1. Current, Resistance,Voltage Electric Power & Energy Series, Parallel & Combo Circuits with Ohm’s Law, Combo Circuits with Kirchoff’s Laws
Review for Chapters 19-20

2. 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
CIA C= current I= variable A=amps

3. 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

4. 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)

5. 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 (  )

6. Circuit Schematics Simple Circuit resistor switch ammeter
battery A
With a voltmeter attached outside the loop V A

7. Using Meters in Electricity Labs
Ammeter
measures current in amps or milliAmps
connect wire from negative side of battery to
the black peg. Some clips can go right in the top,
others work better if you loosen the cap and
come in from the side/ bottom.
Connect positive wire to 5A first and see if it
registers, if not move to 500mA peg or to 50mA
To read/measure- Use the scale that matches the
red connector you plugged into
Voltmeter (think VP)
Measures voltage/potential difference
Used in parallel ONLY - it should be plugged in outside the “circle”
If wired into the circuit you will fry its little brain!!
CAN NOT use 5 reading with a 6V battery! Start with 10 (2/10th markers) and go to 15 if needed (each line ~ .35)

8. Circuit with Series of Resistors
Current can only travel through 1 path add resistances
Rs=R1+ R2 +R3 +…
Current (amp) is the same through all parts A R1 R2 R3

10. Voltage Drop Across Resistors
Current (amp) is the same through all parts
Sum of the voltage drop must equal the source voltage
R1 = 1Ω A
R2 = 1Ω
6 V

11. Electric Power (Watts)
Power = Current x Voltage
Watts = amps x volts
60W
60W
60W
120V
120V

12. 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

13. 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. R1 R2 R3 V I

14. Solving a Series Circuit
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

15. 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 R1 R2 R3 V

16. Solving a Parallel Circuit
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.
R1=1Ω
R2=2Ω
R3=3Ω
12V
The total of the branches should be equal to the sum of the individual branches.

17. Combo Circuits with Ohm’s Law What’s in series and what is in parallel?
15V 3Ω 5Ω 7Ω 1Ω 2Ω 6Ω 4Ω 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. D C 3Ω A B 6Ω 4Ω 1Ω C 5Ω 2Ω D 7Ω
15V

18. Combo Circuits with Ohm’s Law Now…again…what’s in series and what’s in parallel?3Ω A B 6Ω 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.

19. Combo Circuits with Ohm’s Law Finding total (equivalent) resistance3Ω A B 6Ω 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…
Then, RBC + 2Ω=2.91Ω and this value is in parallel with the 5Ω branch, so…
Finally RT = RAD =
RT = 11.84Ω

20. Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor
RT = 11.84Ω 3Ω A B 6Ω 4Ω 1Ω C 5Ω 2Ω D 7Ω
15V
IT=1.27A
IT=1.27A
Then…
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…
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

21. Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor (continued)3Ω A B 6Ω 4Ω 1Ω C 5Ω 2Ω D 7Ω
15V
IT=1.27A
I2Ω=0.81A
I5Ω=0.46A
Known values from previous slide.
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.
To calculate the current through the 5Ω resistor…
So…

22. Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor (continued)
I6Ω=I4Ω =0.068A
Known values from previous slide. 3Ω A B 6Ω 4Ω 1Ω C 5Ω 2Ω D 7Ω
15V
IT=1.27A
I2Ω=0.81A
I5Ω=0.46A
I1Ω=0.68A
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.
All we need now is the voltage drop across the 6Ω and 4Ω resistors. So…
THE END!