Voltage drops after power leaves the amplifier

[theMessenjah44 0]

Sorry my drawings are a little crude, but I'm taking an auto electrical class and some of our discussions got me thinking. I did a search and found nothing on it, but would like some input. I figured this would be a better place to post it than ca.com

With series wiring, once power has passed through a load, there is a voltage drop depending on all 3 factors, Voltage, amperage, and ohms of resistance.

With parallel wiring, each leg of the circuit gets the same voltage, only amperage and resistance fluctuate (they SHOULDN'T fluctuate differently in an audio setup, but if all coils are identical, theoretically, they all do the same thing)

Remember, these are all theory, but even with impedance rise, they should still hold true...if I'm correct.

We'll start with a simple parallel diagram

At the final resistance of 1 ohm.... we need to calculate amperage to determine power.

40V / 1 ohm = 40A of current

40V x 40 A = 1600 watts of total power

Each coil sees the full 40 volts from the amp, and provides a 2 ohm load, which would look like this

40V / 2 ohms = 20 Amps of current

20 A x 2 ohms = 40 volts of Voltage drop from each leg.

40V x 20 A = 800 Watts per coil

On to Series circuits

At a final resistance of 4 ohms

40V / 4 ohms = 10 Amps of total current

40V x 10 A = 400 watts of total power

On Coil 1

40V x 10 A = 400 Watts

10A x 2 ohms = 20 Volts of voltage drop

40V (original) – 20 Volts (Voltage drop from first coil) = 20 V remaining for the second coil

On Coil 2

20V x 10 A = 200 Watts

10 A x 2 ohms = 20 V of Voltage drop, leaving 0 to return to the battery.

I did one for Series-Parallel as well, but I think this Series one sums it up well. To me…it looks like when you wire in series, the first coil will get full power, every one after that loses power.

It looks to me like parallel is the way to go unless someone can find something wrong with what I’ve done.

[Adrian_D 520]

Whoa, some very wrong information posted here.

When wiring in parallel, each branch of the circuit will 'see' the same voltage and the sum of the currents on the branches is equal to the current supplied by the power supply. In your example, you calculated correctly for the parallel wiring.

Series wiring, the sum of the voltage drop across each coil is equal to the voltage supplied by the power supply and the current through each coil is going to be the same. In your example the correcte value is 20V and 10A through each coil.

There is no 'first in line' when working on these connections and no '0v returning to the battery'.

Trying not to sound harsh here but if this is how the auto electrical class is teaching you, then I feel very sorry for that school.

Start reading here : http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/ohmlaw.html#c1

[theMessenjah44 0]

Ok, this is from the link you posted. if I'm reading this wrong, would someone care to explain what's going on?

Also, there is 0 Volts returning to the battery. the link you posted even says so.

The diagram on the right shows 12V at the first load, dropping by 9 volts, leaving 3 volts at the second load?

How much voltage would be sitting at the second coil? I don't see how series and parallel would be the same.

Also...I'm at school to learn....I didn't claim I know it all yet. I just said we were discussing circuits and this caused me to think of this It's a class for ASE certification.

Edited September 16, 2009 by theMessenjah44

[theMessenjah44 0]

Also from the link you posted. Under the heading Voltage Law it states:

"The voltage changes around any closed loop must sum to zero. No matter what path you take through an electric circuit, if you return to your starting point you must measure the same voltage, constraining the net change around the loop to be zero."

In series circuits, current stays the same, but voltage drops after each load.

In parallel, Voltage stays the same, but current varies depending on resistance.

Am I right?

[Adrian_D 520]

Ok, this is from the link you posted. if I'm reading this wrong, would someone care to explain what's going on?

Also, there is 0 Volts returning to the battery. the link you posted even says so.

The diagram on the right shows 12V at the first load, dropping by 9 volts, leaving 3 volts at the second load?

How much voltage would be sitting at the second coil? I don't see how series and parallel would be the same.

Also...I'm at school to learn....I didn't claim I know it all yet. I just said we were discussing circuits and this caused me to think of this It's a class for ASE certification.

It is the voltage drop across a load, not after a load that we are interested in. In the first post you assumed that the first resistor has a voltage drop of 40v, exactly the voltage of the battery, which is wrong, as you can see in the picture you posted. 12v battery, 9v across the first resistor, 3v across the second.

In series circuits, current stays the same, but voltage drops after each load.

There is a drop across each component, and the first component it line will not 'see' all the voltage of the battery. Look at the circuit as a whole, not just going from the positive terminal of the battery to the negative and treating components independantly.

To make things easier, wire up something and start measuring voltages. Nothing beats the hands-on approach.

[bigrank916 72]

Those diagrams make my head hurt.

[Adrian_D 520]

The diagrams should be drawn differently. Chose a direction in the loop, write down all the equations and things are easier to understand than writing voltages on each corner of the diagram.

[ncacoates 104]

Adrian_D you nailed it on the head. Good info man.

Edited September 17, 2009 by MemphisMzd

[NB50 206]

The diagrams should be drawn differently.

Yeah the current is flowing the wrong way.

[theMessenjah44 0]

On Coil 1

40V x 10 A = 400 Watts

10A x 2 ohms = 20 Volts of voltage drop

40V (original) – 20 Volts (Voltage drop from first coil) = 20 V remaining for the second coil

On Coil 2

20V x 10 A = 200 Watts

10 A x 2 ohms = 20 V of Voltage drop, leaving 0 to return to the battery.

I never said it would be 40V of voltage drop on the first coil, that would only happen in parallel.

I'm still not completely understanding what you're saying....however I was planning on playing with my multimeter on my system tomorrow.

Thanks for the help

Edited September 17, 2009 by theMessenjah44

[Adrian_D 520]

On Coil 1

40V x 10 A = 400 Watts

10A x 2 ohms = 20 Volts of voltage drop

40V (original) – 20 Volts (Voltage drop from first coil) = 20 V remaining for the second coil

On Coil 2

20V x 10 A = 200 Watts

10 A x 2 ohms = 20 V of Voltage drop, leaving 0 to return to the battery.

I never said it would be 40V of voltage drop on the first coil, that would only happen in parallel.

I'm still not completely understanding what you're saying....however I was planning on playing with my multimeter on my system tomorrow.

Thanks for the help

If you never said the voltage across the first coil is 40v, then why did you calculate the power on the first coil using the voltage of the power supply ? Also, the way you said it first, the first coil 'sees' 400w and the second 200w, all out of a 400w supply.

This is how things should be calculated, not by following the drawing and interpreting one component at a time and ignoring the rest :

[stooch 0]

ALL the diagrams are wrong!!! None of them have the amplifier in them (set-up transformer/inverter) The amp does not have resistance, different equation altogether!

[BKOLFO4 29]

The diagram on the right shows 12V at the first load, dropping by 9 volts, leaving 3 volts at the second load

The loads are 2 ohms and 6 ohms in the diagram you posted, that is why the voltage drop is not even. If the resistors were the same size, it would be a 12 volt source with 6 volts dropping across the first resistor and 6 volts dropping across the second one.