Source transformations
Consider the two circuits below. In particular, look at the current and
voltage of RL in each circuit. Using any of the techniques we seen so far,
it is easy to find IRL and VRL for each case.
RA = 2kΩ
IRL
VS = 5V
RL
=1kΩ
+
vRL
–
VRL = 1.667 V and IRL = 1.667 A
IS = 2.5A
IRL
RB
= 2kΩ
RL
=1kΩ
+
vRL
–
VRL = 1.667 V and IRL = 1.667 A
Interesting: From the point of view of the resistor RL, the series combination of
the voltage source and resistor RA gives the exact same result as the parallel
combination of current source and resistor RB.
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source transformations – 1
The series combination seems to behave identically to the parallel
combination. This suggests that we may, in the right circumstances,
replace one configuration for the other. Making this switch is known as a
source transformation.
RS
IS = VS/RS
IS
VS
RP
RP = RS
RS
IS
RP
VS = ISRP
VS
RS = RP
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source transformations – 2
These are perfectly viable substitutions. From the point of view of
whatever circuitry is attached to the two terminals, the result will be
exactly the same for the two source configurations.
If you are trying calculate something about the two components (V-RS or
I-RP), you cannot transform them. In transforming them, you lose the
ability to calculate a specific property.
This is an example of bigger, more important idea known as the Thevenin
equivalent of circuit. We will introduce this later, and make extensive use
of it in discussing amplifiers, etc.
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source transformations – 3
Example 1
Find iR4 in the circuit below
R1
+
VS –
50 V
25 Ω
R2
50 Ω
iR4
R3
R4
100 Ω
200 Ω
Use a source transformation to put everything in parallel.
IS
= VS/R1
=2A
R1
R2
25 Ω
50 Ω
R3
R4
100 Ω
200 Ω
Then use a current divider:
IR4 =
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iR4
1
25⌦
+
1
100⌦
1
50⌦
+
1
200⌦
+
1
100⌦
(2A) = 0.2667 A
source transformations – 4
Example 2
Find iR2 in the circuit below.
R1
VS +
–
20 V
5 kΩ
R2
iR2
7.5 kΩ
IS
2 mA
Transform the voltage source / resistor combo
IST
4 mA
R1
5 kΩ
R2
7.5 kΩ
iR2
IS
2 mA
Net current divides
between the two resistors.
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Writing a KCL equation shows
that there is a net current of IST IS = 2 mA flowing into the
parallel resistor combination.
=
=
+
(
.
+
)
(
.
)= .
source transformations – 5
Example 3
(An incorrect application)
R1 1 kΩ
VS +
–
iR1
Find iR1 in the circuit at left.
R2
1 kΩ
40 V
IS
15 mA
The previous example worked nicely so use the same method. Transform
VS & R1, and use current divider with total current.
IST
40 mA
=
R1 iR1 R
2
1 kΩ
IS
1 kΩ
15 mA
=
+
+
(
+ )
(
)=
.
It seems nice, but it is wrong because you cannot transform the component
for which you are trying to find voltage or current.
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source transformations – 6
Example 3
(Redo it correctly.)
R1 1 kΩ
iR1
VS +
–
Find iR1 in the circuit at left.
R2
IS
1 kΩ
40 V
15 mA
Transform IS & R2.
R1 1 kΩ
VS +
–
40 V
iR1
R2 1 kΩ
+ V
ST
–
15 V
This is the correct answer.
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Writing a KVL loop equation and
solving for iR1 gives
=
=
+
+
=
.
source transformations – 7
Example 4
20 Ω
R1
VS1 +
–
120 V
1.6 Ω
VS2
R2
– 60 V
+
IS
36 A
R4
R3
6Ω
+
8Ω
R5 vR5
Find vR5 in the
circuit at left.
–
5Ω
Transform two voltage sources to current sources. (Pay attention to polarity.)
IST1
6A
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20 Ω
R1 IST2
12 A
5Ω
R2 IS
36 A
1.6 Ω
R4
R3
6Ω
8Ω
+
R5 vR5
–
source transformations – 8
Example 4 (cont.)
Add the parallel current sources into one. Combine the parallel resistors into one.
1.6 Ω
Req
Ieq
R4
2.4 Ω
30 A
8Ω
+
R5 vR5
–
Ieq = 6A – 12 A + 36 A = 30 A.
Req = 20 Ω || 5 Ω || 6 Ω = 2.4 Ω.
Use voltage divider:
Transform Ieq & Req:
2.4 Ω
Veqt
72 V
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+
–
Req
1.6 Ω
=
+
+
R4
8Ω
R5 vR5
–
=
.
+
+ .
+
(
=
source transformations – 9
)