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Transcript
Current, Resistance,Voltage
Electric Power & Energy
Series, Parallel & Combo
Circuits with Ohm’s Law,
Combo Circuits with
Kirchoff’s Laws
Current (I)
• The rate of flow of charges through
a conductor
• Needs a complete closed conducting
path and a potential difference or
voltage to flow
• Measured with an “ammeter” in
amps (A) named for Ampere –
French scientist
• I = current, A q = charge, C
Δt = time,s (Note: 1 A = 1 C/s)
q
I
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)
• Opposition to the flow of charges.
“electrical friction” of the moving charges.
• Measured in Ohms (Ω)
• Resistance in a wire depends on
- material (low R for copper)
- temperature (as T increases, R increases)
- length (as length increases, R increases)
- cross-sectional area ( as A incr., R decr.)
Resistance (R)
• The “electrical friction” encountered by
the charges moving through a material.
• Depends on material, length, and crosssectional area of conductor
• Measured in Ohms (Ω)
R

A
Where: R = resistance, = length of
conductor, A = cross-sectional
 area of
conductor, ρ = resistivity of conducting
material
Note: Don’t need to know this formula.
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 (  )
V
I
R
V  IR
Voltmeter and Ammeter
• Ammeter
– measures current in amps or mA
– must be placed in series with what is being
measured
• Voltmeter
– measures voltage in volts
– must be placed in parallel with what is being
measured
Electric Power (Watts)
Energy
Power 
time
2
V
P  IV  I R 
• Used for thermal energy R
2
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
E  Pt
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.
Req  R1  R2  R3 ...
VT  V1  V2  V3 ...  VBattery
IT  I1  I 2  I 3 ...
R1
V
I
R3
R2
Solving a Series Circuit
Step 1: Find the equivalent
(total) resistance of the circuit
R1=1 Ω
RT  1  2  3
IT
6V
RT  R1  R2
R2=2 Ω
Remember: The current is the
same throughout a series circuit
so… IT = I1 = I2
Step 2: Find the total current
supplied by the battery
VT 6V
IT 

 2A
RT 3
Step 3: Find Voltage Drop
across each resistor.
V1  I1  R1  2 A 1  2V
V2   I 2  R2  2 A  2  4V
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
1
1
1
1



...
Req
R1 R2 R3
I T  I1  I 2  I 3 ...
V
VBattery  V1  V2  V3  ...
R1
R2
R3
Solving a Parallel Circuit
R2=2Ω
12V
R3=4Ω
R1=1Ω
Step 3: Finding the current through each
resistor. Remember, voltage is the same on
each branch as the battery voltage. Note: Step
1 & 2 were not necessary to do step 3!!
I1 
Step 1: Finding the total resistance of the circuit.

1
R1

1
R2
RT  
1
1

1
2
RT 


1
R3

1

 1.75 
so... RT  0.572
Step 2: Finding the total current from the
battery.
VT
12V
T
RT
.572
I 

 21A
1

VT
R1
 121V  12 A
I 2  VR22  VRT2  122 V  6 A
I3 
1 1
4
V1
R1
V3
R3
 VRT3  124 V  3 A
Step 4: Check currents to see if the
answers follow the pattern for current.
I T  I1  I 2  I 3
I T  12 A  6 A  3 A  21A
The total of the branches should be equal
to the sum of the individual branches.
Solving Combo Circuits – Find RT first
R3 = 2 Ω
R2 = 7 Ω
R1 = 4 Ω
R4 = 3 Ω
V = 15 V
Step 1: Find RS for any series within a
R2-3 = 9 Ω
R1 = 4 Ω
R4 = 3 Ω
parallel branch. R2-3= 7 + 2 = 9Ω
V = 15 V
Step 2: Find RP for any parallel parts.
1
R2 4
1 
 1
 RP  

  2.25
 9 3 
Step 3: Find RT for the entire circuit.
RT  2.25  4  6.25
R1 = 4 Ω
R2-4 = 2.25 Ω
V = 15 V
R1-4 = 6.25 Ω
V = 15 V
Solving Combo Circuits
VT
15V

 2.4 A
RT 6.25
Step 4: Find IT using Ohm’s law.
IT 
Step 5: Find the voltage across any
resistors in series with the battery.
V1  I1R1  IT R1
Step 6: Find the voltage drop across
the whole parallel part. (2 methods
are shown.)
V1  2.4 A  4   9.6V
VT  V1  V2  4
VP  V2  4  15V  9.6V  5.4V
.....or.....
VP  I P RP
VP  2.4 A  2.25   5.4V
Step 7: The voltage drop across
each parallel branch is the same as
the total voltage drop across the
parallel part.
𝑉𝑃 = 𝑉4 = 𝑉2−3
Solving Combo Circuits
Step 8: Find 𝐼4 using Ohm’s Law.
Step 9: Find 𝐼2,3 using Ohm’s Law.
In series, the current is same.
𝐼4 =
𝐼2,3 =
𝑉4
𝑅4
𝑉2,3
𝑅2,3
=
5.4 𝑉
3𝛀
= 1.8 A
=
5.4 𝑉
9𝛀
= 0.6 A
𝐼2 = 𝐼3 = 𝐼2,3 = 0.6 A
{Check: 𝐼𝑇 = 𝐼4 + 𝐼2,3 =1.8 + 0.6 = 2.4 A}
Step 10: Find 𝑉2 and 𝑉3 using Ohm’s Law.
{Check: 𝑉𝑃 = 𝑉2 + 𝑉3 = 4.2+ 1.2 = 5.4 V}
𝑉2 = 𝐼2 𝑅2 = 0.6A (7𝛀) = 4.2 V
𝑉3 = 𝐼3 𝑅3 = 0.6A (2 𝛀) = 1.2V
More Combo Circuits with Ohm’s Law
What’s in series and what is in parallel?
A
3Ω
B
5Ω
15V
1Ω
6Ω
4Ω
7Ω
D
2Ω C
6Ω
4Ω
B
1Ω
3Ω
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
5Ω
15V
C
2Ω
D
7Ω
Combo Circuits with Ohm’s Law
Now…again…what’s in series and what’s in parallel?
6Ω
4Ω
B
3Ω
1Ω
A
C
2Ω
D
7Ω
5Ω
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 Law
Finding total (equivalent) resistance
6Ω
4Ω
B
3Ω
C
2Ω
1Ω
A
D
7Ω
5Ω
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…
1
RBC
11
 101  11  10
so... RBC  10
11  0.91
Then, RBC + 2Ω=2.91Ω and this
value is in parallel with the 5Ω
branch, so… 1  1  1
RAD
2.91
5
so... RAD  1.84
Finally RT = RAD +3 + 7 = 1.84 + 3 + 7
RT = 11.84Ω
Combo Circuits with Ohm’s Law
Solving for current and voltage drops in each resistor
RT = 11.84Ω
6Ω
4Ω
IT 
C
 1115.84V  1.27 A
2Ω
B
1Ω
3Ω
VT
RT
D
A
7Ω
5Ω
IT=1.27A
IT=1.27A
15V
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…
IT  I 3  I 7   1.27 A
Then… V3  I 3  R  1.27 A  3  3.81V
V7   I 7   R  1.27 A  7  8.89V
So… VP  15V  3.81V  8.89V  2.3V
AD
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 Law
Solving for current and voltage drops in each resistor (continued)
6Ω
4Ω
B
1Ω
3Ω
A
C
2Ω
I2Ω=0.81A
D
7Ω
5Ω
IT=1.27A
Known values from
previous slide.
RT  11.84
I T  1.27 A
V3  3.81V
V5  8.89V
VPAD  2.3V
15V
To calculate the current
through the 5Ω resistor…
I 5  VR5  25.3V  0.46 A
IT=1.27A
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… I 2  1.27 A  0.46 A  0.81A
V2  I 2  R  0.81A  2  1.62V
Combo Circuits with Ohm’s Law
Solving for current and voltage drops in each resistor (continued)
I6Ω=I4Ω =0.068A
C
B
6Ω
I1Ω=0.68A
A
2Ω
4Ω
1Ω
3Ω
Known values from
previous slide.
I2Ω=0.81A
D
RT  11.84
7Ω
V3  3.81V
5Ω
IT=1.27A
IT=1.27A
15V
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.
I T  1.27 A
Finally we need to calculate the
current through the 6Ω and 4Ω
resistors and the voltage used by each.
I 6  I 4 
0.68V
( 6 4) 
 0.068 A
All we need now is the voltage
drop across the 6Ω and 4Ω
resistors. So…
V7   8.89V
VPAD  V5  2.3V
I 5  0.46 A
I 2   0.81A
V2   1.62V
VPBC  V1  0.68V
I1  0.68 A
VPBC  V1  2.3V  1.62V  0.68V V6  I 6  R  0.068 A  6  0.41V
I1 
V1
R

0.68V
1
 0.68 A
V4  I 4  R  0.068 A  4  0.27V
THE END!
Kirchoff’s Laws
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.
V  0 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).
I  0
or
ΣIin = 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.
Kirchoff’s Law of Nodes
3Ω
IT
I1
I2
2Ω
2Ω
10V
IT
1Ω
I2
1Ω
Node
IT=I1+I2
The current entering one node is equal
to the sum of the currents coming out
Kirchoff’s Laws of Voltage
writing the equations
Draw current loops so that at least one
Use Σ Vinput = Σ Vdrops
loop passes through each resistor.
for each current loop to
Current loops must NOT have branches.
R1
R4
R2
V
IA
IB
IC
R5
R6
R7
R3
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.
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
Kirchoff’s Law of Voltage
putting numbers in the equations
3Ω
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)
5Ω
15V
IA
1Ω
IB
IC
6Ω
4Ω
7Ω
2Ω
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
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
3Ω
Beginning with the system of equations
we wrote on the previous slide, we need
to express these in matrix form to solve
15V
for the 3 unknowns
15 = 15 IA - 5 IB + 0 IC
0 = -5 IA + 9 IB - 1 IC
0 = 0 IA -1 IB +11 IC
IA
15
15 -5 0
-5 9 -1 * IB = 0
IC
0
0 -1 11
coefficients
A
unkowns answers
*
x
= B
Create matrix A and B in your calculator.
(Matrx> >Edit, then choose A or B )
5Ω
IA
1Ω
IB
IC
6Ω
4Ω
7Ω
2Ω
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
3Ω
IA
15
15 -5 0
-5 9 -1 * IB = 0
IC
0
0 -1 11
coefficients
A
5Ω
15V
x
IC
= B
After 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.
IC
IB
6Ω
unkowns answers
*
1.23
IA
0.69 = I
B
0.063
IA
1Ω
So now we know that
IA = 1.23A, IB=0.69A
and IC = 0.063A
Now what?
4Ω
7Ω
2Ω
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!
3Ω
So now we
IA
1.23
know that
0.69 = IB
IA = 1.23A,
5Ω
1Ω
15V
0.063
IB=0.69A
6Ω
IC
I
I
A
B
IC
and
IC = 0.063A
4Ω
7Ω
2Ω
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 dissipated is P=I5Ω2*R=(0.54A)2*5Ω=1.46 W.