I'm having trouble understanding the concept of a problem I have for homework. The problem has a circuit with a variable resistor, and the resistor is adjusted until the power dissipated is 250w. I'm supposed to find multiple values of the resistance that can equal 250w, but I'm having trouble understanding how. Heres the actual question: The variable resistor in the circuit is adjusted until the power dissipated in the resistor is 250 W. Find the values of R that satisfy this condition. Here's a picture of the circuit: This question is from a section introducing the Maximum Power Transfer. I was thinking maybe finding the Thevenin Equivalent excluding the variable resistor R, and then use the resistor R to act as a load to the Thevenin circuit. With the circuit reduced, stating the maximum power would be = 250 W (and in this case Rth = R for a maximum value), but the Rth I found is not correct according to the book. I'm not sure if I'm approaching this right or not. I don't need an answer, just a general direction of how to solve it. Any help is appreciated.
You are correct that R=Rth to have maximum power transfer. When calculating Rth remember that voltage sources are replaced by short circuits and current sources by open circuits.
So in this case, calculating the Thevenin Eq circuit is the way to approach? I was pretty confident in deriving Vth and Rth, but it seems to be wrong which makes me think I'm going about this the wrong way. Heres the equations I used to find Vth: I opened the circuit where the resistor R is and used the mesh analysis method to find the current in the 20ohm resistor. I made the loops clock wise in both meshes and found these equations: 125i1 - 100i2 = 200 -100i1 + 160i2 = 0 i1=3.2A i2=2A Vth = 20*i2 + 30*i2 = 100v Before I post how I found Rth, does this look correct?
I think I just had an epiphany about this problem. The book says there are 2 possible values for resistance that will give a power rating of 250w as the answers to the question. When looking at the graph of max power vs resistance, the graph should look like this no? Where the local max of the graph is the max power delivered by the circuit. So I can see how there could be 2 values of resistance that give the same power output. So now that I realize that, it looks like I need to reduce the circuit to the Thevenin eq, find what the true max power is, then solve for a resistance that equals 250 W. What do you think?
elimenohpee, The epiphany is realizing that the maximum power theorem has nothing to do with this problem. Calculating the Thevenin equivalent of the circuit to the left the dependent current source we get 160 volts in series with 30 ohms of resistance. This will be supplying the "i" part of the circuit. Now we can easily see that the current through the variable resistor will be 31*i = i(R), and the voltage across the resistor will be 160-30*i = v(R). Multiplying the left and right part of the two equations we get -930i^2+4960i = i(R)*v(R) = 250 . Solving the quadratic equation for "i" we get 0.050888 amps and 5.282445 amps. So the current through the variable resistor will be 31*.050888 = 1.58 amps or 5.282445*31 = 163.76 amps. Using the power formula R*I^2 = 250, we get ---> 0.00932281 ohms and 100.46 ohms for the two variable resistance values. Notice how the 20 ohm resistor is irrelevant. Ratch
Have a look at the symbols here: http://en.wikipedia.org/wiki/Current_source I think the controlled source in this problem is a controlled voltage source, not a controlled current source.
mik3, A dependent voltage source? No sweat. The loop equations are: 160 - 30*I1 -20*(I1 - I2) -30*I1 = 0 30*I1 -20*(I2 - I1) -I2*R = 0 Two equations and two unknowns, I1, and I2. Solving for I2 we get 200/(15+2*R) . Squaring I2, multiplying it by R, and setting it to 250 watts, we get 40000*R/(225+60*R+4*R^2) = 250 . Cleaning up the equation and solving the quadratic we get R = 2.5 ohms and 22.5 ohms. Ratch