Transcript Source Transformation to Solve a Circuit

Source Transformation
to Solve a Circuit
Find the power associated with the 6 V source.
State whether the 6 V source is absorbing or
delivering power to the circuit.
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Which method to solve?
Use mesh-current method and solve for the current in the
6 V source. (3 equations)
Use node-voltage method and solve for the voltage across
the 30Ω resistor and calculate the current delivered by the
6 V source.
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Use Source Transformation to “focus in” on
the branch containing the 6V source.
CAUTION --- Preserve the identity of the branch
containing the 6V source!
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Step 1-- Transform the 40 V source
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Step 2 (a) -- combine the 20Ω and 5Ω resistors into
an equivalent 4Ω resistor.
Req = 4Ω
Step 2 (b) -- transform the equivalent 4Ω resistor and
the 8A current source into a 32 V source in series with
a 4Ω resistor.
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Step 3 -- Transform the 32 V source in series with 20Ω
into a 1.6 A source in parallel with a 20 Ω resistor.
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Req = 12Ω
Step 4 (a) -- Combine the 30Ω and 20Ω resistors into an
equivalent 12 Ω resistor.
Step 4(b) -- Transform the equivalent 12 Ω resistor in
parallel with the 1.6 A source into a 19.2 V source in
series with a 12 Ω resistor.
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i
The final circuit has preserved the branch containing
the 6 V source. The current can be determined by
6  i (4)  i(12)  19.2  0
16i  13.2
13.2
i
 0.825 A
16
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i
The actual current is flowing from right to left, into the
+ terminal of the 6 V source. Therefore, the 6 V
source is absorbing power.
p  vi  (6V )(0.825 A)
p  4.95W
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