I am currently attempting this question and I feel like I am over complicating it and previously done questions much more complex than this. Is there any other method to gain the current through each resistor Than using Kirchhoff current law ( loop analysis)?enter image description here

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    \$\begingroup\$ Did you ever heard about superposition ? \$\endgroup\$ – G36 Apr 17 '17 at 18:24
  • \$\begingroup\$ No not in relation to this type of circuit? \$\endgroup\$ – RJRJSanders Apr 17 '17 at 18:27
  • \$\begingroup\$ For each voltage or current source in turn, short-circuit all other voltage sources, open-circuit all other current sources, and solve for that single source configuration. Then add all the solutions together. \$\endgroup\$ – Neil_UK Apr 17 '17 at 18:28
  • \$\begingroup\$ 1. Loop analysis uses KVL, not KCL. 2. You should be able to solve this problem just by writing and solving the two equations for loop analysis. \$\endgroup\$ – The Photon Apr 17 '17 at 18:29
  • \$\begingroup\$ @Neamus here you have an example electronics.stackexchange.com/questions/299596/… \$\endgroup\$ – G36 Apr 17 '17 at 18:31

Is there any other method to gain the current through each resistor Than using Kirchhoff current law ( loop analysis)?

First, loop analysis uses KVL, not KCL, so it's not clear which method you think is the "default" way to solve the problem.

Some ways to solve this problem include:

  1. Use modified nodal analysis. Since there are 4 different nets you might choose as the ground net, there's 4 different sets of equations you might get for this method.

  2. Use mesh analysis

  3. Just write the KCL equation at the "T" node (calling the "bottom" node ground, and knowing by inspection the voltage at the other two nodes). This is, of course, just MNA but short-circuiting a few steps that are obvious by inspection.

  4. Use parallel/series combinations to find the current due to each of the two sources individually, and use superposition to find the total current.


Find the voltage at the tee using superposition formula where || = Parallel resistance.

\$V_{tee} = 12 * \frac{2||6}{4 + 2||6} + 6 * \frac{4||6}{2+4||6} = 6.54545V\$

Then use ohms law to calulcate the currents from there.

Full description of that formula is G36's answer here.


If all you know is how to compute Thevenin equivalents for voltage divider pairs, if and only if one end is already grounded, then you could follow this recipe:


simulate this circuit – Schematic created using CircuitLab

The first step is to pick a convenient node and to assign it to "ground" as \$0\:\textrm{V}\$. The next step is to continue getting rid of some of the clutter in the schematic, so dumped the voltage sources and just replaced the node end points with voltage values, instead. Nothing has changed at this point, except to simplify what you are staring at visually and to remove unnecessary distractions from the schematic.

At this point, you need to start applying the voltage divider rules. Just break \$R_2\$ from the node for a moment and compute the Thevenin equivalent voltage and resistance for the \$R_1\$ and \$R_3\$ pair, replacing the pair with the new voltage and resistance. When that is done, you now have things made a lot simpler but now you have two node voltages neither of which are grounded. If you already know how to compute the value for \$V_X\$ directly, then you could stop at this point and compute:

$$V_X=\frac{7.2\:\textrm{V}\:\cdot\:2\:\Omega+6\:\textrm{V}\:\cdot\:2.4\:\Omega}{2\:\Omega+2.4\:\Omega}\approx 6.545\:\textrm{V}$$

And you'd be done.

But I went a little further here, in case it helps you, by then subtracting \$6\:\textrm{V}\$ from the two nodes, so that one of them becomes "grounded" again, then computed the voltage divider voltage for that, and then finally added back the \$6\:\textrm{V}\$ I'd previously subtracted. I get the same result this way, as well. Depending on your mental preferences, one may be easier than the other to remember for now.

If you learn nodal analysis, then you would write out the following single equation from merely glancing at \$V_X\$:

$$\begin{align*} \frac{V_X}{4\:\Omega}+\frac{V_X}{6\:\Omega}+\frac{V_X}{2\:\Omega} &= \frac{12\:\textrm{V}}{4\:\Omega}+\frac{0\:\textrm{V}}{6\:\Omega}+\frac{6\:\textrm{V}}{2\:\Omega}\\\\ &\therefore\\\\ V_X &=\left[\frac{12\:\textrm{V}}{4\:\Omega}+\frac{6\:\textrm{V}}{2\:\Omega}\right]\cdot\bigg[4\:\Omega\:\vert\vert\: 6\:\Omega\: \vert\vert\: 2\:\Omega\bigg]\\\\ &=\left[3\:\textrm{A}+3\:\textrm{A}\right]\cdot\bigg[2.4\:\Omega\: \vert\vert\: 2\:\Omega\bigg]\\\\ &= 6\:\textrm{A}\:\cdot\: 1\frac{1}{11}\:\Omega \approx 6.545\:\textrm{V} \end{align*}$$

Solved as shown above.


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