Current Electricity, Ohm’s Law & Circuits
Dec 24, 2015
Current Electricity,Ohm’s Law & Circuits
Current (I)• The rate of flow of charges through a
conductor• Needs a complete closed conducting
path to flow• Must have a potential difference
(voltage)• Measured with an “ammeter” in amps
(A) named for Ampere – French scientist
• I = current, A Q = charge, C
t = time, s So: 1 Amp = 1
t
QI
sec
Coulomb
ond
Voltage (V)• Electric potential difference between 2 points on a
conductor. Equal to the electric potential energy
per charge.
• Sometimes described as “electric pressure” that makes current flow
• Supplies the energy of the circuit• Measured in Volts (V) using a voltmeter• 1 Volt = 1 Joule / Coulomb
PEV
q
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.
...
...
...
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.
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 & C 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!
Voltmeter and Ammeter• Ammeter
– Measures current in amps – Placed in series where current is to be measured
• Voltmeter– Measures voltage in Volts– Placed in parallel across whatever is being
measured