Department of Computer Science and Engineering (CSE)
BRAC University
Fall 2023
CSE250 – Circuits and Electronics
SOURCE TRANSFORMATION
PURBAYAN DAS, LECTURER
Department of Computer Science and Engineering (CSE)
BRAC University
Ideal and non-ideal sources
• An ideal voltage source provides a constant voltage irrespective of the current drawn by
the load, while an ideal current source supplies a constant current regardless of the load
voltage.
• Practical voltage and current sources are not ideal, due to their internal resistances or
source resistances 𝑅𝑠 and 𝑅𝑝. They become ideal as 𝑅𝑠 → 0 and 𝑅𝑝 → ∞.
𝑅𝐿 𝑅𝑝
𝑣𝐿 = 𝑅 𝑣𝑠 , if 𝑅𝑠 ≪ 𝑅𝐿 𝑜𝑟 𝑅𝐿 = ∞, 𝑣𝐿 → 𝑣𝑠 𝑖𝐿 = 𝑅 𝑖𝑠 , if 𝑅𝑝 ≫ 𝑅𝐿 𝑜𝑟 𝑅𝑝 → ∞, 𝑖𝐿 → 𝑖𝑠
𝑠 +𝑅𝐿 𝑝 +𝑅𝐿
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 2
Course Outline: broad themes
Source
Transformation
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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Department CSE,BBRACU
RACU 3
Source Transformation
• A source transformation is the process of replacing a voltage source 𝑣𝑠 in series with a
resistor 𝑅 by a current source 𝑖𝑝 in parallel with a resistor 𝑅, or vice versa.
• The transformation does not affect the remaining part of the circuit but greatly
simplifies circuit analysis.
• Note that the arrow of the current source is directed toward the positive terminal of
the voltage source.
• The source transformation is not possible when 𝑅 = 0 and 𝑅 = ∞ (see next slide),
which are the cases with an ideal voltage and current source respectively. However, for
a practical, nonideal voltage source, 𝑅 ≠ 0, and for a practical, nonideal current
source, 𝑅 ≠ ∞.
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
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Das DEPARTMENT OF CSE, BRACU 4
V in series with a R s
• We recall that an equivalent circuit is one whose 𝐼 − 𝑉
characteristics are identical with the original circuit.
Let's see what conditions make the two circuits to have
the same 𝐼 − 𝑉 relations at terminals 𝑎 − 𝑏.
• Let's say we have a configuration of a voltage source
𝐼
(𝑣𝑠 ) in series with a resistor (𝑅𝑠 ) between terminals 𝑎
and 𝑏 . To determine the configuration's 𝐼 − 𝑉
characteristics, if applying a voltage 𝑉 gives rise to a
current 𝐼, we can write,
𝑉 = 𝑣𝑠 + 𝐼𝑅𝑠 𝒗𝒔
𝑉
1 𝑣𝑠
𝐼= 𝑉− 𝒗𝒔
𝑅𝑠 𝑅𝑠 −
𝑹𝒔
• The equation results in a linear 𝐼 vs 𝑉 plot that
𝑣
intersects the axes at 𝑣𝑠 and − 𝑠 .
𝑅𝑠
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
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Das DEPARTMENT OF CSE, BRACU 5
I in parallel with a R
• For the other configuration: a current source (𝑖𝑝 ) in
parallel with a resistor (𝑅𝑝 ) between terminals 𝑎 and p p
𝑏, if applying a voltage 𝑉 gives rise to a current 𝐼,
using KCL the current through the resistor is,
𝐼 + 𝑖𝑝 𝐼
• So, the voltage across the resistor can be written as,
𝐼 + 𝑖𝑝 𝑅𝑝 = 𝑉
𝒊𝒑 𝑹 𝒑
1 𝑉
𝐼= 𝑉 − 𝑖𝑝
𝑅𝑝
−𝒊𝒑
• The equation results in a linear 𝐼 vs 𝑉 plot that
intersects the axes at 𝑖𝑝 𝑅𝑝 and −𝑖𝑝 .
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
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Das DEPARTMENT OF CSE, BRACU 6
𝐼
𝑉 in series with a 𝑅
Conditions for transformation
𝒗𝒔
• The two configurations will be equivalent to each
𝑉
other if their 𝐼 − 𝑉 characteristics are similar. It can be
said by looking at the two plots, they will indeed be 𝒗𝒔
−
similar if the intersecting points are same, that is, if 𝑹𝒔
𝑣
𝑣𝑠 = 𝑖𝑝 𝑅𝑝 and − 𝑅𝑠 = −𝑖𝑝 . This requires 𝑅𝑠 = 𝑅𝑝 =
𝑠
𝑅. Both the equations result in an ohmic relation, 𝐼
𝑣𝑠 𝐼 𝐼 in parallel with a 𝑅
𝑣𝑠 = 𝑖𝑝 𝑅 𝑜𝑟 𝑖𝑝 =
𝑅
• So, if the sources are turned off, the equivalent
resistance at terminals 𝑎 − 𝑏 in both circuits is 𝑅. 𝒊𝒑 𝑹 𝒑
𝑉
−𝒊𝒑
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 7
Example 1
• Use source transformation to reduce the circuit between terminals a and b
shown to a single voltage source in series with a single resistor.
[PDS]07-Mar-24
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Das DEPARTMENT OF CSE, BRACU 8
Example 1: transforming sources
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
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Das DEPARTMENT OF CSE, BRACU 9
Example 1 (contd … 2)
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
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Purbayan ELECTRONICS
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Problem 1
• Use Source Transformation to find the voltage 𝑣0. Find the power developed by the
250 𝑉 source and 8 𝐴 source.
A resistor in series with a current source is redundant, as is a resistor in parallel with a voltage source. We can remove them; this will
have no effect on the circuit except for the sources. Opening a resistor parallel to a voltage source will reduce the current supplied by
the source. Similarly, shorting a resistor in series with a current source increases the voltage across the current source. We have to
keep in mind those facts while calculating parameters for the sources.
Ans: 𝒗𝟎 = 𝟐𝟎 𝑽; 𝑷𝟐𝟓𝟎𝑽 = − 𝟐. 𝟖 𝒌𝑾; 𝑷𝟖𝑨 = −𝟒𝟖𝟎 𝑾
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 11
Problem 2
• Find 𝑖𝑜 in the circuit using Source Transformation.
Ans: 𝒊𝟎 = 𝟏. 𝟕𝟖 𝑨
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 12
Problem 3
• Determine the voltage 𝑣𝑥 across the 10 𝑘Ω resistor by performing a succession
of appropriate Source Transformations.
Ans: 𝒗𝒙 = 𝟒𝟎 𝑽
Fall'22
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 13
Problem 4
• Use Source Transformation to find 𝑉𝑥.
Ans: 𝑽𝒙 = 𝟐. 𝟗𝟖 𝑽
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 14
Problem 5
• Use Source Transformation to find 𝑖𝑥 in the following circuit.
Ans: 𝒊𝒙 = 𝟏. 𝟔 𝑨
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT OF CSE, BRACU 15
Problem 6
• Reduce the circuit to a 2-mesh circuit using Source Transformation. Then
determine 𝑖0 .
Ans: 𝒊𝟎 = 𝟏. 𝟕𝟖 𝑨
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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RACU 16
Problem 7
• Reduce the circuit to a single loop. If 𝑖 = 2.5 𝐴, determine 𝑣0 .
Ans: 𝒗𝟎 = 𝟐𝟖 𝑽
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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Problem 8
• Reduce the circuit to a single loop using Source Transformation. Then
determine 𝐼𝑦 .
Ans: 𝑰𝒚 = 𝟏 𝑨
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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Department CSE,BBRACU
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Problem 9
• Reduce the circuit to a single loop using Source Transformation. Then
determine 𝑉𝑥 .
Ans: 𝑽𝒙 = 𝟒. 𝟔𝟐 𝑽
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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RACU 19
Practice Problems
• Additional recommended practice problems: here
• Other suggested problems from the textbook: here
[PDS] PURBAYAN DAS CSE250 – CIRCUITS AND ELECTRONICS DEPARTMENT CSE,
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RACU 20
Thank you for your attention
[PDS]07-Mar-24
PURBAYAN DAS CSE250 – CIRCUITS
[PDS] AND
Purbayan ELECTRONICS
Das DEPARTMENT CSE,
OFof
Department CSE,BBRACU
RACU 21
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