Unit One: AC Electronics
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Transcript Unit One: AC Electronics
1
ET115 DC Electronics
Unit Six:
Parallel Circuits
John Elberfeld
[email protected]
WWW.J-Elberfeld.com
2
Schedule
Unit Topic
Chpt Labs
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
1
2
3
3
4
5
6
6
6
7
Quantities, Units, Safety
Voltage, Current, Resistance
Ohm’s Law
Energy and Power
Series Circuits
Exam I
Parallel Circuits
Series-Parallel Circuits
Thevenin’s, Power Exam 2
Superposition Theorem
Magnetism & Magnetic Devices
Course Review and Final Exam
2 (13)
3 + 16
5 (35)
6 (41)
7 (49)
9 (65)
10 (75)
19 (133)
11 (81)
Lab Final
3
Unit 6 Objectives - I
• Identify a parallel resistive circuit.
• Determine total resistance, current, and
power in a parallel resistive circuit.
• State Kirchhoff’s current law.
• Solve for an unknown current in a circuit
using Kirchhoff’s current law (KCL).
• State the general current-divider formula.
• State the current-divider formula for two
branches.
4
Unit 6 Objectives – II
• Apply the appropriate current-divider
formula to circuits to find an unknown
• current.
• Construct basic DC circuits on a
protoboard.
• Use a digital multimeter (DMM) to measure
a predetermined low voltage on a power
supply.
5
Unit 6 Objectives – III
• Measure resistances and voltages in
a DC circuit using a DMM.
• Apply Ohm’s Law, Thevenin’s
theorem, KVL and KCL to practical
circuits.
• Test circuits by connecting
simulated instruments in Multisim.
6
Reading Assignment
• Read and study
• Chapter 5: Parallel Circuits:
Pages 163-198
7
Lab Assignment
• Experiment 9, “Parallel Circuits,”
page 65 of DC Electronics: Lab
Manual and MultiSim Guide.
• Complete all measurements, graphs,
and questions and turn in your lab
before leaving the room
8
Written Assignments
• Complete the Unit 6 Homework sheet
• Show all your work!
• Be prepared for a quiz on questions
similar to those on the homework.
9
Ohms Law
• MEMORIZE: V = I R
• Ohms Law
• If you increase the voltage, you
increase the current proportionally
– 3 times the voltage gives you three
times the current
– Resistance (ohms) is the proportionality
constant and depends on the atomic
structure of the material conducting the
current
10
Power Formula
• Power = Work / time
• P=VI
• Voltage is the work done per
coulomb, and current is the number
of coulombs per second passing by
a point.
• The product of voltage and current
gives the work done per second, or
power.
11
Series Circuit Review
• There is just one path for the current
• The current is the same in every
component in a series circuit
• The voltage of the battery is equal to the
sum of voltage drops in the resistors
• Total resistance is the sum
of the resistors
12
Parallel Circuits
• There are two or more paths for current
flow.
• The voltage is the same across all parallel
branches.
• The total current entering any point is
equal to the current leaving the point
13
Parallel Combination of Resistances
• When all the resistors in a circuit are
connected in parallel, the circuit is called a
parallel circuit.
14
Determine Parallel Resistors
• Resistors are parallel to each other if
both pairs of ends are connected
directly together with no electrical
resistance between the ends
• Parallel has NOTHING to do with
geometry or layout of the resistors
15
Which are Parallel?
16
Parallel Circuit
• When a circuit is connected in parallel, the
potential difference in all the circuit
elements remains the same and the
current is distributed proportionally
among all the circuit elements.
17
Voltage Drop in Parallel Resistive Circuit
• The voltage drop across each component
in a parallel circuit is the same, since all the
resistances are connected in parallel. Thus,
the voltage across each connection is
equal to the supply voltage.
• The voltage drop in a parallel circuit is:
V1 V2 V3 V where, V is the supply
voltage.
18
A Practical Example
• Car circuits (and house circuits) are
all parallel
– You can run the radio without the horn
being turned on – an advantage
19
Find Equivalent Resistance
Voltage = 25 V
Find the total Current_______
Find the equivalent Resistance ______
V = 25V
R2 = 50Ω
R1 = 25Ω
I1 = ?
I2 = ?
20
Calculations
V = 25V
RT =
16.7Ω
IT = 1.5A
I1 = V1/R1 = 25V/25Ω = 1 A
I2 = V2/R2 = 25V/50Ω = 500 mA
IT = I1+I2 = 1 A + 500 mA = 1.5 A
RT = VT/IT = 25V/ 1.5 A = 16.7 Ω
If both resistors are replaced by a single
16.7 Ω resistor, the voltage and current
between the two end points is unchanged
• Notice that the total resistance is LESS
than any one resistor in the parallel
circuit.
•
•
•
•
•
21
Reasoning
• Adding a resistor in parallel provides
another path for current, making it
easier for current to flow
• This is the same as reducing the total
resistance for the circuit
22
Effective, Total Resistance
R1 = 10kΩ
• Ohm’s
law
method:
VT
RT
IT
R2 = 10kΩ
23
Parallel Circuit Nodes
• Two types of nodes or
connections:
– Dividing Node: A junction where
current enters by one connection but
leaves by two or more connections
– Summing connection: A junction where
current enters a junction by two or more
connections but leaves via one
24
Parallel Circuit Nodes
• Junctions A and B have current leaving
(Dividing Node), while junctions D and C
have current entering (Summing Node).
– Look at it from the electrons point of view
Summing
Dividing
25
Parallel Circuit Current
• Current leaving the power supply’s
(–) terminal is the same current
entering the (+) terminal.
• This is referred to as total current (IT).
• Individual electrons may follow a
variety of paths, but every electron
that leaves the battery is balanced by
an electron entering the battery
26
Parallel Circuit Current
• Since the total current in the
individual resistors is equal to
the current supplied by the
source, the total current can be
stated as:
IT = IR 1 + IR 2 … + IR n
27
Total Current
• IT = I1 + I2 + I3
IT
I1
I2
I3
28
Kirchhoff’s Current Law
• Kirchhoff’s current law states
that any current entering a
junction must be equal to the
current leaving the junction.
Iin = Iout
29
Kirchhoff’s Current Law (KCL)
• By definition of KCL, the algebraic sum of the
currents meeting at a point or junction is zero.
• In +
• Out -
• For the given circuit,
I1 I 2 I 3 I 4 I 5 I 6 0
30
Power In Parallel Circuits
1. Summation method
PT = PR 1 + PR 2 … + PR n
2. Formulas for Total Power
PT IT VT
2
PT I RT
2
V
PT
RT
31
Sample Problems
+
VT =
36V
R 1=
2kΩ
-
R 2=
4kΩ
Comp.
Resistance Voltage
R1
R2
R3
Total
2kΩ
4kΩ
6kΩ
Current
R 3=
6kΩ
Power
32
Sample Problems
+
VT =
36V
-
R 1=
2kΩ
R 2=
4kΩ
R 3=
6kΩ
Comp.
Resistance Voltage
Current
Power
R1
R2
R3
Total
2kΩ
4kΩ
6kΩ
1.09 kΩ
18ma
9ma
6ma
33ma
648mW
324mW
216mW
1188mW
36V
36V
36V
36V
33
Check your work
•
•
•
•
Things look good, but..
P = I2R
P = (33ma)2 1.09 kΩ = 1187 mW
The values for I and R work together
to give you the accepted value, so
they are probably correct.
• P = V2/R
• P = (36V)2 / 1.09 kΩ = 1189 mW
34
Sample Problems
+
VT =
24V
R 1=
1kΩ
-
R 2=
2kΩ
A
B
Comp.
Resistance
R1
R2
R3
Total
1kΩ
2kΩ
3kΩ
Current A
Voltage
Current B
R 3=
3kΩ
C
Current
Power
Current C
35
Sample Problems
+
VT =
24V
R 1=
1kΩ
-
R 3=
3kΩ
R 2=
2kΩ
A
B
C
Comp.
Resistance
Voltage
Current
Power
R1
R2
R3
Total
1kΩ
2kΩ
3kΩ
545 Ω
24V
24V
24V
24V
24ma
12ma
8ma
44ma
576mW
288mW
192mW
1056mW
Current A
Current B
Current C
44ma
20ma
8ma
36
Finding Total Series Resistance
• Calculating the total resistance for
resistors in SERIES is easy
• In SERIES, just add up the all the
resistors
37
Finding Total Series Resistance
• Ohm’s Law (V = I R) works for the total
circuit as well as each element in the
circuit
• Using Formulas:
• VT = IT RT and RT = VT / IT
• Find VT/IT and you have RT
• VT = V1 + V2 +V3 +V4…
• So
• VT = I1R1 + I2R2 + I3R3 + I4R4 .. - Series
38
Derivation
•
•
•
•
•
•
•
VT = I1R1 + I2R2 + I3R3 + I4R4 .. - Series
But, in SERIES, IT = I1 = I2 = I3 = I4 = …
VT = ITR1 + ITR2 + ITR3 + ITR4 ..
VT = IT (R1 + R2 + R3 + R4 ..)
VT / IT = (R1 + R2 + R3 + R4 ..)
RT = VT / IT = (R1 + R2 + R3 + R4 ..)
RT = (R1 + R2 + R3 + R4 ..) – Series!
39
Finding Total Parallel Resistance
• Every resistor you add in parallel
actually REDUCES the equivalent
resistance in the circuit
– Each resistor you add in parallel
increases the current, which makes the
resistance appear lower
• Simple addition does NOT work for
resistors in parallel
40
Finding Total Parallel Resistance
•
•
•
•
•
VT = IT RT and RT = VT / IT and IT = VT/RT
and 1/RT = IT / VT
We will find 1/RT in this math
IT = I1 + I2 + I3 + I4 + … - Parallel
IT = V1/R1 + V2/R2 + V3/R3 + V4/R4 + …
41
Derivation
•
•
•
•
•
•
But, in PARALLEL,
VT = V1 = V2 = V3 = V4 + …
IT = VT/R1 + VT/R2 + VT/R3 + VT/R4 + …
IT = VT(1/R1 + 1/R2 + 1/R3 + 1/R4 + …)
IT / VT = (1/R1 + 1/R2 + 1/R3 + 1/R4 + …)
1/RT = IT / VT = (1/R1 + 1/R2 + 1/R3 +
1/R4 + …)
• 1/RT = 1/R1 + 1/R2 + 1/R3 + 1/R4 + …
• Parallel Formula
42
Equation
• The reciprocal of the total resistance is
equal to the sum of the reciprocals of the
resistors in parallel
• 1/RT = 1/1kΩ + 1/2kΩ + 1/3kΩ
+
VT =
24V
-
R 1=
1kΩ
A
R 2=
2kΩ
B
R 3=
3kΩ
C
43
Using Fractions
•
•
•
•
•
•
•
1/RT = 1/1kΩ + 1/2kΩ + 1/3kΩ
LCD is 6KΩ
1/RT = 6/6kΩ + 3/6kΩ + 2/6kΩ
1/RT = 11/6kΩ
RT = 6kΩ / 11
RT = 545 Ω
(Hard to do with 2.2KΩ, 470Ω, 5.6MΩ)
+
VT =
24V
-
R 1=
1kΩ
A
R 2=
2kΩ
B
R 3=
3kΩ
C
44
Calculations
• 1/RT = 1/1kΩ + 1/2kΩ + 1/3kΩ
• 1 EXP 3 [INV] + 2 EXP 3 [INV] + 3 EXP 3
[INV] EQUALS [INV]
• RT = 545 Ω
• Learn how YOUR calculator works!
+
VT =
24V
-
R 1=
1kΩ
A
R 2=
2kΩ
B
R 3=
3kΩ
C
45
Check
•
•
•
•
•
I1 = 24V/1kΩ = 24 ma
I2 = 24V/2kΩ = 12 ma
I3 = 24V/3kΩ = 8 ma
IT = 44 ma
RT = VT/IT = 24V/44ma = 545 Ω
+
VT =
24V
-
R 1=
1kΩ
A
R 2=
2kΩ
B
R 3=
3kΩ
C
46
Own your own!
• Find RT using the inverse functions
on your calculator
+
VT =
36V
-
R 1=
2kΩ
R 2=
4kΩ
R 3=
6kΩ
47
Check
•
•
•
•
•
I1 = 36V/2kΩ = 18 ma
I2 = 36V/4kΩ = 9 ma
I3 = 36V/6kΩ = 6 ma
IT = 33 ma
RT = VT/IT = 36V/33ma = 1.09kΩ
+
VT =
36V
-
R 1=
2kΩ
R 2=
4kΩ
R 3=
6kΩ
48
Problem
• The following resistors are
connected in parallel:
• 1 MΩ, 2.2 MΩ, 4.7 MΩ, 12 MΩ, 22 MΩ
• Use your calculator to determine the
total resistance.
RT = 556 k Ω
49
Problem
• The total resistance of a three branch
parallel circuit is 819 Ω.
• If two of the resistors are 1.5 kΩ and
10 kΩ, find the third resistor.
1
1
1
1
RT R1 R2 R3
50
Problem
1
1
1
1
RT R1 R2 R3
1
1
1
1
819 1.5k 10k R3
1
1
1
1
R3 819 1.5k 10k
R3 2.2k
51
POS : Product over the Sum
• If you have just TWO Resistors in
parallel, the equation reduces to:
•
•
•
•
•
•
•
1/RT = 1/R1 + 1/R2
Multiply every term by (RTR1R2)
(RTR1R2)/RT = (RTR1R2)/R1 + (RTR1R2)/R2
R1R2 = RTR2 + RTR1 Factor out RT
R1R2 = RT(R2 + R1)
RT = R1R2 / (R2 + R1)
Product over the sum!!
52
Practice
+
VT =
36V
-
R 1=
2kΩ
R 2=
4kΩ
• RT = R1R2/(R1+R2)
• RT = (2kΩ 4kΩ) /(2kΩ + 4kΩ) = 1.333k Ω
• 2 EXP 3 x 4 EXP 3 ∕ ( 2 EXP 3 + 4 EXP 3 ) =
• Do NOT omit the (
)
53
More Practice
+
VT =
36V
-
R 1=
1.33kΩ
R 2=
6kΩ
• RT = R1R2/(R1+R2)
• RT = 1.33kΩ 6kΩ /(1.33kΩ + 6kΩ) = 1.09k Ω
• Do NOT omit the ( )
54
Practice
+
VT =
36V
-
R 1=
2kΩ
R 2=
4kΩ
R 3=
6kΩ
• You can evaluate a series of parallel
resistors by doing product over sum
for pairs until you get only one left –
which is RT.
55
Calculations
+
VT =
36V
R 1=
2kΩ
-
R 2=
4kΩ
R 3=
6kΩ
• Combining 6kΩ and 4kΩ using POS
+
VT =
36V
-
R1 =
2kΩ
R23=
2.4kΩ
• Combining 2kΩ and 2.4kΩ
+
VT =
36V
-
R 1=
1.09kΩ
56
POS – Parallel Resistors
R1
Y V Bk Gd
Gn Be Br Gd
4.7 kΩ
27 Ω
1.5 kΩ
10 kΩ
6 MΩ
R2
Gn Be Bk Si
Br Bk R Si
2.2 kΩ
56 Ω
2.2 kΩ
10 kΩ
6 MΩ
RT
57
POS
R1
Y V Bk Gd
Gn Be Br Gd
4.7 kΩ
27 Ω
1.5 kΩ
10 kΩ
6 MΩ
R2
Gn Be Bk Si
Br Bk R Si
2.2 kΩ
56 Ω
2.2 kΩ
10 kΩ
6 MΩ
RT
25.6 Ω
359 Ω
1.5 kΩ
18.2 Ω
892 Ω
5 kΩ
3 MΩ
58
Troubleshooting
Opens in a Parallel Circuit
• There are two types of opens in a parallel
circuit:
– Open in a main line
– Open in a branch line
• A main line connects the power source to
the rest of the circuit.
• A branch line is one of the different
branches in a parallel circuit.
59
Opens in a Parallel Circuit
• Open in a main line:
• Open in a branch line:
60
Main line Opens
+
• An open in the main
line stops all current flow through all elements
• The voltage across the open is equal
to the supply voltage
• There is no current to cause a
voltage drop across any element in
the circuit
• There is no power used in the circuit
since there is no current flowing
61
Effects Of a Branch Open
1. Branch opens makes the current go
to 0 A just in that branch
2. If current decreases to 0, resistance
must have increased to infinity
(definition of an open)
3. Branch voltage remains the same,
both across the open branch and the
other branches
62
An Open Resistor
+
• If a resistor becomes
“open” in a pure
parallel circuit:
1. Current in that branch goes to zero
2. Power in that branch goes to zero
3. Total resistance in the circuit INCREASES
because a path is gone
4. Total current in the circuit DECREASES
because a path is gone
5. Total power in the circuit DECREASES
because a path is gone
63
Adding a Resistor
+
• If a new resistor is
added to a pure parallel circuit:
1. Total resistance in the circuit
DECREASES because a path is added
2. Total current in the circuit
INCREASES because a path is added
3. Total power in the circuit INCREASES
because a path is added
64
A Practical Example
• I1 = I2 = I3 = I4 = 100V/10k = 10ma each
65
Shorts in a Parallel Circuit
• A short in a parallel circuit occurs if both
the terminals of any branch are shortcircuited.
• A short draws almost infinite current
because the resistance of the connecting
wires is zero.
66
Shorts in a Parallel Circuit
• To protect any branch of the circuit
from the short-circuited branch, a
resistance (or fuse) is placed in
between the branch and power
supply
67
Effects of a Short
• This is a dramatic and serious event
because total resistance goes to zero
Ohms.
• The power supply attempts to put out
infinite current ( I = V / R = V / 0 = ∞ )
• Since there is 0 Resistance across the
branches, there is no voltage drop
developed, and current is the maximum
the power supply can produce
• A protective device (fuse, circuit breaker)
is required.
68
Contrasting Series And Parallel Circuits
SERIES
• IT is constant
• KVL is used
• VT = sum of
drops
• RT is sum of
resistors
PARALLEL
• IT is the sum of
IRn
• KCL is used
• VT is constant
• 1/RT is sum of
reciprocals
69
Multiple Sources
• Multiple sources of power are usually connected in
parallel to meet additional power requirement at
constant voltage.
– Voltages must be the same!
• The total current supplied by all the voltage sources is
given by:
IT I1 I 2 I 3 ...... I N
70
Voltage Sources In Parallel
• Parallel sources are used to
increase the amount of total
current available.
• While VT remains the same, the
maximum IT available increases
by the amount of each source
• Actual IT is still controlled by the
resistance in the circuit
71
Current Dividers
• Current divider consists of a parallel resistive
circuit dividing the current, provided by the
voltage source, into the parallel resistors.
• The total current, I, is divided between resistors
R1 and R2.
• Smaller resistance value will draw more current
and vice versa.
72
Sample Problems
+
VT =
?V
-
R 1=
2 kΩ
R 2=
4 kΩ
A
B
Comp.
Resistance
R1
R2
Total
2 kΩ
4 kΩ
Current A
Voltage
PREDICT:
Size of RT
Current
8 mA
Current B
Power
73
Sample Problems
+
VT =
?V
-
R 1=
2 kΩ
R 2=
4 kΩ
A
B
Comp.
Resistance
Voltage
Current
Power
R1
R2
Total
2 kΩ
4 kΩ
1.33 kΩ
32 V
32 V
32 V
16 mA
8 mA
24 mA
512mW
256mW
768mW
Current A
24 mA
Current B
8 mA
74
Formulas
• How can we find the currents in one step?
• For TWO parallel resistors, the voltage is
the same
+
• VT = I R = I1 R1 = I2 R2
• IT = I1 + I2, so I2 = IT – I1
• I1 R1 = I2 R2 = (IT – I1) R2
• I1 R1 = (IT – I1) R2 = R2 IT – R2I1
-
• Adding R2I1 to each side:
• I1 R1 + R2 I1 = R2 IT = I1 (R1 + R2 ) = R2 IT
• I1 = R2 IT / (R1 + R2 )
75
Common Sense Challenge
• Unlike the voltage divider formula, to
find I1, you put R2 , the other resistor,
at the top of the formula.
R2
I1
IT
R1 R2
R1
I2
IT
R1 R2
76
Sample Problems
+
VT =
???
-
R 1=
8kΩ
A = 5ma
R 2=
12kΩ
B
IT = 5ma
I1 = R2 IT / (R2 +R1) = 12kΩ 5ma/ (8kΩ + 12kΩ )
I1 = 3ma
I2 = R1 IT / (R2 +R1) = 8kΩ 5ma/ (8kΩ + 12kΩ )
I2 = 2ma
VT = 24V
77
Current Dividers In A Two-branch Circuit
78
Current Dividers
• In the previous circuit, the current
through each resistor is given as:
R2
I1
.I
R1 R2
R1
I2
I
R1 R2
• where, I1 is the current through resistor
R1, and I2 is the current through resistor
R2.
• Current division is similar to voltage
division, but the TOP RESISTORS are
switched
79
Sample Problems
+
VT =
24V
-
R 1=
8kΩ
R 2=
12kΩ
A
B
Comp.
Resistance
R1
R2
Total
8kΩ
12kΩ
Current A
Voltage
Current B
PREDICT:
Which R has
more current
Current
Power
80
Sample Problems
+
VT =
24V
-
R 1=
8kΩ
R 2=
12kΩ
A
B
Comp.
Resistance
Voltage
Current
Power
R1
R2
Total
8kΩ
12kΩ
4. 8kΩ
24V
24V
24V
3ma
2ma
5ma
72mW
48mW
120mW
Current A
5ma
Current B
2ma
81
Sample Problems
+
VT =
?V
-
R 1=
4 kΩ
R 2=
? kΩ
A
B
Comp.
Resistance
R1
R2
Total
4 kΩ
Current A
50 mA
Voltage
PREDICT:
Size of R2
Size of RT
Current
30 mA
Current B
Power
82
Sample Problems
+
VT =
?V
-
R 1=
4 kΩ
R 2=
? kΩ
A
B
PREDICT:
Size of R2
Size of RT
Comp.
Resistance
Voltage
Current
Power
R1
R2
Total
4 kΩ
6 kΩ
2.4 kΩ
120 V
120 V
120 V
30 mA
20 mA
50 mA
3.6 W
2.4 W
6W
Current A
50 mA
Current B
30 mA
83
Conductance
• Conductance is the reciprocal of the
resistance
• G=1/R
• R=1/G
• G is measured in siemens
• R = 2.2 KΩ
• G = 1 / 2.2 KΩ = 454μS
84
Sample Problems
+
VT =
28V
-
Comp R
R 1=
7kΩ
R 2=
14kΩ
A
B
V
I
R1
7kΩ
R2
14kΩ
Total
Current A
Current B
G
P
85
Sample Problems
+
VT =
28V
-
Comp R
R 1=
7kΩ
R 2=
14kΩ
A
B
V
R1
7kΩ
28V
R2
14kΩ
28V
Total 4.67kΩ 28V
Current A
6ma
I
G
4ma
2ma
6ma
142μS 112mW
71μS 56mW
213 μS 168mW
Current B
2ma
P
86
Always Check Your Answers
•
•
•
•
GT = 1/RT
P = IT2RT = VT2/RT
RT = VT/IT
Your answers must be consistent no
matter what formula you use.
• Try it!
87
Sample Problems
+
VT =
12V
-
R 1=
4kΩ
R 2=
4kΩ
A
Comp R
B
V
I
R1
4kΩ
R2
4kΩ
Total
Current A
Current B
G
P
88
Sample Problems
+
VT =
12V
-
R 1=
4kΩ
R 2=
4kΩ
A
B
Comp R
V
I
G
R1
4kΩ
R2
4kΩ
Total 2kΩ
12V
12V
12V
3ma
3ma
3ma
250μS 36mW
250μS 36mW
500 μS 72mW
Current A
6ma
Current B
3ma
P
89
Sample Problems
+
VT =
12V
-
R 1=
4kΩ
R 2=
4kΩ
A
Comp R
B
V
I
R 3=
4kΩ
C
G
P
R1
R2
R3
Total
Current A
Current B
Current C
90
Sample Problems
+
VT =
12V
-
R 1=
4kΩ
R 2=
4kΩ
A
R 3=
4kΩ
B
C
Comp R
V
I
G
P
R1
R2
R3
Total
12V
12V
12V
12V
3ma
3ma
3ma
9ma
250μS
250μS
250μS
750μS
36mW
36mW
36mW
108mW
4kΩ
4kΩ
4kΩ
1.33kΩ
Current A
Current B
Current C
9ma
6ma
3ma
91
Multiple Resistors
• If you have two resistors that are the
same: 1 1 1 2
RT
RT
R
R
R
R
2
• Three and four identical resistors
1
1 1
1
3
RT R R R
R
1
1 1
1
1
4
RT R R R R
R
R
RT
3
R
RT
4
92
Identical Parallel Resistors
• If you have “n” identical resistors
with R ohms each, the total
equivalent resistance is:
1
1 1
1
1
n
...
RT R R R R
R
R
RT
n
• The more resistor you add in parallel,
the smaller the total resistance
93
Sample Problems
+
VT =
120
V
-
Comp R
R 1=
???
R 2=
????
A
B
V
I
R1
R2
Total
Current A
G
P
60W
75W
Current B
94
Sample Problems
+
VT =
120
V
-
Comp R
R 1=
???
R 2=
????
A
B
V
R1
240Ω 120V
R2
192Ω 120V
Total 107 Ω 120V
Current A
1.125A
I
G
500ma
625ma
1.125A
4.14mS 60W
5.21mS 75W
9.35ms 135W
Current B
625mA
P
95
Sample Problems
+
VT =
??V
-
R 1=
??kΩ
A
Comp R
R1
R2
5kΩ
R3
Total
Current A
V
R 2=
??kΩ
R 3=
??kΩ
B
C
I
G
P
5ma
100μS
1.5W
Current B
Current C
96
Sample Problems
+
VT =
12V
-
R 1=
??kΩ
A
R 2=
??kΩ
R 3=
??kΩ
B
C
Comp R
V
I
G
P
R1
R2
R3
Total
50V
50V
50V
50V
5ma
10ma
15ma
30ma
100μS
200μS
300μS
600 μS
250mW
500mW
750mW
1.5W
10kΩ
5kΩ
3.33kΩ
1.66 kΩ
Current A
Current B
Current C
30ma
25ma
15ma
97
Sample Problems
+
VT =
60V
-
R 1=
3kΩ
R 2=
6kΩ
A
Comp R
R1
B
V
I
R 4=
12kΩ
R 3=
9kΩ
C
G
P
R2
R3
R4
Total
Current A
Current B
Current C
Current D
98
Sample Problems
+
VT =
60V
R 1=
3kΩ
-
R 2=
6kΩ
A
R 4=
12kΩ
R 3=
9kΩ
B
C
D
Comp R
R1
3kΩ
V
60V
I
20ma
G
333μS
P
1.2W
R2
6kΩ
60V
10ma
167μS
600mW
R3
R4
9kΩ
12kΩ
60V
60V
6.67ma 111μS
5ma
83 μS
400mW
300mW
Total
1.44kΩ
60V
41.7ma 694 μS
2.5W
Current A
Current B
Current C
Current D
41.7ma
21.7ma
11.7ma
5ma
99
Observations
• When voltage is constant, like it is in
parallel circuits:
• Current is inversely proportional to the
value of the resistance (V constant!)
• Power is inversely proportional to the
value of the resistance (V constant!)
• Conductance is inversely proportional to
the value of the resistance (Always)
– When resistance goes UP, all three other
values go DOWN
100
Unit 6 Summary
• Identifying parallel circuits
• Calculating total resistance in
parallel
• Solving for unknown current using
KCL
• Solving problems with the currentdivider formula