Problem with equivalent circuits

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#1 21672502 06 Oct 2014 21:07

Kyosuke Yasashi Kyosuke Yasashi

Anonymous

Post #1
21672502 06 Oct 2014 21:07

Hi guys,

I have a problem understanding how did the 1st circuit transformed into its equivalent. This is indicated as superposition problem so I don't think Thevenin and Norton theorems has to do anything here.

The value of the current source in the problem is 1A.

Thanks,

GT
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#2 21672503 08 Oct 2014 23:25

Neil Mula Neil Mula

Anonymous

Post #2
21672503 08 Oct 2014 23:25

Hi Kyosuke,

I think you are dealing a source transformation problem.

from a current source transformed to a voltage source by simplifying [6Ω||3Ω] + 2Ω that will give you 4Ω multiplied to 1A equals 4V. Rearrange the component with 4V source parallels become series vice versa and external at 2Ω which is Vo2, then use voltage divider to solve for Vo2 given 4V source.

Hope it helps.

Best,
Neil
#3 21672504 09 Oct 2014 18:19

John Schmitz John Schmitz

Anonymous

Post #3
21672504 09 Oct 2014 18:19

Another way to look at it is by taking the current source and 4 ohm resistor to ground, convert that to a voltage source with series resistance on the output and the you have the circuit on the right.
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