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# Formula for Impedance/Power per driver in a series circuit???

It's been a while since I studied ohms law and voltage drops/power.

I have a question as to the formula for calculating how much power each driver gets in a series circuit if the impedance is different in each driver.

Say: 1000W into a series circuit> first driver is 8 ohm> second driver is 4 ohms...

Total Impedance is naturally 12 ohms but how is the power calculated? Would it go 1000 / 3 = 333W... 333W going to the 8 ohm driver and 666W going to the 4 ohm driver?

What about 1000W series circuit> first driver 8 ohm> sedond driver has DVC of 1 ohm each (each in series makes it 2 ohm)...

1000W/10 ohms=100W... 200W going to the 8 ohm driver and 800W going to the DVC driver?

(my logic = 4x the impedance... 4 times less power???)

What's the formula to calculate how many watts each driver gets?

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Quote:
Originally Posted by jeff lam

It's been a while since I studied ohms law and voltage drops/power.

I have a question as to the formula for calculating how much power each driver gets in a series circuit if the impedance is different in each driver.

Say: 1000W into a series circuit> first driver is 8 ohm> second driver is 4 ohms...

Total Impedance is naturally 12 ohms but how is the power calculated? Would it go 1000 / 3 = 333W... 333W going to the 8 ohm driver and 666W going to the 4 ohm driver?

What about 1000W series circuit> first driver 8 ohm> sedond driver has DVC of 1 ohm each (each in series makes it 2 ohm)...

1000W/10 ohms=100W... 200W going to the 8 ohm driver and 800W going to the DVC driver?

(my logic = 4x the impedance... 4 times less power???)

What's the formula to calculate how many watts each driver gets?

OK, you start with the power equation > P = I*I*R. Do this for the TOTAL load. For your 1000W example, this would mean that the current is 10A (Square root of 100A).

Keep in mind that current is constant in a series circuit, i.e., each driver will see the same exact mount of current (assuming no parallels).

Now, you apply the power equation again - P = I*I*R - but do it for each individual load. Since you know the current, you can now figure out the power in each driver individually. In this case the power for the 2 Ohm load would be 200W (2 Ohms * 10 A * 10A), and 800W for the 8-Ohm load. You can also use this to determine how much total current and Voltage is being drawn from the amp; use this to determine how close you are to clipping the amp's output stage.

The same basic principle applies for a parallel circuit, but you use Ohm's law - V = I*R - to determine the total voltage drop for each paralleled group. Then you can use V=I*R to get the current through each individual driver in the group; the power equation gets youback to Watts, since you now know both Volts and Amps.

Hope this helps.....
Why are you putting a 4 Ohm and 8 Ohm speaker in series?
With 2 different speakers in series, the frequency response will get strange.
Quote:
Originally Posted by Kevin Graf

Why are you putting a 4 Ohm and 8 Ohm speaker in series?
With 2 different speakers in series, the frequency response will get strange.

Well, I want to wire in 2 bass shakers with my sub. The shakers are 4 ohms each and can be 2 or 8 ohms total. I'm just trying to figure out how to wire everything up so my sub gets most of the power and the shakers get less.

Quote:

OK, you start with the power equation > P = I*I*R. Do this for the TOTAL load. For your 1000W example, this would mean that the current is 10A (Square root of 100A).

Keep in mind that current is constant in a series circuit, i.e., each driver will see the same exact mount of current (assuming no parallels).

Now, you apply the power equation again - P = I*I*R - but do it for each individual load. Since you know the current, you can now figure out the power in each driver individually. In this case the power for the 2 Ohm load would be 200W (2 Ohms * 10 A * 10A), and 800W for the 8-Ohm load. You can also use this to determine how much total current and Voltage is being drawn from the amp; use this to determine how close you are to clipping the amp's output stage.

The same basic principle applies for a parallel circuit, but you use Ohm's law - V = I*R - to determine the total voltage drop for each paralleled group. Then you can use V=I*R to get the current through each individual driver in the group; the power equation gets youback to Watts, since you now know both Volts and Amps.

Hope this helps.....

I thought the less resistance you have the more power... It makes sense that an amp would supply more power with lower impedances so I thought there would be more power going to the lower impedance driver... no?

P=I*E right? Less voltage drop across a lower impedance driver means higher voltage correct? So shouldn't there be more power across the 2 ohm driver as opposed to the 8 ohm?
Quote:
Originally Posted by jeff lam

Well, I want to wire in 2 bass shakers with my sub. The shakers are 4 ohms each and can be 2 or 8 ohms total. I'm just trying to figure out how to wire everything up so my sub gets most of the power and the shakers get less.

I'm not so sure I'd recommend that. The impedance of drivers varies by frequencies. If you wire drivers with different response characteristics in series, the response of one affects the other. The response of the Auras is VERY peaky, around 42hz. I think such a plan could have a negative affect on the sound quality of your sub. On top of that, the Auras don't have a whole lot of room between the how much they need to be felt, to how much they need to hit their excursion limits. The chances of you finding the right balance by luck with some very large adjustment steps (like 4 or 8 ohms) is probably slim. You'd probably be much better off finding a cheap used receiver, or a plate amp that you can modify the HP filter in, and powering them separately.

Quote:

P=I*E right? Less voltage drop across a lower impedance driver means higher voltage correct? So shouldn't there be more power across the 2 ohm driver as opposed to the 8 ohm?

Well, no, less voltage drop is less voltage drop. If a driver had a ZERO impeadance (i.e. a shorted coil), then the voltage at it's termals would be the the same on the + & - leads. Essentially, no voltage drop. As you increase impedance, voltage differential would increase. Since the power consumed by the driver is volts * amps, and since the amps are the the same if the drivers are in series, then the driver with the highest voltage drop is going to be consuming the highest power.

It only SEEMS backwards because we associate low impedance with higher power, because lower impedance allows more current from the amp. But since the current is the same across both drivers, it's the voltage difference across the driver that is left to determine power.
I have a related question regarding connecting in series. I have two identical Speakers(From a Klipsch Quintet) of 4 ohms each. Can they be connected in series to get the Nominal impedance requirement of my Receiver which is 6+ ohms. I am assuming that connecting 4ohm to the receiver might heat up the receiver. So I am trying to avoid it. Any suggestions would be appreciated.
Quote:
Originally Posted by audio_newbee

I have a related question regarding connecting in series. I have two identical Speakers(From a Klipsch Quintet) of 4 ohms each. Can they be connected in series to get the Nominal impedance requirement of my Receiver which is 6+ ohms. I am assuming that connecting 4ohm to the receiver might heat up the receiver. So I am trying to avoid it. Any suggestions would be appreciated.

Yes, you can connect two 4 Ohm speakers in series to create an 8 Ohm load. Make sure you fully understand how a series circuit works; it will take 3 single wires instead of 2 pairs:

1) The Positive (red) lead from the amp goes to the Positive (red) lead of speaker 1.
2) The Negative (black) lead from speaker one goes to the Positive (red) lead of speaker 2.
3) The Negative (black) lead from speaker 2 goes to the Negative (black) lead of the amp.

Make sense?
Yes, it makes sense and thanks for your response. Will there by any Frequency reponse issues? I hope not, since the drivers are identical.
Quote:
Originally Posted by audio_newbee

Will there by any Frequency reponse issues? I hope not, since the drivers are identical.

You might cause some phasing issues with placement. If it sounds weird or 'off' just move one or the other around until it sounds right. [If you have this issue, you'll hear it, and you'll understand immediately.]
Thanks again for your response. This my plan for laying out the speakers.

FR - Use a pair of Klipsch Quintet(4 ohm) connected serially.
FL - Use a pair of Klipsch Quintet(4 ohm) connected serially.
Center - Use the center speaker which came with the Quintet(8 ohm). I also have Polk RM202. I have not decided which to pick.
Surrounds - Pair of Polk RM101
Rear - Pair of Polk RM201
SVS PB10 NSD sub-woofer

I have a Denon 2307CI to go with it.

Any suggestions - Will it work out? Is there a better configuration with
what I have? Any inputs will be much appreciated.
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