Thevenin circuit Max power across Load resistor

  • Thread starter EricSomin
  • Start date
  • #1
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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 RL = Rth. 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. Im 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.
 

Answers and Replies

  • #2
gneill
Mentor
20,901
2,854

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 RL = Rth. 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. Im 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 RL: P(RL) = ....

Find the maximum w.r.t. RL.
 
  • #3
9
0
I think ive figured it out. I can solve for I, input that into

P=I2RL

Since im trying to max. power disipated by RL i can take the first derivative and set it equal to zero.

does this sound incorrect to anyone?
 
  • #4
gneill
Mentor
20,901
2,854
I think ive figured it out. I can solve for I, input that into

P=I2RL

Since im trying to max. power disipated by RL i can take the first derivative and set it equal to zero.

does this sound incorrect to anyone?
Nope. Sounds like a good plan.
 

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