Current, Resistance, Voltage
Electric Power & Energy
Series, Parallel & Combo
Circuits with Ohm’s Law
Current (I)
• The rate of flow of charges through
a conductor
• Needs a complete closed
conducting path to flow
• Measured with an “ammeter” in
amps (A) named for Ampere –
French scientist
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)
• The “electrical friction” encountered by
the charges moving through a material.
• Depends on
– Material
– Length
– Temperature
– 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 (  )
V  IR
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 ...
VBattery  V1  V2  V3 ...
I T  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
IT
6V
R2=2 Ω
Step 3: Find Voltage Drop
across each resistor.
RT  R1  R2
RT  1  2  3
Step 2: Find the total current
supplied by the battery
VBatt 6V
IT 

 2amps
RT
3
V1  I  R  2 A 1  2V
V2  I  R  2 A  2  4V
Note: Voltage adds in series and voltage drops should add to the battery voltage, 3V+3V=6V
Turn in WS
Electric Power (Watts)
P  IV
units
Amps  Volts  Watts
C J J
  W
s C s
using
V  IR
2
V
PI R
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
Example Problem #1
An electric furnace is connected across a 117 V
circuit. The furnace dissipates 3.5 kW of power in
the form of heat.
a) What is the resistance of the furnace?
b) How much heat does the furnace produce in 5 min?
Example Problem #2
Your bedroom lights are four 60 W bulbs.
a) If they are each hooked up to 120V, how much
current do they each draw?
b) What is the resistance of each bulb?
c) If the lights are on for 3 hours a day, how much
energy (in kWh) do they consume in 30 days?
d) If electricity costs $0.12 per kWh, how much does
it cost to operate the lights for a month?
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
Step 1: Find the total resistance of the circuit.
RT

( R1  R1  R1 )
1
RT 
Step 3: Find the current through
each resistor. Remember, voltage is
the same on each branch.
1
2
3
1
1
12
 61  41
 2
Step 2: Find the total current from the battery.
IT 
VT
RT
 122 V  6 A
I1 
V1
R1
12V
 12
  1A
I2 
V2
R2
 126 V  2 A
I3 
V3
R3
 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
R2=6Ω
12V
R1=12Ω
I T  1A  2 A  3 A  6 A
R3=4Ω
The total of the branches should be equal
to the sum of the individual branches.
Example problem
A circuit has 3 resistors in parallel: 3, 3 and 6 attached to
a 10V battery. Draw the circuit, then find the voltage across
and the current through each resistor.
REVIEW
Series
Picture
Total Resistance
Current
Voltage
with more resistorstotal resistance
total current
Parallel
Reading Resistors
Color bands: digit, digit, mulitpler, tolerance
Review Worksheet
Omit 1 and 3 on the front.
Answers online
Due Tues – Parallel WS, Review WS
Solving a Parallel Circuit
Step 1: Find the total resistance of the circuit.
1
RT

1
R1

1
R2
1
RT

1
1

1
2

1
R3

so... R 
1
3


11
6
6
T
11
Step 2: Find the total current from the battery.
IT 
VT
RT

12V
6 
11
 22 A
Step 3: Find the current through
each resistor. Remember, voltage is
the same on each branch.
I1 
V1
R1
 121V  12 A
I2 
V2
R2
 122 V  6 A
I3 
V3
R3
 123V  4 A
Step 4: Check currents to see if the
answers follow the pattern for current.
I T  I1  I 2  I 3
R2=2Ω
12V
R1=1Ω
I T  12 A  6 A  4 A  22 A
R3=3Ω
The total of the branches should be equal
to the sum of the individual branches.
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
R AD
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 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 dissapated is P=I5Ω2*R=(0.54A)2*5Ω=1.46 W.
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
Voltmeter and Ammeter
• Ammeter
– measures current in amps or mA
– used in series
• Voltmeter
– measures voltage
– used in parallel
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.