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simulate this circuit – Schematic created using CircuitLab


marked as duplicate by Eugene Sh., Enric Blanco, Andy aka, CL., Dmitry Grigoryev May 16 '17 at 15:34

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  • 3
    \$\begingroup\$ I am pretty sure I've seen an exact same circuit in a question few days ago... \$\endgroup\$ – Eugene Sh. May 15 '17 at 17:36
  • \$\begingroup\$ Use en.wikipedia.org/wiki/Y-%CE%94_transform \$\endgroup\$ – Anonymous May 15 '17 at 17:41
  • 1
    \$\begingroup\$ @EugeneSh. ...and an exact same one a few days before that too. This comes up every few days. Should we make a canonical question? Or should we flush them all out thus making students think more? \$\endgroup\$ – Nick Alexeev May 15 '17 at 17:41
  • \$\begingroup\$ I'd use an anonymous name too with this question. \$\endgroup\$ – Sunnyskyguy EE75 May 15 '17 at 18:05
  • \$\begingroup\$ I've seen this within the past two weeks. \$\endgroup\$ – StainlessSteelRat May 15 '17 at 18:06

The general answer is: 'Use Kirchhoff's circuit laws to find the currents in the circuit and then calculate the equivalent resistance from the appropriate V/I relationship.'

More specifically, I would set the currents as follows: circuit with generic loops added

Write KVL equations for each of the three loops (as defined by the currents),

Note that some elements will have compound currents eg $$I_{R3}=(I_1-I_3)$$ Solve for each of the currents, and then find the equivalent resistance by using: $$R_{eq}=\frac{V_1}{I_1}$$

An example of this process can be found here.


Thevenin is often taught before mesh, branch, or nodal analysis methods. You can use Thevenin here with some ease. Just follow the steps indicated by the blue arrows below:


simulate this circuit – Schematic created using CircuitLab

  1. Disconnect \$R_5\$, temporarily.
  2. Solve the remaining left and right sides using Thevenin.
  3. Reconnect \$R_5\$ and solve for the series current, all of which must flow through \$R_5\$.
  4. Add this known current to the original schematic.

Now there is added information. You know that there is \$400\:\textrm{mA}\$ of current in \$R_5\$ and you know its direction. So:

$$\frac{V_1-V_X}{R_1}-400\:\textrm{mA}=\frac{V_X}{R_2},\quad\quad V_Y=V_X-400\:\textrm{mA}\cdot R_5$$


$$V_X=\left(V_1-400\:\textrm{mA}\cdot R_1\right)\cdot\frac{R_2}{R_1+R_2}$$

From there you can figure out the pair of currents through \$R_2\$ and \$R_4\$ (or, alternatively, the currents through \$R_1\$ and \$R_3\$), the sum of which is the total current for \$V_1\$. Dividing the voltage of \$V_1\$ by this total current provides the resistance.

For example, just "eyeball" the above equation for \$V_X\$. You can easily see that the first factor is just \$16\:\textrm{V}\$ and that the second factor is just \$\frac{3}{4}\$. So \$V_X=12\:\textrm{V}\$ is very quickly observed without even bothering with a calculator, at all. \$V_Y\$ is now obvious, as well. And the currents are similarly now quite trivial to compute and add up. From there it is just a matter of dividing that sum into the original source voltage and you have the equivalent resistance.

This approach uses only Thevenin and some quite modest algebra and doesn't rely upon techniques or tools used to solve simultaneous linear equations. So it can be all done on a cheap calculator (The US$1 variety) or even approximated quickly in your head (as I just demonstrated.)


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