Blog #38 - Calculating Another Circuit Using Source Transformation

As part of my problem series in this video, I will be using Loop Analysis Using Source Transformation. Once I am finished with this series of problems I will be posting them in my Stan store, at this web address…https://stan.store/GVB

Our task will be to find V0 in this circuit using source transformation.

Recall that when employing source transformation, at a pair of terminals, we can exchange a voltage source V in series with a resistor R subscript v and for a current source, I in parallel with a resistor R subscript I provided that the relationships among these shown parameters exist. 

R subscript V = R subscript I and

I = V/RV…as shown and we can also exchange a current source I in parallel with a resistor R subscript I for a voltage source V in series with a resistor R subscript V. Provided that the relationships among these shown parameters exist…

R subscript V = R subscript I and

V = IR subscript I…as shown

Note that in the original circuit shown here, we have a 12V source in series with a 6kΩ resistor that can be exchanged for a current source in parallel with the resistor. This appears to be a viable exchange since we will then have two current sources in parallel, which we can add algebraically. Performing the exchange yields the above network…

Note that the voltage source was positive at the bottom terminal and therefore the current source points in that direction. 

Replacing the two parallel current sources with one equivalent. Using The Superposition Theorem, we consider the current contribution from each current source…flowing into node A and node B…Node A has 6 minus  2 or 4 A flowing into it  and node B has 6 minus 2 or 4 A flowing out from it… so the two current sources can be replaced by one of 4 A flowing into and out of nodes A & B…shown above..

At this point we can apply current division to obtain a solution for the current in the 4kΩ resistor which is…4/3 mA…then…

V0 = (I4k)(4kΩ) = 16/3 Volts or 5.33 Volts.

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