What’s difference between series and parallel connecting a resistor?

I know this question may be too theoretical. I’ve long known that the conventional way to reduce efficiency of the driver is to apply an l-pad circuit or, easier way, a serial resistor. But I’ve never seen anyone use a single parallel resistor anymore.

Let us talk a little on basic electronics; A Kirchoff’s law, sorry I’m not a professional engineer nor a mathematician, I may be wrong. If so, please help to correct me.

Series resistor; assume we place a resistor with equal resistance to a driver in series, the VOLTAGE across the resistor and the driver will be half. This may be called a voltage divider. Hence, the SPL of the drivers will be lowered. The total impedance will be twice.

Parallel resistor; assume we place a resistor with equal resistance to a driver in parallel, the CURRENT going through the resistor and the driver will be half. Hence, the SPL of the driver will be lowered. The total impedance will be half.

What’s difference between series and parallel connecting a resistor to a driver in order to attenuate the efficiency? Surely, the impedance is. But if I have very high-impedance drivers, can I use parallel resistor instead of series resistor?
 
I ask with formal statements

Using the Filter Wizard in Hornresp will give you the answers to your questions, and allow you to see exactly how the response is affected by inserting a series or parallel resistor between the amplifier and the driver.
 

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For those who discourage me, please have a reason.
danny_66 has given you a reason, but you never seem to respond to explanations!

Lets assume a perfect voltage source and two, identical parallel resistors.

The same voltage will be connected across each resistor. Ohm's law says that the current (I = V/R) will be the same in each resistor. Remove one of these resistors, and the current (I = V/R) in the remaining one will remain exactly the same.

If one of these resistors is a (purely resistive!) loudspeaker, then adding a parallel resistor would not affect the current though it and hence its SPL will remain unaffected.

The big down side in this, as you've been told repeatedly, is that half of the power delivered by the voltage source will be wasted in the parallel resistor.

It apears to be your knowledge of series and parallel circuits and Ohm's law that is letting you down. Back to school time, I would suggest. :)
 
Thank you all for the replies, especially Galu. :)

Galu, if the amplifier deliver a current I to the load; and a purely resistive loudspeaker is connected to an equal value resistor, should the current I be divided into I/2 to each branch of equal resistance? Hence, the SPL of this driver shall be halved due to I/2 going though it instead of I. Is my understanding correct? Sorry to make you confused, this is my intended meaning on the started post. Please help to correct me if I’m wrong.
 
Again incorrect.
Current is not equal to SPL !
Voltage swing is equal to SPL, if your amplifier can deliver the current into that impedance load (loudspeaker), most amplifiers can deliver the current for 4-8 ohm,
probably less at full voltage swing.
 
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Is my understanding correct?
You have failed to understand my explanation because you fail to understand the relationship between voltage and current in electrical circuits.

When connecting the resistor in parallel with the (resistive) loudspeaker, the current in the loudspeaker would not change, but the current drawn from the supply would double.

So, you see, you have it the wrong way round.

The current in the loudspeaker is not halved when you add a parallel resistor, it stays exactly the same.

The current from the supply doubles when the parallel resistor is added, in accord with Kirchhoff's law which states that the current from the supply equals the sum of the currents in the parallel resistors.
 
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Presscot. Simply put it is best if you don't put a resistor where the amp can see it clearly at all frequencies because it will just drain current without helping.
That depends on the impedance of the source feeding it.
Yes. When presscot mentions that an amp will "deliver a current", it might be seen in those terms in anything besides a Voltage amplifier. Parallel components by default wouldn't find effective use with a low source impedance.

However in the thick of a crossover the circuit impedance becomes more inviting to these opportunities.