# Superposition method to get the voltage

I want to know the voltage level at node E. Using superposition, I'm having a hard time calculating when 1 V and 2.5 mA go to 0. When 5 V meets 8 kΩ and 8 kΩ and 0.4 kΩ will be in parallel.

I'm confused that using a voltage divider, E should be 2.5 V when 5 V * 8 / (8 + 8) is applied. but what about the other 0.4 kΩ?

I think there is a concept missing and this seems to be a simple question.

• Just convert the current source and the 400 ohm resistor into its thevenin equivalent. Then all you have are three resistances sharing your node of interest, each driven by separate voltages. Commented Mar 23 at 4:53
• @periblepsis Thevenin will be Rtv = 400 ohm and Vth = 1v. but I was confused because just by the left loop. E should 2.5v Commented Mar 23 at 5:01
• You could convert the left side to its single resistor thevenin, too. That works as well. But keep in mind where the resulting thevenin voltage is at with respect to its thevenin resistance and your node of interest. Commented Mar 23 at 5:05

When there is a voltage source between two nodes, it can be convenient to use the concept os supernode. The idea behind it is to consider the section marked in the figure below as a single node, as if the voltage source made a "short circuit" between $$\V_E\$$ and $$\V_F\$$; apply the KCL to that node and finally use the voltage difference between these two points as $$\ 1 \space \mathrm{ V}\$$ (this will be the additional equation needed to solve that system).

Also, here is convenient to convert the $$\5 \space \mathrm{V}\$$ voltage source to a current source:

Applying the KCL on the supernode:

$$0.625 \space \text{m} - \frac{V_E}{\text{4k}} - \frac{V_F}{\text{0.4k}} + \text{2.5 m} = 0 \qquad (1)$$ The second equation is:

$$V_E - V_F = 1 \qquad (2)$$

Solving $$\(1)\$$ and $$\(2)\$$:

$$\begin{cases} V_E = 2.045 \text{V} \\ \\ V_F = 1.045 \text{V} \end{cases}$$

The simulation in the next Figure can be used to check the calculations:

It is necessary to find the effect on the network only one generator at a time while the effect of all the other generators has been neutralized, that is: once the first generator has been chosen, the remaining voltage generators must be replaced with a short circuit, while the remaining current generators, since they do not have to generate more current, they must be replaced with an open circuit:

Node Analysis based on the Network Topology:
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}