3.5mm Splitter Question

[DoctorSarcastic]

I want to buy a 3.5mm splitter that I can add into my audio setup. I am just wondering if the splitter would degrade the sound in anyway. I do not plan on running the two sources at the same time. Cheers!

[Enderman]

If the splitter is not some cheap dollar store connector it will be fine.

[EmoRarity]

As long as it works, should be fine

[Lutkeveld]

If the splitter is not some cheap dollar store connector it will be fine.

 

Even 1 dollar ones will work fine as long as they work

[Enderman]

Even 1 dollar ones will work fine as long as they work

Yeah but the cheap metals they use in them can have a bad connection and cause static noise when the cable moves a tiny bit.

They also fail very often.

[ShearMe]

Yeah but the cheap metals they use in them can have a bad connection and cause static noise when the cable moves a tiny bit.

They also fail very often.

 

RMA the $1 cable duh

[KitsuneJin]

dont cheap out on it

[KamiKatze]

It will half the current so you will probably have less volume per connected headphones.

[ShearMe]

It will half the current so you will probably have less volume per connected headphones.

 

Actually, current increases when you add additional parallel circuits.

 

(1 / R_total) = (1 / R_A) + (1 / R_B ) + ( 1 / R_C)

With each new resistance added, R_total decreases

 

and

 

V = I * R

As R goes down, I goes up.

[Enderman]

RMA the $1 cable duh

LOL because the dollar store totally takes returns right?

[ShearMe]

LOL because the dollar store totally takes returns right?

 

sad day in hell when the Enderman can't recognize a joke

[KamiKatze]

Actually, current increases when you add additional parallel circuits.

 

Voltage stays the same, current halfs, that's what i learned in physics lessons but i am not an expert and you can proove me wrong. Wouldn't this also mean that the audio is getting louder?

 

To put this mathematical: let's say we have two parallel circuits with the same headphone each. The headphone has a resistance R. Two parallel resitances are added accodring to 1/Rtotal = 1/R1 + 1/R2    which, because we have identical headphones is 2/R So our total resistance R total = R/2

 

And after the formula for resistance R=U*I (where voltage is constant) we can derive that if our resitance halfs, the current has to half aswell.

 

But still, proove me wrong, i am not an expert.

[Enderman]

sad day in hell when the Enderman can't recognize a joke

says the person who didn't get the second joke...

[ShearMe]

Voltage stays the same, current halfs, that's what i learned in physics lessons but i am not an expert and you can proove me wrong. Wouldn't this also mean that the audio is getting louder?

 

To put this mathematical: let's say we have two parallel circuits with the same headphone each. The headphone has a resistance R. Two parallel resitances are added accodring to 1/Rtotal = 1/R1 + 1/R2    which, because we have identical headphones is 2/R So our total resistance R total = R/2

 

And after the formula for resistance R=U*I (where voltage is constant) we can derive that if our resitance halfs, the current has to half aswell.

 

But still, proove me wrong, i am not an expert.

 

When you add a parallel circuit, you have more things that need current, so your overall current goes up. Source voltage remains constant like you already noted. Your second formula is incorrect. R = V / I NOT V * I

 

If you use the cliche water hose analogy, resistance is the hose being clamped down. Add parallel hoses that are also clamped, and the water from the source has more paths to follow. Even if they're small, they're bigger than nothing, so more water comes out the nozzle at the end.

[KamiKatze]

When you add a parallel circuit, you have more things that need current, so your overall current goes up. Source voltage remains constant like you already noted. Your second formula is incorrect. R = V / I NOT V * I

 

If you use the cliche water hose analogy, resistance is the hose being clamped down. Add parallel hoses that are also clamped, and the water from the source has more paths to follow. Even if they're small, they're bigger than nothing, so more water comes out the nozzle at the end.

 

*oh god why, the formula*

 

That explains my wrong result. If R halfs, the current doubles. I walk away in shame, cuddle into my lab coat and cry myself to sleep.

 

Edit: but why does the volume go down anyways? At least that's what happened the last time i used a splitter.

[ShearMe]

Edit: but why does the volume go down anyways? At least that's what happened the last time i used a splitter.

 

The voltage isn't remaining constant. There's a handful of reasons why this'd happen with certain sources/amps, and I'm not sure I'm well versed in them all. Zac of all trades, master of none.

[DoctorSarcastic]

The voltage isn't remaining constant. There's a handful of reasons why this'd happen with certain sources/amps, and I'm not sure I'm well versed in them all. Zac of all trades, master of none.

I don't really care if the volume decreases if I use them at the same time... I won't be using them at the same time. If I use them one at a time, would the volume still be decreased? Thanks!

[ShearMe]

I don't really care if the volume decreases if I use them at the same time... I won't be using them at the same time. If I use them one at a time, would the volume still be decreased? Thanks!

 

Nope!

 

Sorry, we went on a tiny tangent.