P = I2 R dissipated in each resistor.


Example

Example : Four equivalent light bulbs R1 = R2 = R3 = R4 = 4.50 , emf = 9.00 Volts. Find current and power in each light bulb. Which bulb is brightest?

Later, if bulb 4 is removed which bulbs get brighter? Dimmer?Still Bulb 1 is the brightest


Itotal=I1=1.5A,, I2=I3=I4+0.5A., P1=10.1W, P2=P3=P4=1.125W., V2=V3=V4=2.25VP1> P2 , P3 , P4 Bulb 1 is the brightest

• The brightness (intensity) of the bulb is basically the power dissipated in the resistor.– For example A 200 W bulb is brighter than a 75 W bulb, all

other things equal

What is the current through a?1 /Req = (1 /5) + (1 /10) (1 /30) Req =30 / (6+3+1) = 3 What is the current through e?What is the current each branch b-d?

5

10

3030v a

e

Itot = 10A

b

c

d

Same voltage is across each pathb: V= IR 30 = Ix5 I= 6Ac: 30= Ix10 I= 3Ad: 30= Ix30 I= 1A


Example

If all N branches have the same resistance, total resistance is equal to the resistance of one branch divided by the number of branches

Total resistance= 1 /Req = (1 /30) + (1 /30) (1 /30) Req = 30 /3 = 10 Total current= V / Req I total = 30 /10 = 3A Current in b= 30 /3 0 = 1 A


Example

If there are only two branches, the total resistance is equal to the product of the resistances divided by the sum of the resistances

Total resistance= Req = 12 X 4 /( 12 + 4 ) = 3


Example

What is the:o Total resistance?o Total current flow?o Current flow through bo Current flow through co Current flow through do Voltage between b and do Voltage between c and do Voltage between d and e

Example


Total resistance:o In compound circuits, reduce all parallel parts to a single

resistance until you have a simpler series circuit

o The resistance between a and b is 2 o Therefore, total resistance is 4 (2 + 2)

Example


Total current:V = Itotal Req

20v = I X 4 Itot = 5 A


R1eq = 3 X 6 / ( 3 + 6 ) = 2

the total resistance of the series portion of the circuit R eq = 2 + 2 = 4

Find the Total current.o the total resistance of the parallel portion of the circuit

Example

Current flow through bo We need to know the voltage drop across b-do Voltage drop across e-d will be (V= 5A X 2 = 10v)o Therefore, voltage drop across each parallel branch

(c and b) must be ( Vda =20 V -10V = 10 V)o Current flow in b: 10 = I X 3 ; I = 3.33Ao Current flow in c: 10 = I X 6 ; I= 1.67A

Example



2

6

12

3.204.

Example

In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it?The resistor with the largest resistance (30 )Which resistor has the greatest current flow through it?Same for all because series circuitIf we re-ordered the resistors, what if any of this would change?Nothing would change


Example

If we added a resistor in series with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor?Total resistance would increaseTotal current would decreaseVoltage across each resistor would decrease (All voltage drops must still sum to total in series circuit; Kirchhoff’s law of voltages)Current through each resistor would be lower (b/c total current decreased, but same through each one)


Example

In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it?All the same in parallel branchesWhich resistor has the greatest current flow through it?The “path of least resistance” (10)What else can you tell me about the current through each branchThey will sum to the total I (currents sum in parallel circuits; Kirchhoff’s law of current)


Example

If we added a resistor in parallel with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor?Total resistance would decreaseTotal current would increaseVoltage across each resistor would still be VCurrent through each resistor would be higher and would sum to new total I

Example



R1=100 ohmsR2=100 ohmsR3=100 ohmsReq= ? Ohms

1 / Req= 1 / R1 + 1 / R2 + 1 / R3 1 / Req = 1/100 + 1/100 + 1/100

1 / Req =.01 + 01 + .01 = .03

Req = 1/.03 = 33.33 ohms

Example


A parallel circuit is shown in the diagram above. In this case the current supplied by the battery splits up, and the amount going through each resistor depends on the resistance. If the values of the three resistors are: R 1 = 8 Ω R2= 8 Ω R 3 = 4 Ω 1 / R eq = 1 / 8 + 1 / 8 + 1 / 4 = 4 / 8 R eq = 2 Ω

With a 10 V battery, by V = I R the total current in the circuit is: I = V / R = 10 / 2 = 5 A. The individual currents can also be found using I = V / R. The voltage across each resistor is 10 V, so: I1 = 10 / 8 = 1.25 A I2 = 10 / 8 = 1.25 A I3=10 / 4 = 2.5 A Note that the currents add together to 5A, the total current.

Example

• What would V1 read in the illustration?• Ohm’s Law states: • Therefore:


• At this point there is insufficient data because I (amp) is unknown.

• Using Ohm’s Law to solve for the current in the circuit:

• Knowing the amount of current we can calculate the voltage drop.

A 9.09 13.2

V 120=

R

V=I IR=V

V 45.4= 5.0A x .099=IR=V

I

V=R

IR=V

Example

• Determine the readings for A1 and A2 in the illustration. • In a series circuit the electricity has no alternative paths,

therefore the amperage is the same at every point in the circuit.


• The current in the circuit is determined by dividing the voltage by the circuit resistance.

A 1.0= 6.3+4.2+1.5

V 12=

R

V=I IR=V

Example

Determine the readings for amp

meters A1 and A2 in the circuit.


• In a parallel circuit the amperage varies with the resistance.• In the illustration, A1 will measure the total circuit amperage,

but A2 will only measure the amperage flowing through the 6.3 Ohm resistor.

• To determine the total resistance of the circuit must be

Rt =R1 x R2

R1 +R2

=1.5 x 4.2

1.5 +4.2=

6.3

5.7=1.10

1.10 x 6.3

1.10 +6.3=

6.96

7.40=0.94

Example

When the total resistance is known, the circuit current (Amp’s) can be calculated.


Branch current is:A2 = 1.9 A

A1= 12.76 A

Total current is:

When the circuit current (Amp’s) is known, the current for each branch circuit can be calculated.

A 12.76= 0.94

V 12=

R

V=I

A 1.9= 6.3

V 12=

R

V=I

Example

• To calculate the circuit current the total circuit resistance must be known.


R1 x R2

R1 R2

=1.2 V x 5.8 V

1.2 V +5.8 V=

6.96

7.0= 0.99 2.3 = 3.29

Determine the readings for the two volt meters in the illustration.

• Volt meter one (V1) will read source voltage:• Volt meter two (V2) will read the voltage in the circuit after the 2.3

resistor. • To determine this reading, the voltage drop across the resistor must be

calculated. • Before the voltage drop across the resistor can be calculated, the circuit

current must be determined.

12 V

Example

Circuit current is:


Volt meter 1 will read source voltage: 12 VVolt meter 2 will read 3.6 V.

The voltage drop caused by the 2.3 Ohm resistance is:

12 V - 8.39 V = 3.61 V The voltage remaining in the circuit

is:

A 3.65= 3.29

V 12=

R

V=I IR=V

V 8.39= 2.3 x 3.65=IR=V

Example

• Determine the readings for the amp meters in the illustration.• The first amp meter will read circuit amps and the second one will

measure the current in that branch of the circuit.• To determine circuit amperage, the total circuit resistance must be

known. This was calculated in the previous slide as 3.29 .


• Total circuit current was also calculated as 3.65 A.

• A1 = 3.65 A• To determine the reading for A2, the current flowing in

that part of the circuit must be calculated.

I =V

R=

3.61 V

5.8 = 0.62 A

A2 = 0.62 A

• In the previous slide the voltage left after the 2.3 resistor was 3.61 V.

Example


When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R1 (a) increases, (b) decreases, or (c) stays the

same. The brightness of bulb R2 (a) increases, (b) decreases, or (c) stays the same.

R1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because

essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R2.

Example


With the switch in this circuit (figure a) closed, no current exists in R2 because the current has an alternate zero-resistance path through the switch. Current does exist in R1 and this current is measured with the

ammeter at the right side of the circuit. If the switch is opened (figure b), current exists in R2. After the switch is opened, the reading on the ammeter

(a) increases, (b) decreases, (c) does not change.(b). When the switch is

opened, resistors R1 and R2 are in series, so that the total

circuit resistance is larger than when the switch was closed.

As a result, the current decreases.

Example


What is the value of R?

• IR = V / R =. 8 / 12 = .667A

1 / R 1 / R eqeq = 1 / 10 + 1 / 15 + 1 / R =1 50+25 R / 150 R = 1 / 10 + 1 / 15 + 1 / R =1 50+25 R / 150 R R R eqeq = 150 R / 150+ 25R = 150 R / 150+ 25R R R eqeq = 8 / 2 = 4 = 8 / 2 = 4 ΩΩ

44= 150 R /1 50+25 R= 150 R /1 50+25 R 4 (150+ 25R ) = 150 R4 (150+ 25R ) = 150 R100 R + 600 = 150R R = 12 100 R + 600 = 150R R = 12 ΩΩ

Example

• What is the Value of R?


R = Δ V / I = 5 .0 V / ( 100 X 10 -3 )AR = 25Ω

Example

What is the current through the circuit? What are the values of potential at points a-f?


• I: 1A• a:0V • b:12V • c:9V• d: 6V • e: 0V • f: 0V

Example

28.3 Kirchhoff’s Rules• There are ways in which resistors can be connected so

that the circuits formed cannot be reduced to a single equivalent resistor

• Two rules, called Kirchhoff’s Rules can be used instead


Statement of Kirchhoff’s Rules• Junction Rule

– The sum of the currents entering any junction must equal the sum of the currents leaving that junction

• A statement of Conservation of Charge

• Loop Rule– The sum of the potential differences across all the elements

around any closed circuit loop must be zero• A statement of Conservation of Energy

the Junction Rule

• I1 = I2 + I3

• From Conservation of Charge• Diagram b shows a mechanical analog


I1 I3

I2

I4

I5

I1+I2+I3=I4+I5

Setting Up Kirchhoff’s Rules


• Assign symbols and directions to the currents in all branches of the circuit– If a direction is chosen incorrectly, the resulting answer will

be negative, but the magnitude will be correct

• When applying the loop rule, choose a direction for transversing the loop– Record voltage drops and rises as they occur

Electrical Work and PowerElectrical Work and Power

I I+ -

Higher V1 Lower V2

Resistance R

Current I flows through a potential difference V

Follow a charge Q : at positive end, U1 = QV1

at negative end, U2 = QV2

P.E. Decreases: 0U Q V The speed of the charges is constant in the wires and resistor.

What is electrical potential energy converted to?


Electrical resistance converts electrical potential energy to thermal energy (heat), just as friction in mechanical systems converts mechanical energy to heat.

This thermal energy means the atoms in theconductor move faster and so the conductor getshotter.

The average kinetic energy of the electrons doesn’t increase once the current reaches a steady state; the electrons lose energy in collisions with the atoms as fast as it is supplied by the field.


More About the Loop Rule• Traveling around the loop from a to b• In aIn a, the resistor is transversed in the

direction of the current, the potential across the resistor is –IR

• In bIn b, the resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR


• In cIn c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε

• In dIn d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε

Junction Equations from Kirchhoff’s Rules

• Use the junction rule as often as needed, so long as, each time you write an equation, you include in it a current that has not been used in a previous junction rule equation– In general, the number of times the junction rule can be

used is one fewer than the number of junction points in the circuit


Loop Equations from Kirchhoff’s Rules

• The loop rule can be used as often as needed so long as a new circuit element (resistor or battery) or a new current appears in each new equation

• You need as many independent equations as you have unknowns


Problem-Solving Strategy – Kirchhoff’s Rules

• Draw the circuit diagram and assign labels and symbols to all known and unknown quantities

• Assign directions to the currents.• Apply the junction rule to any junction in the circuit• Apply the loop rule to as many loops as are needed

to solve for the unknowns• Solve the equations simultaneously for the unknown

quantities• Check your answers


The emfs and resistances in the circuit have the following values

Starting at point a lets apply the loop rule counter-clockwise and tally the voltages along the way

A

V

24.0 9.6

V 3.3 I

0 I ) (9.6 V 3.3-

0 V 2.1I ) 8.1( I ) 5.5( I ) 3.2( 4.4


Example

What is the potential difference between the terminals What is the potential difference between the terminals of battery 1 ??of battery 1 ??

This is the equivalent of asking what the potential difference is between points b and a.So lets travel from point b to point a clockwise

Vb – i r1 + ε1 = Va

Va - Vb = - i r1 + ε 1

= -(0.24A) (2.3 ) + 4.4V = 3.8V


Example. Multiloop CircuitFind the current in R3 in the figure below.

1 2 3 0I I I Loop 1:

Loop 2: 1 36 2 0I I

2 39 4 0I I 0 R I - R I 33222

0 R I - R I - 33111

Norah Ali Al- Moneef 76AI

I

I

II

II

II

II

III

III

RII

37

21

217

0721

049

06412

049

2)0326(

0226

01)(26

0)(26

3

3

3

32

32

32

32

332

332

331

Label the 3 branch currents I2, I3, and I4.Since VAB across all 3 branches is the same and

is known: V4 = I4R4 = 5A X (4Ω) = 20 Volts,

the currents and εε can be readily solved.


V4 = I4R4 = 5A (4) = 20 Volts I3 = V3 / R3 = 4V / 3= 1.33 A

At junction B, I in = I outI4 = I2 + I3 ; I2 = I4 – I3 = 5A - 1.3A

I2 = 3.7ALoop #1: Vi =0

- + I2R2 + I4R4 = 3.7A (2) + 5A(4) = 27.4 V

Example


Find r and ε .

I in = I out

I = 1 A + 2A = 3 A

Loop 1: - ε +1 A X 1Ω -2A X 3 Ω = 0

ε = -5 VLoop 3 : 12 – 3 A Xr – 2 A X 3 Ω = 0r = 6 V / 3 A = 2 Ω

Example


1) I1= 12 / ( 9 /2 ) = 8/3 AKirchhof 22) V-I1R1-I2R2=0 12-3 I1-3 I2=0Kirchhof 23) V-I1R1-I3R3=0 12-3I1-3I3=0Kirchhof 24) 0-I3R3+I2R2=0 -3I3+3I2=0Kirchhoff I5) I1-I2-I3=0 I1=I2+I3 1) & 2) 12-8-3I2=0 so 4=3I2 and I2=4/3 A1) & 3) 12-8-3I3=0 so 4=3I3 and I3=4/3 AUse V=IR for R1 V1=8/3X3= 8 V for R2 V2=4/3X3= 4 V for R3 V3=4/3X3= 4 V

Given V=12V, what is the current through and voltage over each resistor R eq = 3 +( 3X 3 / 3 + 3 ) = 3 + ( 9 / 6 ) = 4.5 Ω

I=I1

R1 R2

V

R3

R1=3 ΩR2=3 ΩR3=3 ΩV=12V

I3I1

I2

I=I1

Example


• which of the following cannot be correct?

• a) V-I1R1-I3R3-I2R2=0

• b) I1-I3-I4=0

• c) I3R3-I6R6-I4R4=0

• d) I1R1-I3R3-I6R6-I4R4=0

• e) I3+I6+I2=0

NOT a LOOP

I4 R4I3 R3

I2 R2

I1,R1

I5 R5

V

I6 R6

Example


• consider the circuit. Which of the following is/are not true?

1. If R2=R3=2R1 the potential drops over R1 and R2 are the same

2. for any value of R1,R2 and R3 the potential drop over R1 must be equal to the potential drop over R2

3. The current through R1 is equal to the current through R2 plus the current through R3 (I1=I2+I3)

I

R1 R2

V

R3

1) if R2=R3=2R1 then 1/R23=1/R2+1/R3=1/R1 so R23=R1 and I1=I23 and potential of R1 equals the potential over R23 and thus R2 and R3. THIS IS TRUETHIS IS TRUE

2)2) nono, this is only TRUE in the case of 1)3)3) true: true: conservation of current.

Example


Example

12V 12V 12V

A B C D

At which point (A,B,C,D) is the potential highest and at which point lowest? All resistors are equal.

a) highest B, lowest Ab) highest C, lowest Dc) highest B, lowest Dd) highest C, lowest Ae) highest A, lowest B

circuit 1: I=12/(2R)=6/R VA=12-6/R*R=6Vcircuit 2: VB=12Vcircuit 3: I=12/(3R)=4/R VC=12-4/R*R=8V VD=12-4/R*R-4/R*R=4V

1

23


R1 R3

R2

V1 V2

R1=R2=R3= 3 OhmV1=V2= 12 V

what is the current through and voltage over each R?I1 I3

I2

• apply kirchhoff’s rules1) I1+I2-I3=0 (kirchhoff I)

2) left loop: V1-I1R1+I2R2=0 so 12-3I1+3I2=0

3) right loop: V2+I3R3+I2R2=0 so 12+3I3+3I2=0

4) outside loop: V1-I1R1-I3R3-V2=0 so –3I1-3I3=0 so I1=-I3

combine 1) and 4) I2=2I3 and put into 3) 12+9I3=0 so I3= 4/3 A

and I1= - 4/3 A and I2= 8/3 A

Example


• what is the power dissipated by R3?

P=VI=V2/R=I2R

I=I1

R1 R2

V

R3

R1= 1 Ω , R2= 2 Ω , R3= 3 Ω , R4= 4 ΩV= 5V

I3I1

I2

I=I1

R4I4

We need to know V3 and/or I3.Find equivalent R of whole circuit.1/R23=1/R2+1/R3= 1/2+1/3= 5/6 R23= 6/5 ΩR1234= R1+R23+R4= 1+6/5+4=31/5 Ω I=I1=I4= V / R1234= 5 / (31/5)I = 25 / 31 AKirchhoff 1: I1=I2+I3= 25 / 31 Kirchhoff 2: I3R3-I2R2=0 so 3I3-2I2=0 I2= 3 / 2 I3

Combine: 3/2 I3+I3= 25 / 31 so 5 / 2 I3=25 / 31 I3=10/31 AP=I2R so P=(10 / 31)2 X 3=(100/961)X3= 0.31 W

Example


What is Kirchhoff II for the right small loop (with R2 and R3)?

a) I3R3+I2R2=0 b) I3R3-I2R2=0 c) I3R3-I2R2+V=0

What is Kirchhoff II for the loop (with V,R4 and R3)?

a) V-I4R4+I3R3=0 b) V+I4R4-I3R3=0 c) V-I4R4-I3R3=0

R1 R2

V

R3I3

I1 I2

I4R4 What is Kirchhoff I for ?

a) I1+I2-I3-I4=0 b) I1+I2+I3+I4=0 c) I1-I2-I3-I4=0

What is Kirchhoff II for the left small loop( with R4 and R1?)a) I4R4+I1R1=0 b) I4R4- I1R1=0 c) I4R4+I1R1-V=0

Example


= I r + I R = I (r + R)

I = /(r + R) - Ir = V = IR

Terminal voltage = V = IR

Electromotive Force ( Electromotive Force ( ))

The terminal voltage decrease = - Ir as the internal resistance r increases or when I increases

.If r =0 ( no internal resistance ) ( ideal battery ) = V = I R

I = V / R I = / R

Summary:


Series connection

Parallel connection

Resistors connected in a circuit in series or parallel can be simplified using the

following:


Documents

Norah Ali Al- Moneef1 Sources of emf The source that maintains the current in a closed circuit is called a source of emf – Any devices that increase