Transcript Series Parallel Circuits

Lesson 6: Series-Parallel
DC Circuits
1
Learning Objectives
• Apply the rules for analyzing series and parallel circuits to a
series-parallel circuit.
• Compute the total resistance in a series-parallel circuit.
• Analyze series-parallel circuits for current through and
voltage across each component.
• Analyze the power dissipated by each element in a series
parallel circuit and calculate the total circuit power.
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Series - Parallel Circuits
• Topology:
1. Branch:
• Part of a circuit that can be simplified into two terminals (2 nodes).
• A single element such as voltage source or resistor.
2. Node:
• Point of connection between two or more branches.
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Series - Parallel Circuits
•
To analyze a circuit:
•
You need to be able to identify which elements
are in series and which elements are in parallel:
(a)
RT = R1+(R2||R3||R4)+R5
RT = R1||(R2+(R3||R4))
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Rules for Analysis
1. Same current occurs through all series elements.
2. Same voltage occurs across all parallel elements.
3. KVL and KCL apply for all circuits, whether they
are series, parallel, or series-parallel.
4. Redraw complicated circuits showing the source at
the left-hand side.
5. Label all nodes.
6. Solve the problem…
Reduce and Return Approach
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Reduce and Return Approach
1.
2.
3.
4.
5.
Series and parallel elements from (a) must be
combined to establish the reduced circuit of (b).
Then series elements are combined to form the
simplest of configurations in (c). The source
current can now be determined using Ohm’s law,
and we can proceed back through the network as
shown in (d).
The voltage V2 can be determined and then the
original network can be redrawn, as shown in (e).
Since V2 is now known, the voltage divider rule
can be used to find the desired voltage V4.
Because of the similarities between the networks
of (a) and (e), and between (b) and (d), the
networks drawn during the reduction phase are
often used for the return path.
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Hints…
• Develop a strategy:
− Best to begin analysis with components most distant from
the source.
− Simplify recognizable combinations of components.
− Determine equivalent resistance (RT).
− Solve for the total current.
− Label polarities of voltage drops on all components.
− Calculate how currents and voltages split between
elements in a circuit.
− Verify your answer by taking a different approach (when
feasible).
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The Series-Parallel Network
• Given the series-parallel circuit below, how would
you analyze the circuit?
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The Series-Parallel Network
FIRST: Identify elements in series and/or parallel:
•
•
R2 (20Ω), R3 (30Ω), and R4 (8Ω) are in parallel.
This parallel combination is in series with R1 (2Ω) and R5 (6Ω).
R234

 


 

1
1


 4.8

 1  1  1   1  1 1
 R R R   20 30 8 
3
4 
 2
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The Series-Parallel Network
SECOND: Simplify and redraw the circuit from
calculation(s) from step 1.
RT  2  4.8  6  12.8 
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The Series-Parallel Network
• Given the series-parallel circuit below, how would
you analyze the circuit?
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The Series-Parallel Network
Step 1:
• Understand the circuit:
− In this circuit
• R3 and R4 are in parallel.
• Combination is in series with R2.
− Entire combination is in parallel with R1.
 R3 * R4   50*50 
R34  

  25
 R3  R4   50  50 
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The Series-Parallel Network
Step 2:
• Redraw the circuit from step 1.
• Now combine the series elements R34 and R2 for the resultant
R234.
R234  R2  R34  15  25  40
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The Series-Parallel Network
Step 3:
• Redraw the circuit from step 2.
• Now calculate the parallel resistance from elements R1 and
R234 for the resultant RT.


 1 
1
RT 

  8
1
1
1
1
 


R1 R234  10 40 
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Example Problem 1
Determine the Rbc of this network: Rbc = 20+(160||((50||(100+50))+40))
Parallel(
Series (100+50 = 150)
1
1
1
+
50 150
150Ω
= 37.5)
37.5Ω
160Ω
40Ω
Series (37.5+40 = 77.5)
160Ω
52.2Ω
77.5Ω
Parallel (
1
= 52.2)
1
1
+
160 77.5
Series (20+52.2 = 72.2)
15
Rbc = 72.2Ω
Example Problem 2
Determine IT, I1, I2, Vad.
First, simplify the circuit
(deconstruct):
Req
24Ω




1


  16
1
1  1
1 




R2 R3  80 20 
16Ω
12V
1
Substitute the parallel equivalent
resistance for resistors R2 and R3
After the circuit is
simplified, now solve for IT
(from Ohm’s Law):
E  12V 
IT 

  300mA
RT  40 
Now solve for I1 and I2 using CDR:
 16 
 0.3 A 
  60mA
R2
 80 
I1  IT
Req
I 2  IT
Req
 16 
 0.3 A 
  240mA
R3
 20 
Notice the use of Req here. Since we want I1
and I2, we need to figure out the fraction of
IT flowing through the parallel legs.
R1 is not part of the calculations of I1 and I2
except in the sense that it used to verify
through KCL.
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12V
RT
40Ω
Add R’ to R1 to get RT
Now we can solve for Vad (using
Ohm’s Law):
Vad  I1 * R2  60mA *80  4.8V
Vbc  I 2 * R3  240mA * 20  4.8V
You might be wondering what
happened here; why doesn’t Vad = E?
Don’t forget about the voltage drop
that occurs with R1!
VR1  IT * R1  300mA * 24  7.2V
Don’t worry, KVL still holds:
E = V1+Vad=7.2V+4.8V = 12V
Example Problem 3
Determine IT, I1, I2, Vad.
1. Deconstruct the circuit:
a) Realize the R3 and R4
are in series thus an
equivalent resistance of
30Ω can replace these
resistors.
R34 = R3 + R4 = 5Ω+25Ω = 30Ω


b) Now find the parallel


1
1
R

R



  7.5
eq
2||34
resistance of the R34 and
1
1
 1  1 

R2 resistors (AKA Req):
R2 R34  10 30 
c) Complete
deconstruction by adding RT = R1 + R2||34 = 2.5Ω+7.5Ω = 10Ω
the R1 to R2||34
2. Find IT:
3. Reconstruct the circuit:
Again, don’t worry, KVL still holds:
VR1  IT * R1  2.4 A * 2.5  6V
E = V1+Vad=6V+18V = 24V
a) Find I1 and I2 using
CDR:
b) Now find Vad using
Ohm’s Law:
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IT 
E  24V 

  2.4 A
RT  10 
I1  IT
I 2  IT
 7.5 
 2.4 A 
  1.8 A
R2
 10 
Req
Req
 R3  R4 
 7.5 
 2.4 A 
  600mA
 5  25 
Vad  I1 * R2  1.8 A *10  18V
Vbc  I 2 * R34  600mA *30  18V
Solution Steps
•
•
•
•
•
Determine equivalent resistance RT.
Solve for the total current IT.
Label polarities of voltage drops on all components.
Calculate how currents and voltages split between
elements in a circuit.
Verify your answer by taking a different approach
(when feasible).
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Common Mistakes Applying VDR
•
•
•
•
•
•
E in the VDR is the voltage across JUST the series elements.
Va is the voltage ‘left over’ after the voltage drop across R1 (Va = V2= 40V = Vbd).
Rx is the resistor for which you want to determine the voltage drop.
RT refers to the combination of all resistors in the circuit (90Ω).
Req refers to the combination of the resistors that are in series (below this is R3+R4
= 40Ω).
Req’ refers to the combination of the resistors that are in parallel (below that is
(R3+R4)//R2 = 30Ω) that you know the total voltage across.
R 
R
VX  E  X   V1  E  1
 RT 
 RT
+
V2
_

 60 

120
V


  80V
90




 Va  E  V1  120  80V  40V =V2
R
Vbc  V3  Va  3
R
 eq

 10 
  40V 
  10V
40




 R4
Vcd  V4  Va 
R
 eq

 30 
  40V 
  30V
40




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= 40V
Power Calculations
• Again, to calculate the power dissipated by each
resistor, use either:
V I, I2 R, or V2/R
• Total power consumed in a Series-Parallel Circuit is
the sum of the individual powers:
PT = P1+P2+P3+…+PN
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Example Problem 4
Determine the voltage drop across the R4 resistor (Vcd) using the
VDR. Determine power dissipated by each resistor and verify
total power = sum of all power dissipated.
Req  R1  ( R2 || ( R3  R4 ))  60 
IT 
E  120V

Req  90

  1.33 A

1
1
1

120 40
 90
VR1  IT * R1  1.33 A *60  80V
Va  Vb  VT  VR1  120V  80V  40V


R 
R4
30


VX  E  X   Vcd  Vb 
  40V 
  30V
R
R

R
10


30



 T 
4 
 3
PT  E * IT  120V *1.33 A  160W
To confirm total power (PT) calculated, it is the summation of power consumed in the circuit:
PT=P1+P2+P3+…+PN
V32  10V 2 
V12  80V 2 
2
2
P



106.7
W
P


Element power consumed is VI, I R, or V /R


  10W
1
3
R
60
R
10


2
 40V 
V
P2  2  
  13.33W
R2  120 
1
Verify:
PT=P1+P2+P3+P4=106.7+13.33+10+30=160W
2
21


2
 30V 
V
P4  4  
  30W
R4  30 
3
2
Common Mistakes Applying CDR
• REQ refers to the combination of the resistors that are
in parallel that you know the total current through.
• Not using all impedances in the branch.
REQ   20  40  30  20
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 REQ 
I X  IT 

 RX 
 20 
I 2  ITOT 

20

40


 20 
I1  ITOT  
 30 
Example Problem 5
Determine I2 using the CDR. Find Va, Vbc
REQ   20  40  30  20
RT  R1  R e q  R5  10  20  50  80
IT 
E  80V 

  1A
RT  80 
R 
I X  IT  EQ 
 RX 
20


I 2  1A 
  333mA
20


40



R 
VX  E  X   Using a reduced circuit:
 RT 
 R5  Req

50  20


Va  80V 
 80V 

  70V
R R R 
10


20


50



1
eq
5


Vbc  I 2 * R4  333mA * 40  13.32V

Req
Vad  80V 
R R R
eq
5
 1

20


  80V 
  20V
10


20


50




 20 
I1  1A 
  667mA
 30 
VR1  IT * R1  1A *10  10V
VR2  I 2 * R2  667 mA *30  20V
VR3  I 2 * R3  333mA * 20  6.66V
VR4  I 2 * R4  333mA * 40  13.32V
VR5  IT * R5  1A *50  50V
KVL = V1  V2  V5  10V  20V  50V  80V
23
Example Problem 6
Determine IT, I1, Va, Vad.
REQ  ((( 50  150  300)  80)) || 200  100
RT  R25  R e q  R75  25  100  75  200
IT 
E  28V 

  140mA
RT  200 
R 
I X  IT  EQ 
 RX 
 100 
I1  140mA 
  70mA
200



R 
VX  E  X   Using a reduced circuit:
 RT 
 R  R75 
 100  75 
Va  28V  eq
  28V 
  24.5V
R
200




T

 100 
Vad  28V 
  14V
 200 
24
QUESTIONS?
25