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This isn't homework, it's just a review problem. The answer is 4mW. I tried working the problem and came up with the wrong answer. What did I do wrong, and how can I solve this correctly?

My steps:

  1. Remove \$R_L\$ and create an open in its place
  2. Find the voltage at the open, this is the open circuit voltage
  3. I applied KCL at node 2 (what I labeled V2)
    • 6/3k amps in, 2mA out, 2mA in, and thus the current going out through the 6k ohm resistor (center branch) is 2mA. 2mA * 6kohms = 12 V across the 6kohm resistor.
    • Also, there is 4V going across the 2kohm resistor.
    • 12 + 4 = 16, so the open circuit voltage is 16V.
  4. Opening the voltage source and shorting the current source, the internal resistance is 6k ohms (2k ohm is ignored, no current flow, and 3kohm is ignored, no current flow)
  5. The Thevenin equivalent circuit has a 16V source, 6kohm internal resistance, and 6kohm load R_L. The voltage divides evenly. So, 8V/6kohms = 1.3mA through RL
  6. P=IV, 1.3mA * 8V = 10.667 mW

What did I do wrong?

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  • \$\begingroup\$ Find the maximum power...as what parameter is varied? \$\endgroup\$
    – Phil Frost
    Jul 17, 2014 at 0:30
  • \$\begingroup\$ I'm sorry, I do not understand what you are asking @PhilFrost \$\endgroup\$
    – asdf
    Jul 17, 2014 at 0:31
  • \$\begingroup\$ If we are finding a maximum, then it suggests we are trying to find the maximum of some function that has some parameters. What are the parameters? What do we vary to find the maximum? \$\endgroup\$
    – Phil Frost
    Jul 17, 2014 at 0:33
  • \$\begingroup\$ Well, I know that maximum power is delivered when the internal resistance of the circuit is equivalent to the resistance of the load (or vice versa). \$\endgroup\$
    – asdf
    Jul 17, 2014 at 0:37
  • \$\begingroup\$ So we are asking what value of RL results in the maximum power in that load, and what will the power be under those conditions? You can't just ask "what's the maximum power in some load" if you don't specify what the load is. Maybe RL is 100M, and then the maximum power is very nearly 0. So is the minimum power. \$\endgroup\$
    – Phil Frost
    Jul 17, 2014 at 0:40

1 Answer 1

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Your analysis is incorrect. For one, the 6V source is not all across the 3k resistor so the current thru the 3k resistor is not 2 ma. The easiest way to solve this is to first find the thevenin equivalent of the 6v source and the 3k and 6k resistors. The equivalent voltage is simply (6/9)*6 or 4 volts. The equivalent resistance is 3k in parallel with 6k or 2k. So now you have a 4 volt source in series with a 2k resistor. To find the equivalent resistance of the complete circuit, you short the 4v source and open the current source. Then you just have two 2k resistors in series. So the equivalent resistance of the circuit is 4k. To find the equivalent voltage, note that no current can flow through the 4v source and its 2k equivalent since RL is now an open circuit. The only voltage is the 2ma current source through the other 2k resistor which yields 4 volts. So the equivalent circuit that RL sees is 8 volts (4 from the 4 volt source and 4 from the current source through the 2k resistor) in series with a 4k resistor. For maximum power, RL must equal 4k. Half the voltage will be across it or 4 volts. 4 volts across 4k is 4 mw.

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