Impedance of a systems effect on voltage drop.

RSDXzec
10+ year member

CarAudio.com Elite
Will be going from a system at 4 ohms to a system at 1.34 ohms. Since at lower impedance on the same power, most of that power should consist of current draw compared to more voltage draw from a higher impedance system. Would this mean that i'd get less voltage drop with a lower impedance than with a higher impedance system?

I only draw these conclusions from V=IR and P=IV so I'm not sure if they are correct or not.

Cheers.

 
At a lower impedance you are going to be sending more electrons (current = I) through your amp/speakers. The measured voltage will be the same, you are increasing the current (I) to compensate, you will be putting a larger strain on your battery. Think of V as constant and, I and R as changing values.

 
The voltage you are measuring is the voltage from your amp to your speakers set by your gains, not your car's voltage (which is constant). You will have to lower your gains when running at a lower impedance/resistance if you wish to have the same power.

It all depends on your point of reference, your cars or your audio system. Don't think of "drawing" from a cars voltage, think of drawing current from your car.

 
The concept of voltage drop is unrelated to your inquiry.

It is the loss of potential/voltage from your battery to your system due to the inherent resistance of conducting cables.

It has everything to do with the connections you use to link your amps to your battery and to a ground.

 
The concept of voltage drop is unrelated to your inquiry.It is the loss of potential/voltage from your battery to your system due to the inherent resistance of conducting cables.

It has everything to do with the connections you use to link your amps to your battery and to a ground.
ok when I say voltage drop I am talking about the battery voltage. So I'm not sure if when drawing more current (due to the lower impedance) I'll be putting more strain on the system causing more voltage drop at the battery compared to a system drawing the same amount of power, but more current due to a higher impedance.

You will have to lower your gains when running at a lower impedance/resistance if you wish to have the same power.
I set my gains using V = sqrt(Prms*R) so I am aware of this.

Cheers.

 
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