# Nodal and Mesh analysis

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:

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:

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.

• I get the same as your mesh $I_2$ formula using nodal analysis. Commented Feb 19, 2022 at 16:09
• So, if my nodal analysis agrees with your mesh analysis I would say that your K analysis must also have an error somewhere. Commented Feb 19, 2022 at 16:16
• @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? Commented Feb 19, 2022 at 16:24

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

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

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

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)?

• 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. Commented Feb 19, 2022 at 16:59
• @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. Commented Feb 19, 2022 at 17:01
• equation number 3 Commented Feb 19, 2022 at 17:05
• Cheers got it!! and eq 4 fixed too. Commented Feb 19, 2022 at 17:05

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).

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