Can't find error in nodal analysis

Question

For the following circuit I did node analysis in order to find the greatest possible power at R and ended up with the equations (using node analysis):

$$n_1:\frac{v_1-V}{R_1} + \frac{v_1}{R_2} = 0$$ $$n_2:\frac{v_2-V}{R_3} + \frac{v_2}{R_4} = 0$$ $$v_1 - v_2 = v_R$$ $$R_1 = 80 \Omega, R_2 = 20 \Omega, R_3 = 100 \Omega, R_4 = 10 \Omega, V = 40V$$

N1 is the node at the top, N2 is the node at the bottom.

By these equations - setting GND at the node in the "negative side" of voltage source - I found out the drop voltage at R is -34, but the solutions say it is -28V. I can't find what I did wrong.

Answer 1

I found out the drop voltage at \$R\$ which is -34. But the solutions say it is -28V. I can't find what did wrong.

$$\color{red}{\text{Neither answer is correct }}$$

Think about it: How can the voltage across \$R\$ be greater than half the voltage of the original supply source voltage. In fact, by simple manipulation of the circuit, you can find the voltage across \$R\$ to be 14 volts at the maximum power point.

If the supply voltage was 80 volts, the volt drop across \$R\$ would be 28 volts.

Simple manipulation: -

^{simulate this circuit – Schematic created using CircuitLab}

Maximum power transfer occurs when \$R\$ = 25 Ω.

Answer 2

Your equations ignore the current through R. You can't do that.

If you don't believe this, think about what happens when R is 0.

For example, your node1 is missing the term \$ \frac{+v_2 - v_1}{R}\$ on the left side, and node2 is missing a similar term. You can't make believe the current through R doesn't exist without just getting Kirchoff's Current Law all wrong.

Answer 3

Been a while. But I finally see this. I'll give a hack at it. See if it sings.

Fast Attack

First off, redraw the schematic:

^{simulate this circuit – Schematic created using CircuitLab}

This is so much simpler to analyze. All you need to do is Thevenize the left and right legs to get:

^{simulate this circuit}

And from this you know that for maximum power then \$R\$ must be \$25\:\Omega\$. (Load resistance equals source resistance.)

It's actually pretty trivial to find.

Nodal

Let's go back to your schematic and do this the hard way. From what little I can glean from your writing, you have \$V=+40\:\text{V}\$ and you've assigned the voltage supply's negative terminal to \$0\:\text{V}\$. Let's look at what I think you started with:

^{simulate this circuit}

Pretty much the same thing I wrote earlier. Just lots uglier.

To solve for the resistance of the known part of the circuit (the part seen by \$R\$ looking backwards into the circuit), you want the open-circuit voltage (\$V_{_\text{OC}}\$) that happens when \$R=\infty\:\Omega\$ and you want the short circuit current (\$I_{_\text{SC}}\$) through \$R\$ that happens when \$R=0\:\Omega\$. Then you can find that the maximum power value for \$R\$ happens when \$R=\frac{V_{_\text{OC}}}{I_{_\text{SC}}}\$.

With \$R=\infty\:\Omega\$ then:

eq1 = Eq( v/r1 + v/r3 + iv, v1/r1 + v2/r3 )            # KCL for node v
eq2 = Eq( v1/r1 + v1/r2, v/r1 + 0/r2 )                 # KCL for node v1
eq3 = Eq( v2/r3 + v2/r4, v/r3 + 0/r4 )                 # KCL for note v2
ans = solve( [ eq1, eq2, eq3 ], [ v1, v2, iv ] )       # Solve!!!
ans[v1].subs( { v:40, r1:80, r2:20, r3:10, r4:90 } )
8
ans[v2].subs( { v:40, r1:80, r2:20, r3:10, r4:90 } )
36

I'm using SymPy and Sage here. (All free to have.)

The difference is obviously \$\mid\, V_{_\text{OC}}\mid\:=28\:\text{V}\$. If we take the difference, v1-v2, then we find that \$V_{_\text{OC}}=v_1-v_2=-28\:\text{V}\$.

Now, let's compute the short-circuit voltages for both nodes. We'll keep the nodes distinct, but assign them equal to each other as that is what must be if we short-circuit the nodes. I've added a new current (ic) as shown below that proceeds from v1 to v2:

eq1 = Eq( v/r1 + v/r3 + iv, v1/r1 + v2/r3 )                 # KCL for node v
eq2 = Eq( v1/r1 + v1/r2 + ic, v/r1 + 0/r2 )                 # KCL for node v1
eq3 = Eq( v2/r3 + v2/r4, ic + v/r3 + 0/r4 )                 # KCL for note v2
eq4 = Eq( v1, v2 )                                          # v1 = v2
ans = solve( [ eq1, eq2, eq3, eq4 ], [ v1, v2, iv, ic ] )   # Solve!!!
ans[ic].subs( { v:40, r1:80, r2:20, r3:10, r4:90 } )
-28/25

There we are.

So we need to compute \$R=\frac{V_{_\text{OC}}}{I_{_\text{SC}}}=\frac{-28\:\text{V}}{-\frac{28}{25}\:\text{A}}=25\:\Omega\$. That's the source resistance. And again, we know that \$R\$ must be the same. So \$R=25\:\Omega\$, as well, to maximize the power into \$R\$.

If you want to test all this, let's do a new set of nodal equations that now include \$R\$:

eq1 = Eq( v1/r1 + v1/r2 + v1/r, v/r1 + 0/r2 + v2/r )    # KCL for node v1
eq2 = Eq( v2/r3 + v2/r4 + v2/r, v/r3 + 0/r4 + v1/r )    # KCL for note v2
ans = solve( [ eq1, eq2 ], [ v1, v2] )                  # Solve!!!
(ans[v1]-ans[v2]).subs( { v:40, r1:80, r2:20, r3:10, r4:90, r:25 } )              
-14

From this, we'd expect to see the power as \$\frac{\left(-14\:\text{V}\right)^2}{25\:\Omega}=7.84\:\text{W}\$.

Let's see what LTspice says:

Yup! Looks good.

Feel free to slightly adjust \$R\$ up a little bit or down a little bit to see what the resulting power is. You will find that either way it gets a little smaller. The peak power remains only when \$R=25\:\Omega\$.

Nodal works. It just can be a pain at times.

Answer 4

Using nodal analysis we have: $$ n_1:\frac{v_1−V}{R_1}+ \frac{v_1}{R_2}+ \frac{v_1-v_2}{R} = 0 \\ n_2:\frac{v_2−V}{R_4}+ \frac{v_2}{R_3}+ \frac{v_2-v_1}{R} = 0 $$

$$ \begin{bmatrix} G_1+G_2+G & -G\\ -G & G_4+G_3+G \end{bmatrix} \begin{bmatrix} v_1\\ v_2 \end{bmatrix}= \begin{bmatrix} G_1\\ G_4 \end{bmatrix}\cdot V $$ while $$ G_1 = R_1^{-1} = 0.0125 \\ G_2 = R_2^{-1} = 0.05 \\ G_3 = R_3^{-1} = 0.01\\ G_4 = R_4^{-1} = 0.1\\ G = R^{-1} $$ by applying linear algebra rules for equation solving, we get: $$ v_1 = \frac{ \begin{vmatrix} G_1 & -G\\ G_4 & G_4 + G_3 + G \end{vmatrix} \cdot V } {\begin{vmatrix} G_1+G_2+G & -G\\ -G & G_4+G_3+G \end{vmatrix}} $$

$$ v_2 = \frac{ \begin{vmatrix} G_1+G_2+G & G_1\\ -G & G_4 \end{vmatrix} \cdot V } {\begin{vmatrix} G_1+G_2+G & -G\\ -G & G_4+G_3+G \end{vmatrix}} $$

$$ v_1 = \frac{180\cdot G + 11/5}{276\cdot G+11} \cdot V $$

$$ v_2 = \frac{180\cdot G + 10}{276\cdot G+11} \cdot V $$

$$ P_R = (v_2-v_1)^2\cdot G = V^2\cdot (\frac{10-11/5}{276\cdot G+11})^2 \cdot G $$

$$ \frac{d}{dG} P_R= V^2\cdot (7.8)^2 \cdot \frac{(276\cdot G +11)^2-2\cdot 276\cdot(276\cdot G^2+11\cdot G)}{(276\cdot G + 11)^4} = 0 $$ which yields: $$ (276\cdot G +11)=2\cdot 276 \cdot G \Rightarrow R = 276/11 \approx 25.1 \Omega $$

$$ v_2-v_1|_{R=276/11} \approx 14.18 $$

Note: If R3 is considered 90 ohms, then the answers will be the same as Andy's