Gustav Robert Kirchhoff
(12 March 1824 – 17 October 1887)
Developed Theory's in:
German
Physicist
•
•
EM -Energy
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Voltage Laws
Light
Thermal
Electric
Circuits
Current Laws
1
Chapter 5
Parallel Circuits
2
Objectives
Identify a parallel circuit
 Determine the voltage across each parallel
branch
 Apply Kirchhoff’s current law
 Determine total parallel resistance
 Apply Ohm’s law in a parallel circuit
 Use a parallel circuit as a current divider
 Determine power in a parallel circuit

3
Resistors in Parallel


Each current path is called a branch.
A parallel circuit is one that has more than one branch.
4
Parallel Arrangements
Identifying Parallel Circuits
If there is more than one current path (branch)
between two separate points, and if the voltage
between those two points also appears across each of
the branches, then there is a parallel circuit between
those two points (common point).
6
Voltage in Parallel Circuits
The voltage across
any given branch of a
parallel circuit is equal
to the voltage across
each of the other
branches in parallel.
7
Kirchhoff’s Current Law (KCL)
The sum of the currents into a node (total
current in) is equal to the sum of the
currents out of that node (total current out).
IIN(1) + IIN(2) + . . . + IIN(n) = IOUT(1) + IOUT(2) +. . . +IOUT(m)
8
The total current out of a node A is the sum of the
two branch current. Therefore the total current into
node A is?
IT = I1 + I2
IT = 5 mA + 12 mA
IT = 17 mA
9
Kirchhoff’s Current Law
Kirchhoff’s current Law (KCL) can be stated
another way:
IT – IR1 – IR2 – IR3 = 0
The algebraic sum of all the currents entering
and leaving a junction is equal to zero.
10
Total Parallel Resistance
When resistors are connected in parallel, the
total resistance of the circuit decreases.
The total resistance of a parallel circuit is
always less than the value of the smallest
resistor.
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Parallel Resistance Relationship

Conductance (G) is a measure of the ability of a component or
circuit to conduct.

The reciprocal of resistance (1/R) is called conductance and its symbol is G.
The units of conductance are Siemens (S)

The total conductance (GT) in a parallel circuit equals the sum of the
branch conductance values.
GT = G1 + G2 + G3 +…… GN
12
Parallel Resistance Relationship

Conductance (G) is a measure of the ability of a component or
components in a circuit to conduct current.
GT = G1 + G2 + G3 +…… GN
13
Parallel Resistance Relationship


Parallel branches increases the circuit’s ability to conduct.
Therefore, the addition of a branch increases the conductance of a
circuit.
GT = G1 + G2 + G3 +…… GN
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Parallel Resistance Relationship

Conductance and Resistance are inversely proportional.
15
Total Parallel Resistance

Using Conductance we can calculate Resistance.

If
and
then:
16
Total Parallel Resistance
Calculate the total resistance.
RT = (R1-1 + R2-1 + R3-1)-1
RT = (300-1 + 300-1 + 300-1)-1Ω
RT = 100 Ω
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Total Parallel Resistance
Calculate the total resistance.
RT = (R1-1 + R2-1 + R3-1)-1
RT = ( (4.6 x 103)-1 + (7.5 x 103)-1 + (43.6 x 106)-1 )-1Ω
RT = 2.851053224 x 103 Ω
RT = 2.85 kΩ
If you’re not getting this exact value
then you’re doing something wrong.
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For Two Resistors in Parallel
We use Product-Over-Sum- Method

The total resistance for two resistors in parallel is
equal to the product of the two resistors divided
by the sum of the two resistors.
RT = R1R2/(R1 + R2)
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Calculate the total Resistance in
the circuit?
20
Calculate the Total Current
IT = VS/RT
= VS/(R1-1 + R2-1)-1
= 10 V / (100-1 + 56-1)-1 Ω
= 278.57 x 10-3 A
IT = 279 mA
Check: IT = IR1+IR2
IR1=VR1/R1
IR1=10V/100 Ω
IR1=100 mA
IR2=VR2/R2
IR2=10V/56 Ω
IR1=179 mA
IT = IR1+IR2
IT=100 mA + 179 mA
IT=279 mA
21
Fuse Specification
You are designing a circuit for an automobile and
you must calculate the fuse size required for you
circuit.
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Fuse Specification
Component information:
1)
Each head light has a resistance of 4 ohms.
2)
Each tail light has a resistance of 10 ohms.
3)
Voltage source is 12.6 volts.
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Fuse Specification
What do we do first?
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Fuse Specification
Step 1: Calculate total resistance.
RT = (R1-1 + R2-1 + R3-1 + R4-1)-1
RT = (4-1 + 4-1 + 10-1 + 10-1)-1 Ω
RT = 1.4286 Ω
Step 2: Calculate total current.
IT = VS/RT
IT = 12.6 (V) / 1.4286 (Ω)
IT = 8.82 A
The total current through the fuse is 8.82 amps. The recommended fuse
size is a 10 amp fuse.
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Notation for Parallel Resistors

To indicate 5 resistors, all in parallel, we
would write:
RT = R1||R2||R3||R4||R5
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Notation for Parallel Resistors

Write the notation – then the formula –
then and fill in the values and calculate.
RT = R1||R2||R3||R4||
RT = (R1-1 + R2-1 + R3-1)-1
RT = (300-1 + 300-1 + 300-1)-1 Ω
RT = 100 Ω
Notation
Formula
Calculation
27
Application of a Parallel Circuit

One advantage of a parallel circuit over a series
circuit is that when one branch opens, the other
branches are not affected.
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Application of a Parallel Circuit


All lights and appliances in a home are wired in
parallel.
The switches are located in series with the
lights.
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Current Dividers

A parallel circuit acts as a current divider
because the current entering the junction
of parallel branches “divides” up into
several individual branch currents.
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Current Dividers Rule

For two parallel elements of equal value, the current will
divide equally.

For parallel elements with different values, the smaller
the resistance, the greater the share of the input current.

For parallel elements of different values, the current will
split with a ratio to the inverse of their resistor values.

For example, if one of two parallel resistor is twice the other,
then the current through the larger resistor will be half the other.
Current Dividers
Unlike Voltage Dividers, the current division
is inversely proportional to the resistance.
?
What do you think
the currents
through the two
resistors are?
32
General Current-Divider Formula
?
What do you think the currents
through these resistors are?
33
General Current-Divider Formula

The current (Ix) through any branch equals the
total parallel resistance (RT) divided by the
resistance (Rx) of that branch, and then
multiplied by the total current (IT) into the
junction of the parallel branches
Ix = (RT/Rx)IT
This is the general current divider formula that applies to any number of branches
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Current Dividers
Unlike Voltage Dividers, the current division is
inversely proportional to the resistance.
Find IR1 and IR2
R1
Store Value
R2
Step 1: Calculate RT
IRT = R1//R2
IRT = (200-1 + 100-1)-1Ω
IRT = 66.66666667 Ω
IR1 = (RT/R1)IT
IR1 = (66.667 Ω/100 Ω) x 300 mA
IR1 = 200 mA
IR1 = (RT/R2)IT
IR1 = (66.667 Ω/200 Ω) x 300 mA
IR1 = 100 mA
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Current-divider Formulas for Two Branches

When there are two parallel resistors, the
current-divider formulas for the two
R1
branches are:
I1 = (R2/(R1 + R2))IT
R2
I2 = (R1/(R1 + R2))IT
In words, for two parallel branches, the current through either
branch is equal to the product of the other parallel resistor and
the input current divided by the sum (not the total parallel
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resistance ) of the two parallel resistance.
Calculate I1 and I2 using the current divider formula
for two branches
I1 = (R2/(R1 + R2)) x IT
I2 = (R1/(R1 + R2)) x IT
I1 = (47Ω /147Ω) x 100mA
I2 = (100Ω /147Ω) x 100mA
I1 = 32.0 mA
I2 = 68.0 mA
37
Power in Parallel Circuits

Total power in a parallel circuit is found by
adding up the powers of all the individual
resistors, the same as for series circuits
P T = P1 + P 2 + P 3 + . . . P n
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Try this problem
Find: 1) The total resistance in the circuit
2) The total current in the circuit
3) The current through resistor R1 and R2
4) The voltage across resistor R2
5) The total power in the circuit
39
Now try this problem
Find: 1. The total resistance in the circuit
2. The Total current in the circuit
3. The current through each of the branch
4. The voltage across resistor R2
5. The Total Power in the circuit
40
Open Branches

When an open circuit occurs in a parallel
branch, the total resistance increases, the total
current decreases, and the same current
continues through each of the remaining parallel
paths.
41
Open Branches

When a parallel resistor opens, IT is
always less than its normal value.

Once IT and the voltage across the
branches are known, a few calculations
will determine the open resistor when all
the resistors are of different values.
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Summary




Resistors in parallel are connected across the
same two nodes in a circuit
A parallel circuit provides more than one path for
current
The number of current paths equals the number
of resistors in parallel
The total parallel resistance is less than the
lowest-value parallel resistor
43
Summary



The voltages across all branches of a parallel
circuit are the same
Kirchhoff’s Current Law: The sum of the currents
into a node equals the sum of the currents out of
the node
Kirchhoff’s Current Law may also be stated as:
The algebraic sum of all the currents entering
and leaving a node is zero
44
Summary



A parallel circuit is a current divider, so called
because the total current entering a node
divides up into each of the branches connected
to the node
If all of the branches of a parallel circuit have
equal resistance, the current through all of the
branches are equal
The total power in a parallel-resistive circuit is
the sum of all the individual powers of the
resistors making up the parallel circuit
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Summary



The total power for a parallel circuit can be
calculated with the power formulas using values
of total current, total resistance or total voltage
If one of the branches of a parallel circuit opens,
the total resistance increases, and therefore the
total current decreases
If a branch of a parallel circuit opens, there is no
change in current through the remaining
branches
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