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I have done thevenin analysis on a larger circuit and have gotten the result below. I need to find the value of RL.

Would I use thevenin again? But that would give me the same result because I would be finding the same V-open and R-thevenin again. My short circuit current was 2mA, but I'm not sure if that applies anymore if I reduced it to this form below. Can you clarify this for me?

schematic

simulate this circuit – Schematic created using CircuitLab

from my lecture slides: enter image description here

The task is to select RL to maximize the power transfer from source to load. (as found out by @Null in the comments)

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    \$\begingroup\$ There is not enough information here to find R_L, Thevenin or no Thevenin. What other facts are you given in this problem? \$\endgroup\$
    – Greg d'Eon
    Apr 16 '15 at 18:34
  • \$\begingroup\$ As shown in this circuit RL could have any value from 0 to infinity (and possible beyond) - it would simply vary the current from 2mA downwards. \$\endgroup\$ Apr 16 '15 at 18:37
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    \$\begingroup\$ Please post the original large circuit and what part of it you want to analyse, specifically what values you are looking for. Maybe something went wrong in the first step and that's why you are stuck. Give us a chance to understand the big picture problem. =) \$\endgroup\$ Apr 16 '15 at 18:59
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    \$\begingroup\$ It looks like you are being asked to select \$R_L\$ to maximize the power transfer from source to load, in which case it is equal to \$R_{\text{TH}}\$. \$\endgroup\$
    – Null
    Apr 16 '15 at 19:56
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    \$\begingroup\$ Yeah, that was a critical piece of information without which the question is unanswerable. \$\endgroup\$
    – Null
    Apr 16 '15 at 19:58
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The power transferred to the resistor is the voltage times the current through the resistor:

$$P = U \times I$$

The voltage U over \$R_L\$ is (via voltage divider):

$$U = \frac{R_L}{2k\Omega + 4k\Omega + R_L} \times 8V$$

The current I is

$$I = \frac{8V}{2k\Omega + 4k\Omega + R_L}$$

Putting it all together

$$P = \frac{R_L}{2k\Omega + 4k\Omega + R_L}\times 8V \times\frac{8V}{2k\Omega + 4k\Omega + R_L}$$

With \$2k\Omega + 4k\Omega = 6k\Omega\$, we get

$$P = \frac{R_L\times (8V)^2}{(6k\Omega + R_L)^2} $$

The maths people say that at its extreme points, the derivative of P becomes 0. Said derivative \$P'\$ is $$P' = 0 = \frac{(6k\Omega - R_L)\times (8V)^2}{(6k\Omega + R_L)^3} $$ Which is sufficiently equivalent to asking $$P' = 0 = (6k\Omega - R_L)\times (8V)^2$$

There you have it $$R_L = 6k\Omega $$

Admittedly, this could be a maximum or a minimum. The maths people come to the rescue again, saying that it's a maximum if \$P'' < 0\$, which is the case

That's it.

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