Thevenin circuit Max power across Load resistor

[EricSomin]

Homework Statement

One Thevenin theorem result is that the maximum power across the load resistor in the equivalent circuit (and therefore the real circuit) occurs when R_L = R_th. Start with the result that the power dissipated by the load resistor is I²RL and prove this result.

Homework Equations

The Attempt at a Solution

I have really no idea where to begin with this question. I do understand what a Thevenin circuit is, and where i would place a load resistor. I am just looking for some direction as to where to start, or for someone to point me towards what i should be thinking about and working with to get towards an answer.

any help or thoughts would be greatly appreciated.

 

[gneill]

Homework Statement

One Thevenin theorem result is that the maximum power across the load resistor in the equivalent circuit (and therefore the real circuit) occurs when R_L = R_th. Start with the result that the power dissipated by the load resistor is I²RL and prove this result.

Homework Equations

The Attempt at a Solution

I have really no idea where to begin with this question. I do understand what a Thevenin circuit is, and where i would place a load resistor. I am just looking for some direction as to where to start, or for someone to point me towards what i should be thinking about and working with to get towards an answer.

any help or thoughts would be greatly appreciated.

Draw the circuit. Obtain an expression for the power in the load in terms of the given components, yielding the power as a function of R_L: P(R_L) = ...

Find the maximum w.r.t. R_L.

 

[EricSomin]

I think I've figured it out. I can solve for I, input that into

P=I²R_L

Since I am trying to max. power dissipated by R_L i can take the first derivative and set it equal to zero.

does this sound incorrect to anyone?

 

[gneill]

I think I've figured it out. I can solve for I, input that into

P=I²R_L

Since I am trying to max. power dissipated by R_L i can take the first derivative and set it equal to zero.

does this sound incorrect to anyone?

Nope. Sounds like a good plan.

 

[Nickie]

Thank you for your question. This is a common misconception about Thevenin's theorem and the maximum power transfer theorem. The Thevenin equivalent circuit does not determine the maximum power across the load resistor; rather, it is the maximum power transfer theorem that determines this value.

To begin, let's review the Thevenin theorem. It states that any linear circuit can be replaced by an equivalent circuit consisting of a voltage source in series with a resistor. This equivalent circuit will produce the same voltage and current at the load as the original circuit.

Next, let's look at the maximum power transfer theorem. It states that the maximum power will be transferred from a source to a load when the source impedance is equal to the load impedance. In other words, the load resistor will dissipate the most power when its resistance is equal to the Thevenin resistance of the circuit.

Now, let's prove that the power dissipated by the load resistor is I^2RL. We know that the voltage across the load resistor is Vth (from Thevenin's theorem) and the current through the load resistor is I = Vth/(Rth + RL) (from Ohm's law). Therefore, the power dissipated by the load resistor is:

P = I^2RL = (Vth/(Rth+RL))^2 * RL

To find the maximum power, we can take the derivative with respect to RL and set it equal to 0:

dP/dRL = (Vth/(Rth+RL))^2 - 2Vth/(Rth+RL)^3 * RL = 0

Solving for RL, we get RL = Rth. Therefore, the maximum power is transferred when RL = Rth.

In conclusion, the Thevenin equivalent circuit does not determine the maximum power across the load resistor. It is the maximum power transfer theorem that determines this value, and it is equal to I^2RL.