simulate this circuit – Schematic created using CircuitLab

The question for the above circuit asks for the power developed by the 60 Volt power supply. The equations I've written to solve for this value are as follows:

  1. At the super node consisting of V1/V3: (V1-60)/100 + (V1-60)/10 + (V3-V2)/20 + V3/400 -0.625Va = 0

  2. At node V2: (V2-60)/5 - V2/200 + (V2-V3)/20 = 0

  3. KVL around the loop consisting of V1, 175Iq and V3 V1 + 175Iq = V3

Conditional equations: Va = V2-60, Iq = V2/200

The correct answer is given as 1084 watts but so far no matter how I manipulate the above equations the best I can come up with is 1057 watts and that's by making some assumptions that don't agree at all with the polarities given in the schematic. Note: when I see an arrow pointing in the direction of current through a resistor, in this case Iq through R2, I treat it as a voltage drop with the arrow pointing in the direction of the drop...it that correct? (I'm basing all my calc's on conventional current flow.)

As a final note, this is a circuit question out of the 5th edition of 'Electric Circuits' by Nilsson and Riedel. I took a course some years ago (in 1998) that used this text book so I decided I'd see how much I've forgotten since then. Apparently it's quite a bit.


I got P=1084.5 Watt. These are the equations:

1) At supernode, (V1-60)/10 + V1/100 + (V3-V2)/20 + V3/400 - 0.625Va = 0 where Va = V2 - 60

2) For supernode, V3-V1 = 175 Iq where Iq = -V2/200

3) At node V2, (V2-60)/5 + V2/200 + (V2-V3)/20 = 0

Note: Iq flows in opposite direction of voltage drop, so the (-) sign. Also it should be V1/100 (not V1-60/100) at supernode equation.

Solving these, V1 = -60.75, V2 = 30, V3 = -87

Total Current by 60V supply: (60-V2)/5 + (60-V1)10 = 18.075

Power = 60 * 18.075 = 1084.5 Watt

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  • \$\begingroup\$ My thanks to Hassansin. My original calcs included the value 'V1/100'; just forgot to include it in the posted equation for the supernode. My error was my interpretation of sign assignments for terms associated with Iq. \$\endgroup\$ – JRE Jun 11 '14 at 16:49

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