Current, Resistance,VoltageElectric Power & EnergySeries, Parallel & Combo Circuits with Ohm’s Law,

Combo Circuits with Kirchoff’s Laws

Review for Chapters 19-20

Current (I)• The rate of flow of charges

through a conductor

• Needs a complete closed conducting path to flow

• End of the conducting path must have a potential difference (voltage)

• Measured with an “ammeter” in amps (A) named for Ampere – French scientist

qI

t

Voltage (V)

• Electric potential difference between 2 points on a conductor

• Sometimes described as “electric pressure” that makes current flow

• Supplies the energy of the circuit

• Measured in Volts (V) using a voltmeter

Resistance (R)• The “electrical friction” encountered by

the charges moving through a material.

• Depends on material, length, and cross-sectional area of conductor

• Measured in Ohms (Ω)

AR

Where: R = resistance, = length of conductor, A = cross-sectional area of conductor, ρ = resistivity of conducting material

Resistivity (ρ)• Property of material that resists the

flow of charges (resistivity, ρ, in Ωm)• The inverse property of conductivity

• Resistivity is temperature dependent…as temperature increases, then resistivity increases, and so resistance increases.

Ohm’s Law• A relationship between voltage, current,

and resistance in an electric circuit

• used to make calculations in all circuit problems

• V = potential difference (voltage) in volts

• I = electric current in amperes (amps , A)

• R = resistance in ohms ( )

IRV

Electric Power (Watts)

R

VRIIVP

22

• Used for thermal energy

time

EnergyPower

Electric Energy• Electric energy can be measured in

Joules (J) or Kilowatt hours ( kWh )• for Joules use Power in watts and time

in seconds• for kWh use Power in kilowatts and

time in hours

PtE

Series Circuits• Current can only travel through one path

• Current is the same through all parts of the circuit.

• The sum of the voltages of each component of the circuit must equal the battery.

• The equivalent resistance of a series circuit is the sum of the individual resistances.

1 2 3

1 2 3

1 2 3

...

...

...

eq

T Battery

T

R R R R

V V V V V

I I I I

R1

R2

R3

V I

Solving a Series Circuit

ampsV

R

VI

T

BattT 3

2

6

6V

R1=1 Ω

R2=1 Ω

IT

21121

T

T

R

RRR

Step 1: Find the equivalent (total) resistance of the circuit

Step 2: Find the total current supplied by the battery

Step 3: Find Voltage Drop across each resistor. VARIV 3131

Note: Since both resistors are the same, they use the same voltage. Voltage adds in series and voltage drops should add to the battery voltage, 3V+3V=6V

Parallel Circuits• Current splits into “branches” so there is more

than one path that current can take

• Voltage is the same across each branch

• Currents in each branch add to equal the total current through the battery

...

...

...1111

321

321

321

VVVV

IIII

RRRR

Battery

T

eqR1 R2

R3V

Solving a Parallel Circuit

R1=1Ω

R2=2Ω

R3=3Ω12V

Step 1: Find the total resistance of the circuit.

116

611

31

21

111

1111

so...

321

T

R

RRRR

RT

T

Step 2: Find the total current from the battery.

AI VRV

T T

T 2211

612

Step 3: Find the current through each resistor. Remember, voltage is the same on each branch.

AI

AI

AI

VRV

VRV

VRV

4

6

12

312

3

212

2

112

1

3

3

2

2

1

1

Step 4: Check currents to see if the answers follow the pattern for current.

AAAAI

IIII

T

T

224612321

The total of the branches should be equal

to the sum of the individual branches.

Combo Circuits with Ohm’s LawWhat’s in series and what is in parallel?

15V

3Ω

5Ω

7Ω

1Ω

2Ω

6Ω

4Ω

It is often easier to answer this question if we redraw the circuit. Let’s label the junctions (where current splits or comes together) as reference points.

A B

CD

3Ω A

B6Ω 4Ω

1Ω

C

5Ω

2Ω

D 7Ω

15V

Combo Circuits with Ohm’s LawNow…again…what’s in series and what’s in parallel?

3Ω A

B6Ω 4Ω

1Ω

C

5Ω

2Ω

D 7Ω

15V

The 6Ω and the 4Ω resistors are in series with each other, the branch they are on is parallel to the 1Ω resistor. The parallel branches between B & D are in series with the 2Ω resistor. The 5Ω resistor is on a branch that is parallel with the BC parallel group and its series 2Ω buddy. The total resistance between A & D is in series with the 3Ω and the 7Ω resistors.

Combo Circuits with Ohm’s LawFinding total (equivalent) resistance

3Ω A

B6Ω 4Ω

1Ω

C

5Ω

2Ω

D 7Ω

15V

To find RT work from the inside out. Start with the 6+4 = 10Ω series branch. So, 10Ω is in parallel with 1Ω between B&C…

91.0 so... 1110

1011

11

1011

BC

R

RBC

Then, RBC + 2Ω=2.91Ω and this value is in parallel with the 5Ω branch, so…

84.1 so...

51

91.211

AD

R

RAD

Finally RT = RAD +3 + 7 = 1.84 + 3 + 7

RT = 11.84Ω

Combo Circuits with Ohm’s LawSolving for current and voltage drops in each resistor

3ΩA

B

6Ω 4Ω

1Ω

C

5Ω

2Ω

D7Ω

15V

RT = 11.84Ω AI VRV

T T

T 27.184.1115

IT=1.27A IT=1.27A

The total current IT goes through the 3Ω and the 7Ω and since those are in series, they must get their chunk of the 15V input before we can know how much is left for the parallel. So…

AIIIT 27.173

Then…

VVVVV

VARIV

VARIV

ADP3.289.881.315

89.8727.1

81.3327.1

77

33

So…

Since parallel branches have the same current, that means the voltage across the 5Ω resistor V5Ω=4.84V and the voltage across the parallel section between B&C plus the 2Ω is also 4.84V

Combo Circuits with Ohm’s LawSolving for current and voltage drops in each resistor (continued)

VV

VV

VV

AI

R

ADP

T

T

3.2

89.8

81.3

27.1

84.11

5

3

Known values from previous slide. To calculate the current

through the 5Ω resistor…

AI VRV 46.05

3.25

5

To calculate the top branch of the parallel circuit between points A & D we need to find the current and voltage for the series 2 Ω resistor. Since the current through the resistor plus the 0.92A for the bottom branch must equal 1.3A.

So…

VARIV

AAAI

62.1281.0

81.046.027.1

22

2

3ΩA

B

6Ω 4Ω

1Ω

C

5Ω

2Ω

D7Ω

15VIT=1.27A IT=1.27A

I2Ω=0.81AI5Ω=0.46A

Combo Circuits with Ohm’s LawSolving for current and voltage drops in each resistor (continued)

AI

VVV

VV

AI

AI

VVV

VV

VV

AI

R

BC

AD

P

P

T

T

68.0

68.0

62.1

81.0

46.0

3.2

89.8

81.3

27.1

84.11

1

1

2

2

5

5

7

3

Known values from previous slide.

AI

VVVVV

VRV

PBC

68.0

68.062.13.2

168.0

1

1

1

3ΩA

B 6Ω 4Ω

1Ω

C

5Ω

2Ω

D7Ω

15VIT=1.27A IT=1.27A

I2Ω=0.81A

I5Ω=0.46A

Next we need to calculate quantities for the parallel bunch between points B&C. The voltage that is left to operate this parallel bunch is the voltage for the 5Ω minus what is used by the series 2Ω resistor. The 1Ω resistor gets all of this voltage.

Finally we need to calculate the current through the 6Ω and 4Ω resistors and the voltage used by each. AII V 068.0)46(

68.046

All we need now is the voltage drop across the 6Ω and 4Ω resistors. So…

VARIV

VARIV

27.04068.0

41.06068.0

44

66

I1Ω=0.68A

I6Ω=I4Ω =0.068A

THE END!

Kirchoff’s LawsLaw of Loops ( or Voltages) treats complex circuits as if they were several series circuits stuck together. So…the rules of series circuit voltages allows us to write equations and solve the circuit.

0V or ΣVinput = ΣVdrops

Law of Nodes (or Currents) The total of the currents that enter a junction (or node) must be equal to the total of the currents that come out of the junction (or node).

0I or ΣIin = ΣIoutWe use this law already in general when we add currents in the branches of a parallel circuit to get the total before it split into the branches.

Kirchoff’s Laws of Voltage writing the equations

R6

Loop A V= IA R1+( IA- IB) R2+ IA R7

Loop B 0 = (IB- IA) R2+( IB- IC)R4+ IB R3

Loop C 0 = (IC- IB) R4+ ICR5+ IC R6

V IA IB IC

R1

R2

R3

R4

R5

R7

Use Σ Vinput = Σ Vdrops for each current loop to write these equations. Remember that current is a vector so if multiple currents pass through a resistor, the total is the vector sum of the currents assuming the current loop you are writing the equation for is positive.

Draw current loops so that at least one loop passes through each resistor. Current loops must NOT have branches.

Kirchoff’s Law of Voltageputting numbers in the equations

Loop A 15V = IA (3Ω)+( IA- IB) (5Ω) + IA (7Ω) 15 = 3IA+5IA-5IB+7IA 15 = 15 IA - 5 IB + 0 IC

Loop B 0 = (IB- IA) (5Ω) +( IB- IC)(1Ω) + IB (2Ω)

0 = 5IB – 5IA +1IB -1IC+2IB 0 = -5 IA + 9 IB - 1 IC

Loop C 0 = (IC- IB) (1Ω) + IC (6Ω) + IC (4Ω) 0 = 1IC-1IB+6IC +4IC 0 = 0 IA -1 IB + 11 IC

1. Draw current loops so that at least one loop passes through each resistor. Current loops must NOT have branches. 2. Write an equation for each loop. 3. Solve the system of equations for all of the unknowns using a matrix (next slide)

15VIA IB IC

3Ω

5Ω

7Ω

1Ω

2Ω

6Ω

4Ω

Note: you must have coefficients for each unknown (even if it is zero) in every current loop equation.

Kirchoff’s Law of Voltage setting up and solving the matrix for IA, IB, and IC

15VIA IB IC

3Ω

5Ω

7Ω

1Ω

2Ω

6Ω

4Ω

15 = 15 IA - 5 IB + 0 IC

0 = -5 IA + 9 IB - 1 IC

0 = 0 IA -1 IB +11 IC

Beginning with the system of equations we wrote on the previous slide, we need to express these in matrix form to solve for the 3 unknowns

* =15 -5 0 -5 9 -1 0 -1 11

IA

IB

IC

15 0 0

coefficients unkowns answers

A * x = BCreate matrix A and B in your calculator. (Matrx> >Edit, then choose A or B )

In a normal algebra equation Ax=B, the solution is x = B/A, however matrix operations do not allow for division so instead, after you create the matrices, you will use them in the following operation.x=A-1B. The answer will be in matrix form containing all of the unknowns in the order they were set up.

Kirchoff’s Laws of Voltage Interpreting the answers to the matrix problem

15VIA IB IC

3Ω

5Ω

7Ω

1Ω

2Ω

6Ω

4Ω

* =15 -5 0 -5 9 -1 0 -1 11

IA

IB

IC

15 0 0

coefficients unkowns answers

A * x = BAfter performing the operation x=A-1B, the calculator will give you a matrix answer (the number of decimal places will depend on the calculator settings) like below.

=IA

IB

IC

1.230.690.063

So now we know that IA = 1.23A, IB=0.69A and IC = 0.063A

Now what?

Using these current loop values we can now evaluate current, voltage, and power through any resistor in the circuit.

Example: for the 3Ω resistor, only IA passes through it so the I3Ω = 1.23 A, the voltage is V=IR=1.23A*3Ω=3.69V, and power, P=I2R= (1.23)2*3Ω = 4.54 W

Kirchoff’s Laws of Voltage But what if the resistor you ask me about is shared by two current loops?

Yikes!

15VIA IB IC

3Ω

5Ω

7Ω

1Ω

2Ω

6Ω

4Ω

=

IA

IB

IC

1.230.690.063

So now we know that IA = 1.23A, IB=0.69A and IC = 0.063A

Let’s evaluate the 5Ω resistor:Since it is shared by current loops A and B, the current is the vector sum of the two. In this case IA & IB pass through the resistor in opposite directions so…I5Ω = IA-IB=1.23A-0.69A=0.54A .

The voltage drop is calculated V5Ω=I5ΩR=0.54A*5Ω=2.7V.

The power dissapated is P=I5Ω2*R=(0.54A)2*5Ω=1.46 W.

10V

1 Ω

2 Ω

1 Ω

IT

IT I2

I1

I2

Node

Kirchoff’s Law of Nodes

IT=I1+I2

The current entering one node is equal to the sum of the currents coming out

2 Ω

3 Ω

Voltmeter and Ammeter

• Ammeter– measures current in amps or mA– used in series

• Voltmeter– measures voltage– used in parallel