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When i am using Superposition Theorem i am getting 80W but when i using Source transformation getting 0W.Why getting different answers?

Is there anything Hidden in source transformation that I need to know.

Circuit Discription

I am sure about applying Superposition Theorem. For Source transformation I am converting both current source in voltage source with resistor in series. After that I am applying KVL that gives me zero current. Am I doing something wrong? I am new in circuit Analysis. Please Guide

Thanks

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  • \$\begingroup\$ Could you elaborate on how you used superposition and did source transformation for this problem? In one of those you got the wrong answer, but we can't tell what you did wrong if we don't know what you did at all. \$\endgroup\$
    – Hearth
    May 10 '18 at 14:22
  • \$\begingroup\$ @Felthry I am sure about applying Superposition Theorem. For Source transformation I am converting both current source in voltage source with resistor in series. After that I am applying KVL that gives me zero current. Am I doing something wrong? \$\endgroup\$
    – Anurag Rag
    May 10 '18 at 14:28
  • \$\begingroup\$ Are you performing source transformation on the resistor you're trying to measure power in? Zero current is correct, but that only applies to the resistors you don't touch in your source transformation, i.e. the 10Ω one. \$\endgroup\$
    – Hearth
    May 10 '18 at 14:30
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    \$\begingroup\$ Please edit your question to show your work. How did you do the source transformation withou eliminating the 5 ohm resistor? \$\endgroup\$
    – The Photon
    May 10 '18 at 14:32
  • \$\begingroup\$ I think you proceeded by finding current in 5 ohm resistor after transforming 5ohm + current source into a voltage source. That 's wrong calculations. \$\endgroup\$ May 10 '18 at 14:33
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If you used source transformation to transform the combination of current source and parallel resistor in the left side and the right side into voltage sources, then you can find current, power and voltage only in the components which are external to the sources.

schematic

simulate this circuit – Schematic created using CircuitLab

Here its only 10 ohms resistor. Calculations on 5 ohm resistor or 4 ohm resistor which are part of the sources, will yield you wrong results.

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  • \$\begingroup\$ Thanks for your answer. I understand your explanation. Suppose I want to calculate power dissipation in 5ohm using Source Transformation only how can I proceed? \$\endgroup\$
    – Anurag Rag
    May 11 '18 at 5:23
  • \$\begingroup\$ Not useful. If 5 ohms is the concern, tranform only the source on right side. Left side should be left alone and use superposition theorem . \$\endgroup\$ May 11 '18 at 5:25
  • \$\begingroup\$ Thanks Leis. Actually I am new in circuit analysis and I m making lot of error while solving problems.I know I need lot of practice. Can you please suggest me a good book or website having lot of problems. \$\endgroup\$
    – Anurag Rag
    May 11 '18 at 5:30
  • \$\begingroup\$ I use one book by Theraja and schaum's outline series. \$\endgroup\$ May 11 '18 at 5:32
  • \$\begingroup\$ eng.harran.edu.tr/~msuzer/files/edt/edt.pdf : Good one \$\endgroup\$ May 11 '18 at 5:34
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If you take a step back and ask yourself what the voltages will be on the 2 top nodes if the 10 k resistor is removed, you will see that on both nodes, the voltage is 20 volts. If you added any value resistor in place of the 10 k clearly there is no voltage across it so zero current flows through it hence the 5 ohm resistor has 20 volts across it and dissipates 80 watts.

It's just a sanity check.

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  • \$\begingroup\$ Now I have a bit understanding of source Transformation. Thanks @Andy for your Answer. \$\endgroup\$
    – Anurag Rag
    May 10 '18 at 23:58

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