Engineering Understanding Maximum Power in Circuit: Comparing Load and Circuit Approaches

[Tekneek]

Homework Statement

What is the difference between maximum power in load and maximum power by load and circuit?

The Attempt at a Solution

I know max power in load is when Rth (Thevenin) = RL(load). And the equation to find power in load is (Vth^2)/(4*Rth). Not sure how to find max power by load and circuit, is it just P=I*V?

 

[gneill]

Suppose you consider the simple case where there's a voltage supply with some internal resistance and a load resistance:

The only source of power is the voltage supply. But that power can be dissipated in two places.

Can you determine an expression for:
1) the power produced by the voltage supply
2) the power dissipated by ("Delivered to") the load
3) the power lost in the source

Suppose that the source resistance and voltage are both of fixed value. What value of R_L will draw the most power from the voltage supply? What value of R_L will deliver the most power to the load?

 

[Tekneek]

Below

 

[Tekneek]

Suppose you consider the simple case where there's a voltage supply with some internal resistance and a load resistance:

The only source of power is the voltage supply. But that power can be dissipated in two places.

Can you determine an expression for:
1) the power produced by the voltage supply
2) the power dissipated by ("Delivered to") the load
3) the power lost in the source

Suppose that the source resistance and voltage are both of fixed value. What value of R_L will draw the most power from the voltage supply? What value of R_L will deliver the most power to the load?

1) P=I*(Rs+RL)
2) P=I*RL
3) P=I*Rs

Since P=V^2/R, The lower the RL the higher the power from the voltage supply
Rs=RL will deliver the most power to the load.

 

[gneill]

1) P=I*(Rs+RL)
2) P=I*RL
3) P=I*Rs

You'd need to square the current in those expressions. $P = I^2 R$ ; $P = V^2 / R$ ; $P = I V$ .

You might want to write them in terms of the given fixed values V and Rs.

Since P=V^2/R, The lower the RL the higher the power from the voltage supply
Rs=RL will deliver the most power to the load.

So, if the load RL heads toward zero the power in the (overall) circuit gets larger. Will this be larger or smaller than the maximum power deliverable to the load?

 

[Tekneek]

You'd need to square the current in those expressions. $P = I^2 R$ ; $P = V^2 / R$ ; $P = I V$ .

Write might want to them in terms of the given fixed values V and Rs.



So, if the load RL heads toward zero the power in the (overall) circuit gets larger. Will this be larger or smaller than the maximum power deliverable to the load?

Yeah forgot to square the current.
The power should be larger than the maximum power deliverable to the load.

 

[gneill]

Yeah forgot to square the current.
The power should be larger than the maximum power deliverable to the load.

Right. So in general, when there are unavoidable losses (such as the resistance associated with a power supply), the total power in a circuit will be larger than that delivered to the load. That's in a DC circuit.

When you start looking at AC circuits, you'll find that there are other ways that power delivered to a load can differ from the power generated by the power supply.

 

[Tekneek]

Thank you very much for your help! :)