Download Parallel Circuits

Lesson 5: Parallel DC Circuits
and Kirchhoffs Current Law
(KCL)
(Chapter 6)
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Learning Objectives
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Restate the definition of a node and demonstrate how to measure voltage and
current in parallel circuits.
Solve for total circuit resistance of a parallel circuit.
State and apply KCL in the analysis of simple parallel circuits.
Demonstrate how to calculate the total parallel resistance given various
resistors connected in parallel.
Evaluate why homes, businesses and ships are commonly wired in parallel
rather than series.
Demonstrate how to calculate the total current and branch currents in a parallel
circuit using the current divider equation.
Determine the net effect of parallel combining voltage sources.
Compute the power dissipated by each element in a parallel circuit, and
calculate the total circuit power.
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Parallel Circuits
• Two or more elements are in parallel if they are
connected to the same two nodes and consequently
have the same voltage across them.
2-, 3-, and 2A
are in parallel
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Parallel Circuits
• Remember: nodes are connection points between
components.
• Notice each component has two terminals and each is
connected to one of the nodes above.
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Parallel Circuits
• House circuits contain parallel circuits.
• The parallel circuit will continue to operate even
though one component may fail open – an advantage
over the series circuit.
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Parallel Circuits
vs. Series Circuits
• In a series circuit, failure of
a single components can
disable all components in
the circuit.
Series Circuit
• In a parallel circuit, failure
of one component will still
allow other components to
operate.
Parallel Circuit
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Circuit Breaker Panel
• Homes and ships are usually wired in parallel instead of
series.
• All components can operate at rated voltage independent of
other loads when wired in parallel.
AC
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Series - Parallel Circuits
• Circuits may contain a combination of series and
parallel components.
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Parallel Circuits
• To analyze a particular circuit
− First identify the node.
− Next, label the nodes with a letter or number.
− Then, identify types of connections.
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Example Problem 1
Determine which elements are connected in parallel and which
are connected in series.
(A||B) || (C+(D||F)+E)
(A+B)+ C||F+(D+E)
A+ (B||C||D)
A||B||C||D
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Kirchhoff’s Current Law (KCL)
• Kirchhoff’s Current Law states that the
algebraic sum of the currents entering
and leaving a node is equal to zero.
I
I  0
0
• Currents entering a node are positive
and those leaving a node are negative.
N
I
n 1
n
 I1  ( I 2 )  ( I 3 )  ( I 4 )  I 5  0
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Kirchhoff’s Current Law (KCL)
• KCL can also be stated as “the sum of currents
entering a node is equal to the sum of currents
leaving the node.”

Iin   I out
I1  I 5  I 2  I 3  I 4
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Parallel vs. Series Current Flow
• Understanding how fluid flows may help with your
understanding of KCL.
• When water flows in a pipe, the amount of water entering a
point is equal to the amount that leaves that point.
• However, it should be noted that more water flows down the
pipe with the lowest resistance.
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Kirchhoff’s Current Law (KCL)
• In technology, the term node is commonly used to
refer to a junction of two or more branches.
Two-node configuration
Four-node configuration
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Kirchhoff’s Current Law (KCL)
• Determine I3 and I5 using KCL.
KCL states:  Iin   I out
Therefore, at node a:
I1  I 2  I 3
4 A  3 A  I3
7 A  I3
And at node b:
I3  I 4  I5
I5  I3  I 4
I 5  7 A  1A
I5  6 A
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Kirchhoff’s Current Law (KCL)
• Determine Is using KCL.
Parallel Circuit
I s  I1  I 2  I 3
I s  8mA  10mA  2mA  20mA
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Direction of Current
• Assume a current direction and draw current arrows.
• If this assumption is incorrect, calculations will show
that the current has a negative sign.
• Negative sign simply indicates that the current flows
in the opposite direction to the arrow you drew.
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Example Problem 2
Determine the magnitude and direction of each current:
Ohm's Law: I1 
I1 
v1
R1
5V
 2mA
2.5k 
I 2  I1  I 3
I 3  I 2  I1
I 3  4mA  2mA
I 3  2mA
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Battery Cells
Series vs Parallel
Cells connected in series
increases available voltage.
Cells connected in parallel
increases available current.
Recall that voltages are
additive (ET = E1+E2+…+EN) in
series.
However, in parallel circuits,
voltages will be equal across
the parallel elements.
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Voltage Sources in Parallel
• When two equal sources are connected in parallel
− Each source supplies half the required current
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Voltage Sources in Parallel
Demonstrating the effect of placing two ideal supplies of the same
voltage in parallel.
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Voltage Sources in Parallel
• Because the voltage is the same across parallel
elements, voltage sources can be placed in parallel
only if they have the same voltage.
• The primary reason for placing two or more batteries
or supplies in parallel is to increase the current rating
above that of a single supply.
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Voltage Sources in Parallel
• Voltage sources with different potentials should never
be connected in parallel.
• Large currents can occur and cause damage.
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Voltage Sources in Parallel
• If, for some reason, two batteries of different voltages are
placed in parallel, both will become ineffective or damaged
because the battery with the larger voltage will rapidly
discharge through the battery with the smaller terminal
voltage.
Examining the impact of placing two leadacid batteries of different terminal
voltages in parallel.
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Example Problem 3
Determine the currents I2 and I3.
Knowing that voltage is equal across parallel elements we can first find v1…
v1  I1 R1
v1  2 A *101  20V
v2  20V
v3  20V
Now solve for each parallel current:
I2 
v2 20V

 500mA
R2 40
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I3 
v3
20V

 200mA
R3 100
Resistors in Parallel
• For only two resistors connected in parallel, the
equivalent resistance may be found by the product of
the two values divided by the sum:
R1 R2
RT 
R1  R2
• Often referred to as “product over the sum” formula.
(98)(2)
RT 
 1.96
98  2
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Resistors in Parallel
• For a circuit with 3 or more resistors:
IT  I1  I 2  I 3
E V1 V2 V3

  
RT R1 R2 R3
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Resistors in Parallel
• Since voltage across all parallel elements in a circuit
are the same (E = V1 = V2=V3):
E
E E E
 

RT R1 R2 R3

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1
1
1
1
 

RT R1 R2 R3
Resistors in Parallel
• Total resistance of any number of resistors in
parallel:
1
RT 
1
1
1

 ... 
R1 R2
Rn
1
1 1 1
 1
RT  
     20
180 90 60 60 
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Resistors in Parallel
• Sometimes you can solve resistive problems
involving parallel circuits through inspection.
• Below are a few examples of how this can be done:
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Current Divider Rule
• Similar to the Voltage Divider Rule, we can use the Current
Divider Rule to calculate current across a resistive element
without knowing the voltage drop across that element:
E = I T RT
 E 
Ix 

R
 X 
 IT RT
Ix 
 RX
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 RT 

    IT  I x

 Rx 
Example Problem 4
Use the current divider rule to determine all unknown
First, find an equivalent resistance (RT):
currents:
1
RT 
1
1
1
1

 
R1 R2 R3 R4
1
RT 
1
1
1
1



4.7 k  3.3k  1.0k  2.2k 
RT  507
 507 

I1 
 60mA  6.5mA
 4.7k  

 IT

 507 
I2 
 60mA  9.2mA
3.3
k



 507 
I3  
 60mA  30.5mA
1.0
k



 507 
I4 
 60mA  13.8mA
2.2
k



Now use CDR:
 RT
Ix 
 Rx
Verify (KCL): IT  I1  I 2  I 3 +I 4  6.5mA  9.2mA  30.5mA  13.8mA  60mA
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Analysis of Parallel Circuits
• Voltage across all parallel branches is the same as the
source voltage.
• Determine current through each branch using Ohm’s
Law.
• Find the total current using Kirchhoff’s Current Law.
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Example Problem 5
a. Determine all unknown currents and total resistance.
b. Verify KCL for node a.
I
I
6
5
First: understand
your circuit:
I1
270V
I2
5.6kΩ
I4
I3
3.9kΩ
4.3kΩ
2.7kΩ
IS
Now, find an equivalent resistance (RT):
RT 
RT 
Next, find total current (IT):
1
IT =
1
1
1
1

 
R1 R2 R3 R4
1
1
1
1
1



4.7 k  3.3k  1.0k  2.2k 
E
270V
=
 280mA
RT
963.5
Finally, use CDR and find the individual currents (Ix):
 E 
Ix 

 RX 
RT  963.5
 270V 
I1  
  48mA
 5.6k  
 270V 
I3  
  62.8mA
 4.3k  
 270V 
 270V 
I2 
  69.2mA I 4   2.7 k    100mA


 3.9k  
Verify (KCL): IT  I1  I 2  I 3 +I 4  48mA  69.2mA  62.8mA  100mA  280mA
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Power Calculations
• As before, to calculate the power dissipated by each
resistor, use:
VI, I2R, or V2/R.
• Total power consumed is the sum of the individual
powers:
PT = P1 + P2 +…+PN
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Example Problem 6
a)
b)
c)
Solve for indicated currents.
Determine power dissipated by each resistor.
Verify total power = sum of all power dissipated.
Find an equivalent resistance (RT):
RT 
1
1

1 1
1
1
1
1
30V




R1 R2 R3 2k  3k  1.8k 
I1
I2
I3
IS
RT  720
Next, find total current (IT):
IT =
Now, use CDR and find Ix:
 E 
Ix 

 RX 
Finally, find Px:
E
30V
=
 41.67 mA
RT
720
 30V 
I1  
  15mA
 2k  
 30V 
I2 
  10mA
 3k  
 30V 
I3  
  16.67 mA
 1.8k  
2
Px  I x
2
P1  (15mA) * 2k   450mW
2
* RX P 2  (10mA) * 3k   300mW
2
P 3  (16.67 mA) *1.8k   500mW
2
2
P T  I T * RT  (41.7 mA) * (720)  1.25W
P T  P1  P2  P3  450mW  300mW  500mW  1.25W
Verify (KCL): IT  I1  I 2  I 3  15mA  10mA  16.67mA36 41.67mA
QUESTIONS?
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