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I need to find the currents on each resistor of this circuit:

Where \$\beta\$ is a constant with units of resistance. Applying Kirchhoff's laws the equations are: $$ I_1+I_3-I_2 = 0 $$ $$ V_0-I_1R_1-I_2R_2 = 0 $$ $$ \beta I_1-I_3R_3-I_2R_2 = 0 $$ Obtaining $$ I_1 = -\frac{(R_2+R_3)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$ $$ I_2 = -\frac{(\beta+R_3)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$ $$ I_3 = \frac{(R_2-\beta)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$

I want to solve it using also Nodal and Mesh Analysis, using Mesh analysis with the mesh currents as shown:

enter image description here

Applying Kirchhoff voltage law on each loop, I get: $$ V_0-i_1R_1-R_2(i_1-i_2)=0 $$ $$ R_2(i_1-i_2)-i_2R_3-\beta i_1=0 $$ Obtainig $$ I_1 = \frac{(R_2+R_3)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$ $$ I_2 = \frac{(\beta+R_3)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$ $$ I_3 = -\frac{(R_2-\beta)V_0}{R_2(\beta+R_1)+R_3(R_1+R_2)} $$ Certainly, there are some annoying signs. For Nodal analysis using the three nodes as shown: enter image description here

Applying Kirchhoff's current law at node \$V_2\$ gives: $$ \frac{V_0-V_2}{R_1}-\frac{V_2}{R_2}+\frac{V_3-V_2}{R_3}=0 $$ $$ V_3 = \beta I_1 $$ Which gives, for \$I_2\$: $$ I_2=-\frac{(\beta+R_3)V_0}{R_2(\beta+R_1)-R_3(R_2-R_1)} $$ At this point, I know there is something wrong. I have been struggling with this to find that the answers are the same. Maybe I'm missing something, any help is appreciated. Thanks.

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  • \$\begingroup\$ I get the same as your mesh \$I_2\$ formula using nodal analysis. \$\endgroup\$ Commented Feb 19, 2022 at 16:09
  • \$\begingroup\$ So, if my nodal analysis agrees with your mesh analysis I would say that your K analysis must also have an error somewhere. \$\endgroup\$ Commented Feb 19, 2022 at 16:16
  • \$\begingroup\$ @Andyaka I double-check my K analysis, and yes, there was a minus sign wrong in the calculation, so now it gives me the same as the mesh analysis. Now, I don't know what is wrong with the nodal analysis, Can I see your nodal analysis? \$\endgroup\$ Commented Feb 19, 2022 at 16:24

2 Answers 2

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Now, I don't know what is wrong with the nodal analysis, Can I see your nodal analysis?

enter image description here

$$\dfrac{V_0-V_2}{R_1} + \dfrac{\beta I_1 - V_2}{R_3} = \dfrac{V_2}{R_2}\tag{1}$$

$$\dfrac{V_0-V_2}{R_1} + \dfrac{\beta \left(\frac{V_0-V_2}{R_1}\right) - V_2}{R_3} = \dfrac{V_2}{R_2}\tag{2}$$

Collecting terms...

$$V_0\cdot \left(\dfrac{1}{R_1}+\dfrac{\beta}{R_1R_3}\right) = V_2\cdot \left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{\beta}{R_1R_3}\right)\tag{3}$$ $$$$

Equating to \$V_0\$ and dividing by \$R_2\$ to get \$I_2\$...

$$I_2 = \dfrac{V_0}{R_1}\cdot \dfrac{1 + \dfrac{\beta}{R_3}}{\dfrac{R_2}{R_1}+\dfrac{R_2}{R_2}+\dfrac{R_2}{R_3}+\dfrac{\beta R_2}{R_1R_3}}\tag{4}$$ $$$$

$$I_2 = \dfrac{V_0(\beta + R_3)}{R_2R_3 + R_1R_3 +R_1R_2 + \beta R_2}\tag{5}$$

Can you take it from here now (just a few trivial steps to the right answer)?

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  • \$\begingroup\$ I think you are missing \$R_3^{-1}\$ inside the parentheses of the right-hand side in the second equation. But yes, I found the same answer, thanks a lot. \$\endgroup\$ Commented Feb 19, 2022 at 16:59
  • \$\begingroup\$ @Sjang oops I added another equation line to make it clearer so I'm not sure which you mean - I've added numbers now so maybe you can repeat. \$\endgroup\$ Commented Feb 19, 2022 at 17:01
  • \$\begingroup\$ equation number 3 \$\endgroup\$ Commented Feb 19, 2022 at 17:05
  • \$\begingroup\$ Cheers got it!! and eq 4 fixed too. \$\endgroup\$ Commented Feb 19, 2022 at 17:05
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There might be other things wrong, but as a quick sanity check, I1 + I3 should equal I2. But your I1 + I3 (modulo V0/common denominator) is 2R2 + R3 - B, while I2 is just B + R3. If you switch the sign of I3 though, it works out (though it doesn't prove it's the right answer).

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  • \$\begingroup\$ I typed my answers incorrectly, indeed the firsts \$I_2\$ have opposite signs. Now is corrected, thanks. But yes, it is still wrong. \$\endgroup\$ Commented Feb 19, 2022 at 14:44

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