I'm trying to figure out the Thevenine resistance of this circuit using loops. Circuit

Perhaps loops is the wrong way to go? I've made the circuit using a breadboard and found out that the answer should be \$\frac{3R}{5}\$. Using loops I get that

\$V_S-IR-i_1R=0\$ (1)

\$i_1R-2i_2R=0 \Leftrightarrow i_1=2i_2\$ (2)

And from looking at the current, I can see that \$I=i_1+i_2=2i_2+i_2=3i_2\$

Then I rewrite the first equation as

\$V_S=(i_1+i_2)R+i_1R=2i_1R+i_2R \Leftrightarrow 2i_1+i_2=\frac{V_S}{R}=2(2i_2)+i_2=5i_2\$

Solving for \$i_2\$, I get \$i_2=\frac{V_S}{5R}\$, and which gives me that

\$I=3i_2=\frac{3V_S}{5R}\$. Then using Ohms law, I get that \$R_{eq}=\frac{5R}{3}\$, which isn't correct.

Where do I go wrong? Is this an example where loops can't be used to determine the equivalent resistance?

Thanks in advance

  • \$\begingroup\$ On a test, you might be asked to use a specific method, rather than your method-of-choice. In this case, if you only need \$ R_T \$, then it might be wise to use Thevenin method of shorting \$ V_s \$ and solve the resistor network. Otherwise, a hard-marker may give you grief. \$\endgroup\$ – glen_geek Oct 20 '18 at 14:47
  • \$\begingroup\$ Turn the source to zero and you have a simple series/parallel connection of resistors. Easy. \$\endgroup\$ – Chu Oct 20 '18 at 23:23
  • \$\begingroup\$ Thank you. This is the method that our teacher has tought us. But there are some cases where I'm not sure which to add first and the result ends up incorrect. Therefore I would like to have a more "reliable" way of calculating. I guess the way to add parallells and series are reliable as well, I just need to practice it more. \$\endgroup\$ – Derek K Oct 21 '18 at 10:17

Everything is correct and your loops method work.

The current \$I = \frac{3 V_S}{5 R}\$ therefore the \$R_{Th}\$ resistane is:

$$R_{Th} = \frac{V_S}{I} = \frac{V_S}{\frac{3 V_S}{5 R}} = \frac{V_S}{1} \cdot \frac{5R}{3V_S} = \frac{5R}{3} $$

Ans this is the resistance seen from the input voltage source point of view

enter image description here

And this is what you have found because you solve for the \$I\$ current (drawn from the voltage source).

But we can also find the resistance seen between \$A\$ and \$B\$ terminals:

enter image description here

\$R_{th} = (0.5R+R)||R = 1.5R||R = 0.6R\$

But this resistance can be found as I have shown here:

Why is the voltage of a capacitor equal to the voltage of a battery connected it?

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  • \$\begingroup\$ Thank you, but trying this out on the breadboard and on a simulator shows that that is incorrect. The correct answer should be \$\frac{3R}{5}\$ \$\endgroup\$ – Derek K Oct 21 '18 at 10:18
  • \$\begingroup\$ You have to distinguish on which resistance you are talking about? See my update answer. \$\endgroup\$ – G36 Oct 21 '18 at 10:40

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