if u have 8 ohm load resistance(speaker) in series (voltage divider circuit) with another resistor(?)

what resistance do u choose same 8 ohm as load or less or high
i have heard by many to choose same but i dont know reason.plz explain me detially.thank you in advance

 

Why do you want another resistor in series?

 

Maximum power transfer to what? The speaker? The added resistor?

Still, this has some educational value. The speaker is a complex load and the load changes with frequency. Would a person try to get equal power in both parts by using the DC resistance of the speaker or the vector sum of the inductance and the resistance?

 

if u have 8 ohm load resistance(speaker) in series (voltage divider circuit) with another resistor(?)

what resistance do u choose same 8 ohm as load or less or high
i have heard by many to choose same but i dont know reason.plz explain me detially.thank you in advance

Rewrite:
if you have 8 Ohm load resistor in series with another resistor, forming a voltage divider.

Here is the part that did not translate: what resistance do you chose?

My question: choose for what? are we adding a third resistor? if we are talking about third resistor, is it connected in series or in parallel. if third resistor is connected in parallel, then we need to know which resistor from the voltage divider it is in parallel with. or are we talking about the resistor that is part of voltage divider and whose resistance we are not told?

 

If you have a source with a given resistance, then the maximum power is delivered to a load with the same resistance (maximum power theorem). But (here's the kicker that many people miss) if you can adjust the source impedance, then you pick a value of zero ohms (or as low as possible), since that delivers the most power to the load.

 

thanks for your response.first clear one thing,in voltage divider bias.if you want to transfer maximum power to load resistance(simply a resistor)what you do.does you choose the source resistance small or same as load resistance(simply resistor).thanks in advance

 

Do some math and figure this one out yourself.

Start with an imaginary battery and load.

Make the battery have 1 volt of emf. Make the battery have 1 ohm of internal resistance.

Calculate power lost in a load resistor AND in the battery resistance for load resistance of 10 ohm; 1 ohm; and .1 ohm.

In which case does the load resistance show the largest percentage of power lost?

In which case does the battery internal resistance show the largest percentage of power lost?

What did you discover about power transfer when the load resistance changes and the source resistance remains constant?

 

What did you discover about power transfer when the load resistance changes and the source resistance remains constant?

upto before this line everything understand.now what i understand is the source resistance is the one that we have in voltage divider other than load resistance.
if we have load resistance higher than source power transfer would be high in load.and in opposite way the power transfer would be low in load.its what i discovered.IS THAT ANYTHING MORE I WANT TO UNDERSTAND

 

Do you want it to be harder than that?

 

now i got it according to maximum power transfer theory internal resistance should match with load to have max power on load.i am clear.thanks
"but in case of speaker load changes with frequency.then how do we come over that

 

now i got it according to maximum power transfer theory internal resistance should match with load to have max power on load.i am clear.thanks

As I noted, that is incorrect. If you can adjust the source internal impedance, then you want that impedance to be as low as possible to transfer maximum power. The maximum power transfer theorem applies to changing the load resistance, not the source resistance.

 

What is the scope of the problem?

Are we dealing with textbook problem? Where you have a dc voltage source and two resistors? If yes, then talking about loads that change with frequencies and source internal resistance is pointless and waste of time.