Finding load resistance to achieve max power transfer

parallelresistance

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1) So, to find Thenevin equivalent resistance, R-Th,
   I disconnect R_L and disable the voltage source
   by shorting it out.

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2) Now I just look back into the network and
   calculate R_Th, which I know R_L will 
   have to equal to since max. power transfer
   is achieved when R_L == R_Th.

3) Now this is where I'm having trouble.  I see:

   [(2k + 2k) || 2k] + 2k
   (4k || 2k) + 2k
   8/6k + 2k
   10/3 k = 3.333k ohms

   But my book says it's:

   [(2k || 2k) + 2k ] || 2k
   (1k + 2k) || 2k
   3k || 2k
   6/5 k = 1.2k ohms

So, can someone explain to me why my equivalent resistance is incorrect? Starting at the right I see 2k in series with 2k, making 4k. Now, I see that 4k in parallel with the 2k in the middle branch, making 4/3k. And now since I'm left with a single loop, I see the 4/3k and remaining 2k (top left) in series, as they would carry the same current.

Best Answer

To calculate the equivalent resistance between those two terminals, I assume that we are connecting a voltage source V across it. Let I be the current drawn by the circuit, then V/I will give the equivalent resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and R2 are in parallel and same current flows through R3 and R1||R2. so R3 and R1||R2 are in series. Now the current through R4 is V/R4 and current through R3 is V/(R3+R1||R2).

or I = V/R4 + V/(R3+R1||R2).

So the resultant resistance is V/I = (R1||R2 + R3)||R4.

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