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Review for Chapters 19-20

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

Review for Chapters 19-20

- 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

- 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

- 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)

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

- 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.

- 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 ( )

- Used for thermal 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

R1

R2

R3

- 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.

V

I

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

R1=1 Ω

IT

6V

R2=1 Ω

Step 2: Find the total current supplied by the battery

Step 3: Find Voltage Drop across each resistor.

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

R1

R2

V

R3

- 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

R2=2Ω

12V

R3=3Ω

R1=1Ω

Step 1: Find the total resistance of the circuit.

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

Step 2: Find the total current from the battery.

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

The total of the branches should be equal to the sum of the individual branches.

A

B

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.

5Ω

1Ω

3Ω

6Ω

15V

D

C

4Ω

7Ω

2Ω

6Ω

4Ω

C

2Ω

B

1Ω

D

7Ω

A

3Ω

5Ω

15V

6Ω

4Ω

2Ω

C

B

The 6Ω and the4Ω 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.

1Ω

D

A

7Ω

3Ω

5Ω

15V

6Ω

4Ω

2Ω

C

B

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…

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

1Ω

D

A

7Ω

3Ω

5Ω

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

RT = 11.84Ω

15V

RT = 11.84Ω

IT=1.27A

IT=1.27A

6Ω

4Ω

2Ω

C

Then…

B

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…

1Ω

D

A

7Ω

3Ω

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

5Ω

15V

I2Ω=0.81A

I5Ω=0.46A

IT=1.27A

IT=1.27A

Known values from previous slide.

6Ω

4Ω

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.

2Ω

C

To calculate the current through the 5Ω resistor…

B

1Ω

D

A

7Ω

3Ω

5Ω

15V

So…

I6Ω=I4Ω =0.068A

Known values from previous slide.

I1Ω=0.68A

I2Ω=0.81A

I5Ω=0.46A

IT=1.27A

IT=1.27A

2Ω

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.

C

B

6Ω

4Ω

1Ω

D

A

7Ω

3Ω

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

5Ω

15V

THE END!

Law 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.

orΣVinput = 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).

orΣIin = ΣIout

We 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.

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

Use ΣVinput = ΣVdropsfor 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.

R1

R4

R2

V

R5

IA

IB

IC

R6

R3

R7

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

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)

3Ω

15V

5Ω

1Ω

IA

IB

IC

6Ω

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

4Ω

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

7Ω

2Ω

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

3Ω

15 = 15 IA - 5 IB + 0 IC

0 = -5 IA + 9 IB - 1 IC

0 = 0 IA -1 IB +11 IC

15V

5Ω

1Ω

IA

IB

IC

6Ω

IA

IB

IC

15

0

0

- -5 0
- -5 9 -1
- 0 -1 11

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.

4Ω

*

=

7Ω

2Ω

coefficients

unkowns

answers

A * x = B

Create matrix A and B in your calculator.

(Matrx> >Edit, then choose A or B )

3Ω

15V

5Ω

1Ω

IA

IB

IC

6Ω

A * x = B

After performing the operationx=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

15

0

0

- -5 0
- -5 9 -1
- 0 -1 11

4Ω

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

*

=

7Ω

2Ω

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

Now what?

1.23

0.69

0.063

IA

IB

IC

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

=

coefficients

unkowns

answers

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

IA

IB

IC

1.23

0.69

0.063

3Ω

=

15V

5Ω

1Ω

IA

IB

IC

6Ω

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 dissapatedis P=I5Ω2*R=(0.54A)2*5Ω=1.46 W.

4Ω

7Ω

2Ω

IT

I2

I1

2 Ω

10V

1 Ω

IT

I2

1 Ω

Node

3 Ω

Kirchoff’s Law of Nodes

2 Ω

IT=I1+I2

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

- Ammeter
- measures current in amps or mA
- used in series

- Voltmeter
- measures voltage
- used in parallel