I am having a problem with calculating the amount of current flowing through a branch, after deriving potential of its nodes using the Node Voltage Method.

I understand how to calculate voltage drop on a resistor and the voltage of an ideal voltage source as well:

enter image description here

The branch I am having problems with is sketched below:

enter image description here

The current I2 should be equal to $$I_2 = \frac{U_{02}}{R_2} = \frac{(E_2-V_2)}{R_2}$$ but I do not understand why this would be the case.

In the expression $$I_2 = \frac{U_{02}}{R_2}$$ I tried to calculate the voltage U02 the way I thought was correct:

$$U_{02} = U_{0A} + U_{A2} = -E_2 + I_2R_2$$ substituting in the previous equation:
$$I_2 =\frac{(-E_2 + I_2R_2)}{R_2}$$

If we multiply this with R2 we will get:
$$I_2R_2 = -E_2 + I_2R_2$$
$$-E_2 = 0$$ ???

and if we don't substitute anything, but rather continue with the expression for U02 we get:

$$U_{02} = U_{0A} + U_{A2} = -E_2 + I_2R_2$$
$$I_2 = \frac{U_{02}+E_2}{R_2}$$, which would bring us to the suggested solution if it was true that:

$$U_{02} = V_0 - V_2 = -V_2$$

which shouldn't be the case, since the voltage between 0 and 2 isn't determined only by potential difference between the nodes, but also the voltage generator that contributes as well.

I do not understand this at all and I feel like everything I've learned so far is not valid anymore. Things are literally making no sense. Where did I go wrong?

EDIT: To rephrase and recap: How to calculate the current I2 (second sketch) if values R2, E2 and V2 are known (V0 = 0) ?

  • \$\begingroup\$ What do "U02", "V2", "U0A", "UA2", etc., mean? \$\endgroup\$
    – The Photon
    Commented Jan 8, 2016 at 21:24
  • \$\begingroup\$ With U02, UA2 etc. I denoted the voltage between points 0 and 2, A and 2 respectively, that is, U02 = V0 - V2. V0 = 0 since that is the node of reference (in Node Potential Method). \$\endgroup\$
    – 0lt
    Commented Jan 8, 2016 at 21:27
  • \$\begingroup\$ Non-Ideal voltage sources are limited in their output. To properly measure this, simulations must be done with a resistor after the source. This simulates the ESR. To answer your question, the calculations are the same as seen by a simulator, but you must be aware of the extra ESR inherent a non-ideal source. \$\endgroup\$
    – mcmiln
    Commented Jan 8, 2016 at 21:30

1 Answer 1


The current I2 should be equal to I2 = U02/R2 = (E2-V2)/R2, but I do not understand why this would be the case.

This is not correct.

The current through the resistor is determined by the voltage across the resistor. Since the resistor is not connected to node 0, then the voltage at node 0 has no role in determining the current through the resistor.

If I understand your notation, you should have

$$I_2 = \frac{V_A-V_2}{R_2} = \frac{U_{A2}}{R_2}$$

  • \$\begingroup\$ current I2 is equal to voltage UA2? \$\endgroup\$
    – 0lt
    Commented Jan 8, 2016 at 21:34
  • \$\begingroup\$ @banananina, sorry, fixed. \$\endgroup\$
    – The Photon
    Commented Jan 8, 2016 at 21:43
  • \$\begingroup\$ UA2 = VA - V2 , VA can be deduced from E2 , E2 = VA - V0 = VA . Finally, UA2 = E2 - V2 , which yields the suggested solution. This makes sense. I don't know if things would be this easy If we had 2 or more resistors in series with the voltage source. Nonetheless, thank you for your input. \$\endgroup\$
    – 0lt
    Commented Jan 8, 2016 at 21:53
  • \$\begingroup\$ @banananina, OK, but how did you get I2 = U02/R2? \$\endgroup\$
    – The Photon
    Commented Jan 8, 2016 at 22:19
  • \$\begingroup\$ UA2 = E2 - V2, so we get to the proposed solution, but that is not equal to U02... at least your solution is correct. \$\endgroup\$
    – 0lt
    Commented Jan 8, 2016 at 22:21

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