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I'm struggling with a problem our professor gave us, which states:

"In the following circuit, find the value of the resistor R_x so that the power consumed by the resistor R1 is the maximum."

schematic

simulate this circuit – Schematic created using CircuitLab

I'm trying to solve this with the nodal method as follows:

KCL for node A: $$-I_{AC}+I_{GND-A}+I_{AB} = 0$$ KCL for node B: $$-I_{AB}-I_{GND-B}+I_{CB} = 0$$ KCL for node C: $$-I_{GND-C}+I_{AC}+I_{CB} = 0$$

I've expressed the currents as follows: $$I_{GND-A}=\frac{E_A}{2} A$$ $$I_{AB}=1 A$$ $$I_{GND-B}=E_B$$ $$I_{AC}=-\frac{E_A-E_C}{3}$$ $$I_{CB}=\frac{E_C-E_B}{x}$$

The problem is that I can't find a way to express the current $$I_{GND-C}$$ as a function of the unknown node voltages. Any ideas?

Thanks in advance.

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  • \$\begingroup\$ You don't need to know the current through V1 voltage source. In this case. Because you already know the voltage at point C. \$V_C = 2V\$. If the GND is at a different point then you need to use a supernode youtube.com/watch?v=NA_zlZTDiKU \$\endgroup\$
    – G36
    Sep 5 '18 at 15:33
  • \$\begingroup\$ Yes, but these representations need to be substituted into KCL. How does knowing the voltage gives me what to substitute for $$I_{GND-C}$$ in the equation of node C? \$\endgroup\$ Sep 5 '18 at 15:39
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    \$\begingroup\$ Can you find the Thevenin circuit between the B and GND points (which implies removing \$R_1\$) and find \$R_x\$ such that \$R_{TH}=1\Omega\$? That's what it means to me when they talk about maximum power. Google maximum power transfer \$\endgroup\$
    – Big6
    Sep 5 '18 at 15:45
  • \$\begingroup\$ Simply you don't need the equation for node C. You have only two unknowns Va and Vb. Hence Va nodal equation is \$ \frac{V_A}{R_2}+ \frac{V_A - 2V}{R_3} + I_1 = 0\$ and we can solve that \$V_A = -0.4V \$ \$\endgroup\$
    – G36
    Sep 5 '18 at 15:58
  • \$\begingroup\$ @sixcab In this circuit \$Rth = R_x \$ . Therefore the max power at R1 you will get only if Rx = 0 Ohms and P_max = 2V^2/1R = 4W . \$\endgroup\$
    – G36
    Sep 5 '18 at 16:03
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Cheers to all of the fellow users who helped me in this! I'm writing this answer based on their comments for more visibility.

ANSWER: We don't need to use the third equation, actually. We substitute the current representations for \$I_{AC}, I_{GND-A}, I_{GND-B}, I_{AB}\$ and \$I_{CB}\$ in the respective equations for the nodes A and B. Which means:

Node A: \$-\frac{E_A-2}{3}+\frac{E_A}{2}+1=0\$

Node B: \$-1-E_B+\frac{2-E_B}{x}=0\$ (\$E_C = 2V\$)

Solving the equation for node A we get \$E_A\$, and solving the equation of node B we get \$E_B\$. The potential difference between the node B and the ground (V_1) is equal to \$E_B\$, which is a function of x. Therefore:

\$V_1=\frac{2-x}{x+1}\$, and because \$R_1 = 1Ω, I_1=V_1\$.

Finally, we get that: \$P_1 = (\frac{2-x}{x+1})^2\$ and now we need to find the value of x which maps to the maximum of this function for \$x >= 0\$, which is trivial if you know a bit of Calculus I. (Spoiler: it's \$x = 0\$)

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You cannot determine the current through the voltage source the way you want.

This is why the nodal analysis (strictly defined) cannot be used with circuits containing voltage sources. You should have been taught this when you learned nodal analysis.

What you can do instead is use the modified nodal analysis.

This means basically you eliminate the node equation that depends on the current through the voltage source and substitute a new equation based on the properties of the voltage source:

$$E_C = 2\ {\rm V}.$$

It is slightly more involved if one of the voltage source's terminals is not connected to ground. Then you have to combine the node equations for the two nodes connected by the voltage source to form a supernode equation.

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  • \$\begingroup\$ Actually, you do NOT need modified nodal analysis to do the work. You can use simple nodal analysis. You just assign a current through the voltage source as a variable and the number of equations required will work out fine without going to supernodes. Though in this case, I'm confused about why the subject even comes up. It looks too simple for even that nuance. \$\endgroup\$
    – jonk
    Sep 5 '18 at 18:27
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You don't need equations for this at all, if you can choose any value for X, the best case scenario would be a direct short across R_x (or where R_x=0) because that is the highest voltage you can get.

The first thing would be to test the extreme cases, like when R_x is zero or open.

Maximizing the power across R1 also means maximizing the voltage across it.

if R_x is zero, then the voltage V_c = V_b and becomes 2V which will be the highest voltage attainable for V_b which also means the most power for R_1 with the power being 2V^2/1Ω=4W .

if R_x is open, then the current source takes over which means 1A through R would be 1A^2/1Ω=1W

Any other resistance value will be in between these two cases.

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The schematic can be re-drawn into the following:

schematic

simulate this circuit – Schematic created using CircuitLab

The above schematic is much, much simpler to work on. \$R_2\$ and \$R_3\$ are really pointless. You have a current source, which is effectively an infinite impedance. So everything to the left of it really doesn't matter, unless for some reason you care about the voltage, \$V_\text{A}\$. (Which you shouldn't, for this problem.)

I think you can easily see that the right side is now much, much easier to work with. The nodal equation is:

$$\begin{align*} \frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_x}&=\frac{2\:\text{V}}{R_x}+1\:\text{A}\\\\&\therefore\\\\ V_\text{B} &=\frac{R_1}{R_1+R_x}\cdot\left(2\:\text{V}+I_1\cdot R_x\right) \end{align*}$$

Once you have this value, you know that the power in \$R_x\$ is \$\frac{\left(2\:\text{V}-V_\text{B}\right)^2}{R_x}\$. If you plug in the equation for \$V_\text{B}\$ and then take the derivative of this with respect to \$R_x\$ and solve for that slope being zero, you can get the value of \$R_x\$. It results in a very, very simple answer.

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