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I learned today Mesh Analysis and Nodal Analysis principe (with KVL and KCL) for a computer project and I am not certain about one thing.

Can we use nodal analysis with only the voltage of the source and resistances of components in the circuit to know all current information at any point of the circuit? And if it's impossible like I think currently, it's correct that I need to use the Loop current Analysis (who is a general Mesh Analysis if I understand correctly) to solve all circuits, planar and non-planar?

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  • \$\begingroup\$ The voltage of the source and component values are providing the full information for any type of analysis. \$\endgroup\$ – Eugene Sh. Jun 19 '18 at 21:03
  • \$\begingroup\$ So by using nodal analysis and I=U/R i can know all intensities in my circuit ? Because in all example that i saw there was always at least one branch with information about amperage \$\endgroup\$ – Erend Jun 19 '18 at 21:30
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Can we use nodal analysis with only the voltage of the source and resistances of components in the circuit?

Yes. In fact, you may also use it for Mesh Current Analysis as well, the choice is yours. However, one analysis might be easier than another analysis. As long as there is a source of any type at anywhere, you can solve for what you're looking for.

The way I described how to find the current at a node is:

\$i_{x}= (V_{1}-V_{2})/R_{between} \$

Where \$i_x\$ is an arbitrary variable for current at the node, \$V_{1}\$ is the voltage at the branch directly onto the node, \$V_2\$ is the voltage at the branch next to the resistor next to \$V_1\$ and \$R_{between}\$ is a resistor value between the nodes. If there is a voltage source at \$V_1\$ or \$V_2\$, then it will become a numerical value rather than a variable.

Consider the following example.

schematic

simulate this circuit – Schematic created using CircuitLab

Easy right? We'll just solve for one node as you can repeat the process.

\$ \displaystyle i_1=\frac{V_1-V_2}{R_1}=\frac{5-V_2}{100}\$

You can do the same with \$V_2\$ and \$V_3\$.

\$ \displaystyle i_2=\frac{V_2-V_3}{R_2}=\frac{V_2-V_3}{100}\$

You now have a system of equations to solve for \$V_2\$ and \$V_3\$. If you have a resistor between your node and ground, then treat your ground as a separate node with the voltage being zero instead.

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  • \$\begingroup\$ You're welcome! :) Glad I was able to effectively communicate it to you. \$\endgroup\$ – KingDuken Jun 19 '18 at 21:41
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    \$\begingroup\$ As drawn, you just have V2 = V3 = 0 V. \$\endgroup\$ – The Photon Jun 19 '18 at 21:48
  • \$\begingroup\$ I don't fully understood finally .... I tried to practice on an exemple like this so i have as equation : $$ [\frac{10-V_{2}}{100}]-[\frac{V_{2}}{300}]-[\frac{V_{2}-V_{3}}{600}]=0 $$ But i can't resolve this, i maybe missed one subtilty for this analysis \$\endgroup\$ – Erend Jun 20 '18 at 15:49
  • \$\begingroup\$ @Erend Separate your \$i_1\$, \$i_2\$, and \$i_3\$ into separate equations. You're trying to solve everything at once. If you separate them into equations, you can have your program solve for the system of equations. \$\endgroup\$ – KingDuken Jun 20 '18 at 15:59
  • \$\begingroup\$ Ok i found the equations and i fully (and i'm certain) understood, i had trouble with the definition of a node voltage, thanks for helping me! \$\endgroup\$ – Erend Jun 20 '18 at 16:45

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