Post on 12-Jan-2016
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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
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
t
QI
CIA C= current I= variable A=amps
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)
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
Circuit Schematics• Simple Circuit
ammeter
resistor
switch
battery
With a voltmeter attached outside the loop
A
V
A
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
Using Meters in Electricity Labs
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)
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
R1
A
R2 R3
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
Electric Power (Watts)
IVP Power = Current x Voltage
Watts = amps x volts
60W
120V
60W 60W
120V
Electric Energy• Recall that Power = Work Time and that it
requires Energy to do work so we could say
Power = ∆Energy Time • 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
Power, Current, Electric Potential, and Resistance
• Power = ∆ Energy Time
• Icurrent = Q time
• ∆V = ∆PE Q
• I = V R
so . . .
P = V x I
P = V2 R
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.
...
...
...
321
321
321
IIII
VVVV
RRRR
T
Battery
eq
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.
Comparing Series & Parallel
Series• Current Same• Voltage Adds
V1 + V2 + . . .
RT = R1 + R2 + . . .
Parallel• Current Adds• Voltage Same
I1 + I2 + I3 + . . .
R1 = 2ΩA
R2 = 3Ω
R1=1Ω
R2=2Ω
R3=3Ω
...1111
321 RRRReq
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!