...

 

What have you been able to do on the problem so far?

hgmjr

 

You will need to solve the circuit twice: once assuming the load is a short, and once assuming it is open.

 

secondhandloser,

I've tried reducing it down to a Thevenin equivalent, but I don't think I did it correctly, and tried setting up node and mesh equations and just plugging stuff in.

No need to Thevenize or Nortonize. Just analyze and reason it out. Let's make the voltage at the junction of the three resistors 8 volts. That will absolutely guarantee that the voltage across the load will be 8 volts or less, right? So if the load is infinity or disconnected, we get 12*R3/(R1+R3) = 8 volts . The sum of R1 + R3 = 6 ohms from the power value. Solving for R1 & R3 from the two equations we easily get R1 = 2 ohms and R3 = 4 ohms. Now if the load is less than infinity, the voltage at the junction of the three resistors will be less than 8 volts. So if we assume the voltage is 8 volts, and make the calculations on that basis, we guarantee that the values will be less than what we calculated for if the resistor is a finite value less than infinity. Continuing on, R2 = 8/.250 = 32 ohms. This will guarantee that the value of the current through the load can never be above 0.25A, even if the load is a short. Now the maximum power principle says that if a 8 volt source with a R2 = 32 ohms internal resistance drives a load of 32 ohms, then the maximum power will be transferred. So assuming the load is 32 ohms, the total of resistance of the load and R2 will be 64 ohms, for a current of 1/8 amp. That makes the power dissipated by the load less than 32/64 = 1/2 watts, which satisfies all the conditions.

Ratch

 

I dont understand how you get this equation

 

hitmen,

I dont understand how you get this equation

To which one of us are you referring, and to which equation?

Ratch