Remind me to stop arguing with woo-woos.

[MrBogosity]

(On a side note, 400th post for me!!! )

So, what's so special about your 0x190th post?

[Travis Retriever]

So, what's so special about your 0x190th post?

Funny.

[Travis Retriever]

Or he slept when they covered Ohm's Law. The batteries wired in parallel increase in amperage, not voltage; the batteries wired in series increase in voltage, not amperage. But you STILL have the same amount of power increase. It's just that in parallel, you burn through it faster. So he really shot himself in the foot with that one.

If memory from my Physics class serves: connecting batteries in series doubles the voltage.
Connecting the batteries in parallel keeps the voltage constant and the current constant, however, the lifetime of the batteries (capacity in Amp Hours) is doubled http://www.zbattery.com/Connecting-Batteries-in-Series-or-Parallel

[MrBogosity]

Well, okay, two batteries wired in parallel doubles the amperage IF the circuit draws it.

[Travis Retriever]

Well, okay, two batteries wired in parallel doubles the amperage IF the circuit draws it.

...Also, Amperage?
You mean current, right?
If so, because I = V/R (total current equals applied voltage divided by the equivalent resistance of the circuit); if the voltage is still the same, wouldn't the only way to double the current be to half the equivalent resistance?
If not, never mind.

[Travis Retriever]

Regardless, the person's original point is a weak analogy.
If I take 100 pills (regardless of whether they're homeopathic, or not), that's the equivalent as a series circuit:  the pills, like the current only have one path to take.  Why does he think people are not supposed to exceed recommended dosages?
Or why people die from overdose?
I mean, come on!
This is common freaking sense...

[MrBogosity]

...Also, Amperage?
You mean current, right?
If so, because I = V/R (total current equals applied voltage divided by the equivalent resistance of the circuit); if the voltage is still the same, wouldn't the only way to double the current be to half the equivalent resistance?
If not, never mind.

Think of an old cassette player when the batteries are running low. It plays the tape slowly, and everyone on it sounds like Darth Vader, but you can flip it over to the radio and play it just fine. The radio circuit is set up to use the same number of volts as the cassette player, but it's far better at handling low-amp situations. The reason why is that the motor in the cassette player acts like a resistor in a way; it needs the electrical current to do work, but if the juice isn't there it can't do as much. So it'll pull more amps off of a full battery, whereas the radio will just keep pulling as much as it always does. At lower amps, the motor acts as if it's a larger resistor in the circuit.

Now wire up two batteries in parallel to the cassette player, and the radio will play just as it always has but take longer to run down the two batteries, but with the cassette player, it just might be the case that every song will be by The Chipmunks, and it'll drain the batteries at the same rate.

Make sense?

[Travis Retriever]

Think of an old cassette player when the batteries are running low. It plays the tape slowly, and everyone on it sounds like Darth Vader, but you can flip it over to the radio and play it just fine. The radio circuit is set up to use the same number of volts as the cassette player, but it's far better at handling low-amp situations. The reason why is that the motor in the cassette player acts like a resistor in a way; it needs the electrical current to do work, but if the juice isn't there it can't do as much. So it'll pull more amps off of a full battery, whereas the radio will just keep pulling as much as it always does. At lower amps, the motor acts as if it's a larger resistor in the circuit.
Now wire up two batteries in parallel to the cassette player, and the radio will play just as it always has but take longer to run down the two batteries, but with the cassette player, it just might be the case that every song will be by The Chipmunks, and it'll drain the batteries at the same rate.

Make sense?

Let me see if I get you:
Cassette player:  p = v^2/R (electric power equation) small power from a small voltage because of a large resistance.
Radio:  P = v^2/r normal power because of the smaller resistance (or something like that).
I got you on the radio part with two batteries in parallel; it sounds like what I said.
But WHY is it different for the cassette player?

[MrBogosity]

Because the faster the motor for the cassette player runs, the more power it drains.

[Travis Retriever]

The same reasoning would apply to the current as well in my last post.
Also, would these be new batteries in parallel in your example?

OK, I'm currently (no pun intended. ) reading this:  http://www.allaboutcircuits.com/vol_6/chpt_3/3.html
I'm so dang tired I can barely get it...
Though it sounds like it has to do with the internal resistance of the batteries, and because each battery when connected in parallel has half as much current going through it, so not as much power (and voltage) is wasted on the batteries; about half as much power if memory serves because of the square with current, coupled with the addition of the two powers.
I suppose the total resistance of the radio, because it doesn't need as much current (if I get you correctly) wouldn't make as big a difference...but then why wouldn't you get the same thing for the radio; of it being louder, or something.
I'm confused...

[Travis Retriever]

Because the faster the motor for the cassette player runs, the more power it drains.

Well I figured that.
And I also figured why.  But now the problem has become why it doesn't have the same effect of more power delivered to the radio.

[MrBogosity]

Because the circuits in the radio don't use as much power. They'll only draw what they're going to use.

[Travis Retriever]

Because the circuits in the radio don't use as much power. They'll only draw what they're going to use.

Why do they only draw what they need?
How do they have that cut off?
Do they have a variable resistor for the detection of current or something?

[MrBogosity]

No...you use as much power as the work you do. The more work, the more the power requirements, the more the batteries are drained.

[Travis Retriever]

No...you use as much power as the work you do. The more work, the more the power requirements, the more the batteries are drained.

I know that.
I'm asking why from an Ohm's Law perspective.