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As the title mentioned - I am not sure why exactly the maximum power delivered to the load will be max when R_L is 50 ohms.

If I guess why, it's because if the resistance was greater than 50 ohm then the current will be lower, but if it was less than 50 ohm (e.g. 25 ohm), then the constant 50 ohm resistor would deliver the majority of the power instead of going to the load.

Why does the maximum power transfer happen at 50 ohms?

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    \$\begingroup\$ Write the equation for the power delivered vs. Rl, differentiate it and find the zero crossing. What do you get? \$\endgroup\$ – John D Oct 6 at 3:47
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    \$\begingroup\$ Your gut feeling is correct. Considering the -50 ohm case may mess with your gut feeling...current goes to infinity. \$\endgroup\$ – glen_geek Oct 6 at 4:57
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    \$\begingroup\$ Please update this question's title to describe the actual problem, so that people searching for this in future can find the answers. The point of this site is to help all people with a specific kind of problem, not just the first person to encounter it. \$\endgroup\$ – iono Oct 6 at 19:28
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    \$\begingroup\$ @iono: I went ahead and edited the OP's title to "Why exactly does maximum power transfer happen at 50 ohms (matched impedance)?" which actually states a self-contained question (assuming the source impedance is 50Ohm). \$\endgroup\$ – smci Oct 7 at 22:15
  • \$\begingroup\$ And by the way, the generalized form of the Max power transfer theorem is for a complex impedance Z = R + jX. For the resistance-only case, X = 0. I supposed you could simplify the title to "Matched resistance" \$\endgroup\$ – smci Oct 7 at 22:19
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The power delivered to the load is from the Joule heating effect:

\begin{equation} P=\dfrac{\Big(\dfrac{R_L}{R_L + 50}G\,V_{IN}\Big)^{2}}{R_L} \end{equation}

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

\begin{equation} \dfrac{dP}{dR_L} = (G\,V_{IN})^2 \dfrac{((R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50))}{(R_L+50)^4} \end{equation}

Finally by making \$ \dfrac{dP}{dR_L} = 0 \$ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

\begin{equation} (R_L+50)^2 - 2\cdot R_L\cdot(R_L + 50) = 0 \implies \, (R_L)^{2} = 2500 \implies R_L = 50\,\Omega \end{equation}

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    \$\begingroup\$ @NeilP consider formally accepting it and voting for it \$\endgroup\$ – Iron Maiden Oct 6 at 4:28
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    \$\begingroup\$ @Iron: I think you'll find that the power law, \$ P = \frac {V^2}{R} \$ was discovered by Joule, not Ohm (\$ V = IR \$). \$\endgroup\$ – Transistor Oct 6 at 8:49
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    \$\begingroup\$ @Transistor Omg indeed...Shame on me after that.Thank you! \$\endgroup\$ – Iron Maiden Oct 6 at 16:11
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    \$\begingroup\$ It doesn't change the result, but the denominator of dP/dRL should be (RL+50)^4 \$\endgroup\$ – Thierry Lathuille Oct 6 at 16:47
  • \$\begingroup\$ @ThierryLathuille You are right, I'll fix that. \$\endgroup\$ – Iron Maiden Oct 6 at 16:59
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Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).

So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)

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As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.

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    \$\begingroup\$ Just to prevent confusion: when the output resistance is fixed, the maximum power output is achieved when the internal resistance is as low as possible. It doesn't make sense to increase the internal resistance to increase output power. \$\endgroup\$ – Nathan Oct 7 at 8:04
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    \$\begingroup\$ This is the true "intuitive" answer. I assume this isn't an RF class, but the OP should keep this in the back of their head, as maximum power transfer and impedance matching is a pretty important concept. Actually, it's a pretty important concept in amplifier design too (as the question is framed). \$\endgroup\$ – yhyrcanus Oct 7 at 15:41
  • \$\begingroup\$ I think it's more of an issue in "oldschool" amplifier design. When lowering the output impedance of your amplifier came at a cost of signifiantly increasing idle power consumption, impedance matching made sense. Nowadays, at least for audio-frequency amplifiers it makes more sense to design the amplifier with a very low impedance output. \$\endgroup\$ – Peter Green Oct 25 at 16:20
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Another way to solve this, that doesn't involve actually doing any differentation.

Let \$R_S\$ be the source resistance, \$R_L\$ be the load resistance, \$V_{RL}\$ be the voltage accross the load, \$V_{RS}\$, be the voltage across the source resistance, \$P_{RL}\$ be the power delivered to the load resistance and \$I\$ be the current. \$G\,V_{IN}\$ and \$R_S\$ are outside of our control. We control \$R_L\$ and out goal is to maximise \$P_{RL}\$.

$$G\,V_{IN} = V_{RS} + V_{RL}$$

$$P_{RL} = IV_{RL} = \frac{V_{RS}}{R_S}{V_{RL}}= \frac{V_{RS}V_{RL}}{R_S}$$

So to maximise the power to the load we need to maximise \$V_{RS}V_{RL}\$, since this is a quadratic it has exactly one turning point, and by symmetry that turning point must be at \$V_{RS} = V_{RL}\$, which in turn implies \$R_S=R_L\$.

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