This document provides an overview of series-parallel circuits. It explains that a series-parallel configuration is one formed from a combination of series and parallel elements. The document outlines the general approach to analyzing such circuits, which involves identifying series and parallel elements, reducing the circuit to its simplest form, and then working back to solve for unknown currents and voltages. It also discusses Kirchhoff's current and voltage laws and provides examples of using these laws to analyze series-parallel circuits and solve for unknown values.

1. ELECTRICAL CIRCUITS

2. Lecture 4
Series-Parallel Circuits

3. The Series-Parallel Network
 A series-parallel configuration is one
that is formed by a combination of
series and parallel elements.
 A complex configuration is one in
which none of the elements are in
series or parallel.

4. The Series-Parallel Network
 Branch
◦ Part of a circuit that can be simplified into
two terminals
 Components between these two
terminals
◦ Resistors, voltage sources, or other
elements
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5. General approach to circuit
analysis
1. Using Reduce and Return Approach:
 Identify elements in series and elements in
parallel.
 Reduce the circuit to its simplest form across
the source
◦ Determine equivalent resistance RT
◦ Solve for the total current and determine the
source current (Is).
◦ Label polarities of voltage drops on all
components
 Return by using the resulting source current
(Is) to work back to the desired unknown.
◦ Calculate how currents and voltages split
between elements in a circuit
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6.  In this circuit
 R2, R3, and R4
are in parallel
 This parallel
combination is in
Series with R1 and
R5
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7.  In this circuit
◦ R3 and R4 are in
parallel
◦ Combination is in
series with R2
 Entire combination
is in parallel with
R1
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8. 8

9. Circuit analysis
2. For Complex Networks:
 Rules for analyzing series and parallel
circuits still apply
 Try to Reduce and Return for some branches
 Same current occurs through all series
elements
 Same voltage occurs across all parallel
elements
 KVL and KCL apply for all circuits
◦ Whether they are series, parallel, or series-
parallel
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10. Hints
 Redraw complicated circuits showing
the source at the left-hand side
 Label all nodes
 Develop a strategy
◦ Best to begin analysis with components
most distant from the source
 Simplify recognizable combinations of
components
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11. Hints
 Verify your answer by taking a
different approach (when feasible)
 Voltages
◦ Use Ohm’s Law or Voltage Divider Rule
 Currents
◦ Use Ohm’s Law or Current Divider Rule
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12. Example 1
12
RT = 3K Ω+ 6KΩ = 9kΩ

13. Example 2
 Find the current I4 and the voltage V2 for
the following network
RT = R8 // (R1 + R2//R3) = 3.4Ω
I4 = 1.5A

14. KCL (Kirchhoff’s Current Law)
 The sum of the currents entering a node
equals the sum of the currents exiting a
node.
I1 = 2A
I2 = 3A
I3 = 5A

15. I1 = ?
I2 = 150 mA
I3 = 250 mA
KCL

16. KVL (Kirchhoff’s Voltage Law)
 The sum of the potential differences
around a closed loop equals zero.

17. Closed Loop #1

18. Closed Loop #2

19. Closed Loop #3

20. Closed Loop #4

21. Not Closed #1

22. Not Closed #2

23. What KVL Really Means
 Sum of the Voltage drops across
resistors equals the Supply Voltage in
a Loop.
 Even not necessary but try to always
choose a loop which contains Voltage
Supply

24. Example
10 KΩ
10 KΩ
10 KΩ

25. Step 1 choose current directions
and loops
10 KΩ 10 KΩ
10 KΩ
L1 L2
L3

26.  KCL in A or B: I1 + I2 = I3
 KVL:
 Loop 1 is given as :
10 = R1 x I1 + R3 x I3 = 10k I1 + 10K
I3
 Loop 2 is given as :
20 = R2 x I2 + R3 x I3 = 10k I2 + 10k
I3
 Loop 3 is given as :
10 - 20 = 10k I1 - 10k I2

27. rearrange
 As I3 is the sum of I1 + I2 we can rewrite the
equations as;
 10 = 10k I1 + 10k (I1 + I2)
10 = 20k I1 + 10k I2
 20 = 10k I2 + 10k (I1 + I2)
20 = 10k I1 + 20k I2
 We now have two "Simultaneous
Equations" that can be reduced to give us
the value of both I1 and I2
 I2 = (20 – 10k I1)/ 20 k
 I2 = 1 mA
 I1 = 0 A

28.  As : I3 = I1 + I2
 The current flowing in resistor R3 is
given as : -1m + 0 = 1 mA
 and the voltage across the resistor R3
is given as : 1mx 10k = 10 volt
 and the voltage across the resistor R1
is given as : 0 x 10k = 0 volt
 and the voltage across the resistor R2
is given as : 1m x 10k = 10 volt

29. Question
 Can we use KCL & KVL in simple
circuits?

30.  Thank you for your attention
 Any question?
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