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Why exactly is the answer 50 Ohms?

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Why exactly is the answer 50 Ohms?

How to use Recom's R-785-0.5 switching DC-DC converters properly?Basics physics question about calculating resistanceLoad resistance determination in Zener DiodeIn calculation of Maximum power transferred for reactive circuits why is this done?Transistor as a switch in active or saturation region?Why do two transfer functions of a circuit have same denominator?Ultra low power voltage followerFinding R for max power deliverable to R and determining the max powerFinding Norton voltage and resistance and max powerTo find the maximum power that can be delivered to resistor that is specified

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3

1

$begingroup$

As the title mentioned - not sure why exactly the maximum power delivered to the load will be max when R_L is 50 Ohms.

If I'm guess why, its because if the resistance was greater than 50 then the current will be less, but if it was less than 50 (eg 25), then the constant 50 Ohm resistor would be delivered the majority of the power instead of going to the load.

Why does maximum power transfer happen at 50 ohms?

operational-amplifier circuit-analysis homework

share|improve this question

edited 14 hours ago

Voltage Spike

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Neil PNeil P

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Write the equation for the power delivered vs. Rl, differentiate it and find the zero crossing. What do you get?
  $endgroup$
  – John D
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your gut feeling is correct. Considering the -50 ohm case may mess with your gut feeling...current goes to infinity.
  $endgroup$
  – glen_geek
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please update this question's title to describe the actual problem, so that people searching for this in future can find the answers. The point of this site is to help all people with a specific kind of problem, not just the first person to encounter it.
  $endgroup$
  – iono
  36 mins ago
  

  
  
  
  

add a comment
 | 

3

1

$begingroup$

As the title mentioned - not sure why exactly the maximum power delivered to the load will be max when R_L is 50 Ohms.

If I'm guess why, its because if the resistance was greater than 50 then the current will be less, but if it was less than 50 (eg 25), then the constant 50 Ohm resistor would be delivered the majority of the power instead of going to the load.

Why does maximum power transfer happen at 50 ohms?

operational-amplifier circuit-analysis homework

share|improve this question

edited 14 hours ago

Voltage Spike

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asked 16 hours ago

Neil PNeil P

1611 bronze badge

New contributor

Neil P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Write the equation for the power delivered vs. Rl, differentiate it and find the zero crossing. What do you get?
  $endgroup$
  – John D
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your gut feeling is correct. Considering the -50 ohm case may mess with your gut feeling...current goes to infinity.
  $endgroup$
  – glen_geek
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please update this question's title to describe the actual problem, so that people searching for this in future can find the answers. The point of this site is to help all people with a specific kind of problem, not just the first person to encounter it.
  $endgroup$
  – iono
  36 mins ago
  

  
  
  
  

add a comment
 | 

$begingroup$

As the title mentioned - not sure why exactly the maximum power delivered to the load will be max when R_L is 50 Ohms.

If I'm guess why, its because if the resistance was greater than 50 then the current will be less, but if it was less than 50 (eg 25), then the constant 50 Ohm resistor would be delivered the majority of the power instead of going to the load.

Why does maximum power transfer happen at 50 ohms?

operational-amplifier circuit-analysis homework

share|improve this question

edited 14 hours ago

Voltage Spike

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asked 16 hours ago

Neil PNeil P

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Neil P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|improve this question

edited 14 hours ago

Voltage Spike

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asked 16 hours ago

Neil PNeil P

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share|improve this question

edited 14 hours ago

Voltage Spike

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asked 16 hours ago

Neil PNeil P

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edited 14 hours ago

Voltage Spike

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edited 14 hours ago

Voltage Spike

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Neil PNeil P

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3 Answers
3

active

oldest

votes

8

$begingroup$

The power delivered to the load is from the Joule heating effect:

beginequation
P=dfracBig(dfracR_LR_L + 50G_VINBig)^2R_L
endequation

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

beginequation
dfracdPdR_L = G_VIN^2 dfrac((R_L+50)^2 - 2cdot R_Lcdot(R_L + 50))(R_L+50)^4
endequation

Finally by making $ dfracdPdR_L = 0 $ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

beginequation
(R_L+50)^2 - 2cdot R_Lcdot(R_L + 50) = 0 implies , (R_L)^2 = 2500 implies R_L = 50,Omega
endequation

share|improve this answer

edited 2 hours ago

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NeilP consider formally accepting it and voting for it
  $endgroup$
  – Iron Maiden
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Iron: I think you'll find that the power law, $ P = frac V^2R $ was discovered by Joule, not Ohm ($ V = IR $).
  $endgroup$
  – Transistor
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Transistor Omg indeed...Shame on me after that.Thank you!
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't change the result, but the denominator of dP/dRL should be (RL+50)^4
  $endgroup$
  – Thierry Lathuille
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThierryLathuille You are right, I'll fix that.
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  

 | 
show 3 more comments

3

$begingroup$

Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).

So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)

share|improve this answer

answered 15 hours ago

Glenn WillenGlenn Willen

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$endgroup$

add a comment
 | 

1

$begingroup$

As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.

share|improve this answer

answered 14 hours ago

Voltage SpikeVoltage Spike

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$endgroup$

add a comment
 | 

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8

$begingroup$

The power delivered to the load is from the Joule heating effect:

beginequation
P=dfracBig(dfracR_LR_L + 50G_VINBig)^2R_L
endequation

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

beginequation
dfracdPdR_L = G_VIN^2 dfrac((R_L+50)^2 - 2cdot R_Lcdot(R_L + 50))(R_L+50)^4
endequation

Finally by making $ dfracdPdR_L = 0 $ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

beginequation
(R_L+50)^2 - 2cdot R_Lcdot(R_L + 50) = 0 implies , (R_L)^2 = 2500 implies R_L = 50,Omega
endequation

share|improve this answer

edited 2 hours ago

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NeilP consider formally accepting it and voting for it
  $endgroup$
  – Iron Maiden
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Iron: I think you'll find that the power law, $ P = frac V^2R $ was discovered by Joule, not Ohm ($ V = IR $).
  $endgroup$
  – Transistor
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Transistor Omg indeed...Shame on me after that.Thank you!
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't change the result, but the denominator of dP/dRL should be (RL+50)^4
  $endgroup$
  – Thierry Lathuille
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThierryLathuille You are right, I'll fix that.
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  

 | 
show 3 more comments

8

$begingroup$

The power delivered to the load is from the Joule heating effect:

beginequation
P=dfracBig(dfracR_LR_L + 50G_VINBig)^2R_L
endequation

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

beginequation
dfracdPdR_L = G_VIN^2 dfrac((R_L+50)^2 - 2cdot R_Lcdot(R_L + 50))(R_L+50)^4
endequation

Finally by making $ dfracdPdR_L = 0 $ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

beginequation
(R_L+50)^2 - 2cdot R_Lcdot(R_L + 50) = 0 implies , (R_L)^2 = 2500 implies R_L = 50,Omega
endequation

share|improve this answer

edited 2 hours ago

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NeilP consider formally accepting it and voting for it
  $endgroup$
  – Iron Maiden
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Iron: I think you'll find that the power law, $ P = frac V^2R $ was discovered by Joule, not Ohm ($ V = IR $).
  $endgroup$
  – Transistor
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Transistor Omg indeed...Shame on me after that.Thank you!
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't change the result, but the denominator of dP/dRL should be (RL+50)^4
  $endgroup$
  – Thierry Lathuille
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThierryLathuille You are right, I'll fix that.
  $endgroup$
  – Iron Maiden
  3 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

The power delivered to the load is from the Joule heating effect:

beginequation
P=dfracBig(dfracR_LR_L + 50G_VINBig)^2R_L
endequation

So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus:

beginequation
dfracdPdR_L = G_VIN^2 dfrac((R_L+50)^2 - 2cdot R_Lcdot(R_L + 50))(R_L+50)^4
endequation

Finally by making $ dfracdPdR_L = 0 $ we only need to care about the numerator since the denominator does not make the equation be equal to zero from any real value. Thus we have:

beginequation
(R_L+50)^2 - 2cdot R_Lcdot(R_L + 50) = 0 implies , (R_L)^2 = 2500 implies R_L = 50,Omega
endequation

share|improve this answer

edited 2 hours ago

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

$endgroup$

share|improve this answer

edited 2 hours ago

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

answered 15 hours ago

Iron MaidenIron Maiden

50611 silver badge99 bronze badges

3

$begingroup$

Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).

So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)

share|improve this answer

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).

So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)

share|improve this answer

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power).

So which direction power goes, up or down, depends on which effect is stronger. And as it happens, they cross over at 50 ohms (that is, when load resistance is equal to source resistance.)

share|improve this answer

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

$endgroup$

share|improve this answer

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

answered 15 hours ago

Glenn WillenGlenn Willen

1,71611 gold badge66 silver badges1313 bronze badges

1

$begingroup$

As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.

share|improve this answer

answered 14 hours ago

Voltage SpikeVoltage Spike

40k1212 gold badges4444 silver badges116116 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.

share|improve this answer

answered 14 hours ago

Voltage SpikeVoltage Spike

40k1212 gold badges4444 silver badges116116 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

As a rule, maximum power transfer for the load with a series and load resistor always happens when the resistances are equal. Use this rule as a shortcut when designing anything from antennas or transmission lines.

share|improve this answer

answered 14 hours ago

Voltage SpikeVoltage Spike

40k1212 gold badges4444 silver badges116116 bronze badges

$endgroup$

share|improve this answer

answered 14 hours ago

Voltage SpikeVoltage Spike

40k1212 gold badges4444 silver badges116116 bronze badges

answered 14 hours ago

Voltage SpikeVoltage Spike

40k1212 gold badges4444 silver badges116116 bronze badges

answered 14 hours ago

Voltage SpikeVoltage Spike

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