Thévenin resistance and Load resistance vs maximum power

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SUMMARY

The discussion focuses on determining the ideal load resistance for maximum power output in a circuit using Thévenin's theorem. The calculated Thévenin resistance (R_th) is 1.66 ohms, and the Thévenin voltage (E_th) is 8 volts, resulting in a maximum power output of 9.36 watts when the load resistance matches the Thévenin resistance. The user seeks clarification on demonstrating that power output decreases with resistances lower or higher than the ideal value. The confusion arises from interpreting the load resistor's configuration in relation to the Thévenin equivalent circuit.

PREREQUISITES
  • Understanding of Thévenin's theorem and equivalent circuits
  • Knowledge of power calculations in electrical circuits
  • Familiarity with series and parallel resistor configurations
  • Ability to manipulate algebraic expressions for optimization
NEXT STEPS
  • Learn how to derive the power expression for a load resistor in a Thévenin equivalent circuit
  • Study the concept of maximum power transfer theorem in electrical engineering
  • Explore the effects of varying load resistance on power output in practical scenarios
  • Investigate the implications of series versus parallel configurations in circuit analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone studying power transfer in electrical circuits will benefit from this discussion.

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Homework Statement


RL.PNG


This is the network.

The problem states that a new resistor is put between A and B, in parallel with R4.
I should find the ideal resistance for this load resistor and show that this is indeed the ideal resistance for maximum power output.


Homework Equations



R_th = R_L-max
P=\frac{E_(th)^2}{4 \times R_th}

The Attempt at a Solution


I have determined the Thévenin resistance and voltage in an earlier problem and found them to be:
Rth = 1.66 ohm
Eth = 8 Volts
I_R4 (before load resistor is connected) = 4 ampere
This is double checked with the solution.

I have also double checked that the load resistor should be 1.66 ohm to provide maximum power.

The power output is found using the P=formula stated above which gives me 9.36 W.
This is also correct according to the solution.


My problem is showing that with a lower or higher resistance, the power output will drop.
I can't simply put another resistance value into the formula above as the power will increase if I insert a lower resistance.

Can someone kick me a step in the right direction?
 
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Use the Thevenin equivalent circuit to drive a load resistor R of indeterminate value (it's a variable). Write the expression for the power transferred to R. How might you going about maximizing the power expression with respect to R?
 
Thanks for your reply, gneill.

When i draw the thevenin circiuit with load resistor connected i will have an 8V dc suppy in parallel with 1,66 ohm thevenin resistor in parallel with 1.66 ohm load resistor.

Using these numbers, i get that the effect in the load resistor is 38W (P=U^2 / R where u=8V and R = 1,66ohm)

So what am I not taking into concideration in this situation?
 
A Thevenin equivalent consists of a voltage source in SERIES with a Thevenin resistance.
 
Oh my god.

Thank you.
 
That implies that even when the text said "a load resistor is connected to R4 in parallel", the load resistor is in series when doing the calculations based on the thevenin equivalents.

The text fooled me.
 

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