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Author Topic: Resistors and Ohm's law  (Read 1404 times)

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kahlid74

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Resistors and Ohm's law
« on: June 06, 2005, 12:09:02 pm »
I have a set of PC speakers I want to hook up to a computer PSU.  The DC power adapter for the speakers has the following Output values.
9volt
1amp

By those readings you can determine that the wattage is 9 and the resistance is 9.
To figure out the resistor needed you can perform the following equation.
Speaker voltage/Speaker resistance = Resistor voltage/Resistor value
9/9 = (12-9)/R
1= (12-9)/R
R= 12-9
R=3

By this equation I would need a 3 Ohm resistor to take 12 volts to 9 volts.  No way in heck is that right.  I tried a 10 Ohm resistor and it did nothing.  Where is the failure in my equation/logic?

I used a voltmeter and the speakers used a maximum of 110 Mili-amps when the volume was maxed, so I assume the 1 amp on the power plug just means a max of 1 amp can travel through.

SirPoonga

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Re: Resistors and Ohm's law
« Reply #1 on: June 06, 2005, 12:18:11 pm »
Hmmmm, not sure.  I just wanted to say you could use 4 diodes to get it down to 9v (assuming DC of course).  Diodes drop voltage by 0.7V.

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Re: Resistors and Ohm's law
« Reply #2 on: June 06, 2005, 12:27:44 pm »
And of course, 4x the amount of components that could fail, if you use 4 diodes instead of the right resistor.

SirPoonga

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Re: Resistors and Ohm's law
« Reply #3 on: June 06, 2005, 12:39:00 pm »
I was just saying, as a backup plan :)

SirPoonga

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Re: Resistors and Ohm's law
« Reply #4 on: June 06, 2005, 12:39:55 pm »

Speaker voltage/Speaker resistance = Resistor voltage/Resistor value
9/9 = (12-9)/R

You have a 9ohm speaker?  most pc speakers are like 6ohm or 8ohm.

kahlid74

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Re: Resistors and Ohm's law
« Reply #5 on: June 06, 2005, 12:48:50 pm »

Speaker voltage/Speaker resistance = Resistor voltage/Resistor value
9/9 = (12-9)/R

You have a 9ohm speaker?

ShinAce

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Re: Resistors and Ohm's law
« Reply #6 on: June 06, 2005, 01:25:40 pm »
The problem is current(or amps).

Your adapter is in fact 9 volts, but only when you are drawing the full 1 amp. Otherwise, the quivalent resistance of the amp can be anywhere between 9 ohms and infinite ohms.

You could use a 9 volt regulator. You could build a voltage divider using 2 resistors, but that will waste a lot of power. Lastly, you could use diodes. 5-6 of them in series will drop the voltage around 9, assuming you're using a regulated 12 volt supply like a PC power supply.

If you are using a computer power supply, you could likely supply that straight to the speakers and never have a problem. Plug in that 9 volt adapter and read what kind of voltage it gives off with a multimeter. With no load, you'll see it's likely close to 13-14 volts.

edit: diodes are not likely to fail. The only way to blow them is too much current, which means a flaw somewhere else in the design. You would need about a 100V, 3 amp or more diode. You can use anything with higher voltage and current specs.
« Last Edit: June 06, 2005, 01:29:36 pm by ShinAce »
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Re: Resistors and Ohm's law
« Reply #7 on: June 06, 2005, 02:17:59 pm »
Don't forget that for something like a computer PSU, the values that you read on them are usually going to be what is called "peak" or "RMS" output.  That means, those are the maximum values that the current or voltage will ever reach.

Craig

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Re: Resistors and Ohm's law
« Reply #8 on: June 06, 2005, 03:09:58 pm »
I have a set of PC speakers I want to hook up to a computer PSU.  The DC power adapter for the speakers has the following Output values.
9volt
1amp

By those readings you can determine that the wattage is 9 and the resistance is 9.
To figure out the resistor needed you can perform the following equation.
Speaker voltage/Speaker resistance = Resistor voltage/Resistor value
9/9 = (12-9)/R
1= (12-9)/R
R= 12-9
R=3

By this equation I would need a 3 Ohm resistor to take 12 volts to 9 volts.  No way in heck is that right.  I tried a 10 Ohm resistor and it did nothing.  Where is the failure in my equation/logic?

I used a voltmeter and the speakers used a maximum of 110 Mili-amps when the volume was maxed, so I assume the 1 amp on the power plug just means a max of 1 amp can travel through.

If you wanted to convert 12 V to 9 V at 1 amp, you work it out like this:

12V-9V=3V
then
3 divided by 1 = 3


The formula:

R =
(VIN-VOUT)
/
IOUT


Now we need to work out the required wattage of the resistor.
P = V X I
So, 12V X 1 = 12


You would need a 12 watt resistor for the job.

But you should be able to scrap all that. You would likely find like on most unregulated power supplies that 9 Volt output could even be as high as 12-14 volts! Get a voltmeter out and test the plug pack. You'll be surprised!

A 12 volts PSU should be OK.
« Last Edit: June 06, 2005, 03:15:34 pm by Craig »
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