Problem involving Source Transformation and Nodal Analysis

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The discussion revolves around a homework problem involving source transformation and nodal analysis. The user is attempting to transform a voltage source into a current source but is struggling to calculate the current correctly. After some calculations, they believe the current should be -0.5 A. A response clarifies that the total resistance of three parallel resistors is 1 ohm, leading to a current of -1 A and a voltage of -1 V across them. The user finds the explanation helpful and acknowledges the solution.
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Homework Statement


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Homework Equations

The Attempt at a Solution


I have attempted and completed all questions. However, I am now trying part (b) with transforming the voltage source into a current source. I am unable to work out the current and I am unsure what to do next.

I have included my workings

I believe the answer should be -0.5 A

I appreciate any help. Thanks!

Attempt at part (b):
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Completed Solution:
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In my opinion, all three solutions are correct.
 
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Babadag said:
In my opinion, all three solutions are correct.

Thanks for checking my solution. I really appreciate it.

However, what about my attempt at the source transformation of the voltage source? I haven't completed that part and I am unsure what to do. Any advice?
 
Conor_B said:
Thanks for checking my solution. I really appreciate it.

However, what about my attempt at the source transformation of the voltage source? I haven't completed that part and I am unsure what to do. Any advice?
You have 3 resistors in parallel. Find the current through that equivalent "resistor", which will give you the voltage across that resistance.
 
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upload_2017-11-10_11-59-32.png
Source transformation is o.k.
 

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scottdave said:
You have 3 resistors in parallel. Find the current through that equivalent "resistor", which will give you the voltage across that resistance.

Thanks Scott! That's makes sense
  • The total resistance of the three resistors in parallel is 1 ohm.
  • The resulting current flowing into the resistors is -1 A.
  • The voltage across the three resistors is then -1V
  • And I = -V/2 = -1/2 = -0.5 A
 
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