Electronic – Nodal analysis: how can I represent the current i2 in terms of voltage over R2

Your logic is impeccable! :-)
You just need to trust the answer and see where it leads you..

3 Choices

The less you take from me nd the more you do yourself the more you will learn and the happier you will be afterwards. So, I have given you a progressive answer. Use as little as possible.
Check answer once youv'e got it correct.

• (1) Single clue
  (2) Progressive clues
  (3) Answer

(1) You can take it from me that you are correct and that

• For

  3 amps x 'X' volts to equal zero

  then X must equal zero.
  So V1 must equal zero

And go away and work it out.

OR

(2) Try these few progressive clues:

Stop after each clue and think.

• If V1 = 0 then what is the current from the 18V supply?

Stop reading. Think about it.

Then clue 2 - stop reading now to avoid clue 2. .
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• Clue 2:
  The current from the 18V supply and the 3A current both flow into Vi.
  Where do they go to?

Stop reading. Think about it.

Then clue 3 - stop reading now to avoid clue 3. .
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• Clue 3
  For the currents in clue 2 to go where they must go, to what is the necessary voltage of Vs?

Stop reading. Think about it.

OR - answer below - Then clue 2 - stop reading now to avoid answer. .
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(3) ANSWER:

• For power in current source to be zero, voltage across source must be zero. Call voltage at top of current source Vi.
  So Vi = 0

• If Vi=0 then 18V supply sees a load of 3+2 = 5 ohms.
  Current from 18V supply is i=V/R = 18/5 = 3.6A.

• Node at Vi has current from current source + current from V18 flowing into it.
  So Current in = 3A + 3.6A = 6.6A.

  • Current into Vi MUST go somewhere.
    Only path is via Vs.
    Voltage across V2 + 8R + 6R = 0

  • For 6.6A to flow in 6+8 = 14 ohms via Vs the voltage must be V = IxR = 6.6 x 14 = 92.4V.

  Required polarity to "suck' current via 14 ohms as shown is -92.4V

  • QED.

You can solve this circuit by more or less the same method you've given in the question; however you need to plug in one more equation (\$V_1=V_2+12\$) into the system and introduce an unknown current variable. So I'm not sure if we can call it a 'pure' nodal analysis.

This is what you've got to do:

• Write KCL on the left node (\$I_s\$ is the current through the voltage source): $$6A=\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } +\frac { { V }_{ 1 } }{ { R }_{ 3 } } +{ I }_{ s }$$

• Do it again on the node on the right side: $$4A={ I }_{ s }-\frac { { V }_{ 2 } }{ { R }_{ 2 } } +\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } $$

• So far, we're in line with the method described in the question. As a last step, write down this one: $$V_1=V_2+12$$

Now we're left with 3 equations in three unknowns, which you can easily solve to get \$V_1, V_2\$ and \$I_s\$