I desperately needed help on this question, I have to calculate the value of the RL load,
I'm guessing I have to find the resistance of the whole circuit and what ever resitance is that is the resistance of RL,
I have to do this using kirchoffs,
where have I gone wrong?
I need to calculate the value of RL then calculate the maximum power transfer. Any help please?
I got a Rth of 8.688 Ω so the RL should not be 32 ohms for maximum power transfer. The Max Power pdf could not open so I have no idea what the OP has for RL.
Joe is right; Ratch=Bzzzzt.
MikeMl, ...The problem should not be worked with ... a simulator. ...
JoeJester,
Yes, you are correct. That is the Thevenin value, but the instructions were not to use Thevenin's theorem. I was wrong in assuming that the power dissipated in a network is the same as the the power of the Thevenin equivalent. For the example given, the total power supplied by the 15 volt source is 12.70 watts, whereas the power dissipated by the equivalent load is only 2.85 watts. Don't use equivalent resistances for power calculations. What I should have done was calculate the open circuit voltage with RL = ∞, and the short circuit current with RL = 0 using loop equations. Then I could get the circuit impedance looking back from RL.
So, from my previous post:
I1 = -(120*(96+7*RL))/(11112+1279*RL)
I2 = -(375*(24+RL))/(11112+1279*RL)
I3 = -9000/(11112+1279*RL)
Setting RL to ∞ in I2 and multiplying by R10=24 gives the open circuit voltage of 7.04 volts.
Setting RL to 0 in I3 gives the short circuit current of 0.81 amps.
Dividing 7.04 by 0.81 gives an impedance of 8.69 ohms.
Ratch
...What do you mean by set RL to ∞ and set RL to 0?
He is trying to find the current that the network will put out if RL=0 (i.e. replaced with a short) and the voltage across RL if RL=∞ (i.e. the network is open circuited). That can be done using only KCL. However, you must have studied Thevenin to know that the network can be replaced with a new voltage source of the same voltage as the open circuit voltage, in series with a resistor of Reff=Voc/Ish.
Having those two values, you have to know apriori that the max power transfer occurs when RL is set to Reff. If you dont know that apriori, you have to set up a voltage divider consisting of Reff and RL, calculate the power in RL as a function of RL, differentiate that equation and find where it crosses zero...
All of the things that Spice knows how to do intrinsically...
I just knew you were going to say that, which is why I said "Simulation is not what you should be doing right now, but you should take away...
At least I didn't just work the entire problem for the student; where is the learning in that?
All of the things that Spice knows how to do intrinsically...
This equation 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24ΩI₂ is wrong. try 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24Ω(I₂-I3)
It helps to actually draw the loop currents.
Kiss,
Which post # shows the wrong equation? Doesn't post #7 show the correct equation? It helps to specify the post # where the error is.
Ratch
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