So I have this node voltage equation problem. Seems really easy, actually.


simulate this circuit – Schematic created using CircuitLab

In fact, this appears to be only one equation, barring constants. $$\frac{1}{110}v_{12}+(2)\frac{20}{10}+\frac{1}{100}v_1=0$$ with the fairly obvious addition of \$v_2=20\$.

This works out to be \$v_1=200\space\mathrm{v}\$.

However, the book says the answer is 219 volts. What is the error in my approach?

  • \$\begingroup\$ Where is ground? \$\endgroup\$
    – mkeith
    Oct 29 '14 at 6:04

Not sure where you made a mistake. It seems like you only have three terms in your sum, but there are four legs. Is that part of the problem? I actually think this is a hard problem, but I graduated from college over 10 years ago. Let me walk you through it.

I am going to sum all the currents going into V2. Just have to calculate them first, and write them with the correct polarity (going INTO V2).

i_1 is -2A by inspection. So 2*(i_1) is -4A. Current in the Voltage source is V1/100 +2A (going up... this itself is a sum) Current in the 110 Ohm is (V1-20)/110 (going to right in 110 Ohm resistor)

So, summing currents into node, -2A -4A +2A + V1/100 + (V1-20)/110 = 0. V1/100 + (V1-20)/110 = 4 V1/100 + V1/110 - 20/110 = 4 V1/100 + V1/110 = 4+20/110 V1 (1/100 + 1/110) = 4 + 20/110 V1 = (4 + 20/110)/(1/100 + 1/110) V1 = 219.05.

Hope that helps.


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