seires / parallel wattage question

[R.G.]

If there's anything I can't stand, it's someone with no tolerance!!

[the fatch]

Two 8 ohm 100 watt dummy loads. I think when in parallel it makes 4 ohm 200 watt, but about in 16 ohm series mode?

Late to the table as usual, but here’s my two centsworth in answer to your original question. In parallel the total resistance is half that of each 8Ω resistor, i.e. 4Ω. In series, the total resistance is twice that of each 8Ω resistor, i.e. 16Ω. Power output is the product of the output voltage and output current. In parallel the whole voltage appears across both resistors but the current is shared equally between them. In series the whole of the current flows in both resistors but the voltage is shared equally. The whole voltage and half the current results in the same power as half the voltage and the whole current. In parallel or series each resistor dissipates half the applied power. If the amp was outputting 4A at 50V i.e. 200W, the series resistors each see 4A x 25V = 100W, a total of 200W. Parallel resistors each see 2A x 50V = 100W, a total of 200W.

[Chuck H]

If there's anything I can't stand, it's someone with no tolerance!!

(for other cultures)

And the Dutch

[the fatch]

If it is my original question some are unsure of, I apologise if I was unclear. I know a 100 watt resistor is a 100 watt resistor all day, that seems obvious enough. I was thinking in terms of dummy loads as well as speakers in cabinet combinations. I am looking at building a dummy load array, and wanted to know what the power handling ability would be for the different series/parallel pairings of 100 watt resistors. Also, was wondering about the series/parallel wiring in a 16 ohm Marshall cab loaded with four 16 ohm 25 watt greenbacks connected to a 100 watt Marshall head. By the look of RG's answer, whether wired in series or parallel, those four 25 watters are handling 25 watts each for a total of 100 watts. Right?

A pair of 25W 16Ω speakers wired in series becomes a 32Ω 50W speaker. Another pair wired in series becomes another 32Ω 50W speaker. The two pairs wired in parallel become an 16Ω 100W speaker. Alternatively, a pair of 25W 16Ω speakers wired in parallel becomes an 8Ω 50W speaker. Another pair wired in parallel becomes another 8Ω 50W speaker. The two pairs wired in series become an 16Ω 100W speaker. Either way, each speaker is handling 25W.

[the fatch]

One wrench in the works is if the resistances are different values. Then you can't just add the wattages, you have to do more calculations. (ex. 4 ohm 100W in series with 8 ohm 100W does not give 200W power handling)

If a pair of different value resistors of the same power rating are connected in series, the power handing of the pair is determined by the power rating of the larger value resistor. Consider a 4Ω 100W resistor in series with an 8Ω 100W resistor. Assume an input voltage of 48V and a current of 4A, i.e. 192W. The voltage across the 4Ω is 16V and the voltage across the 8Ω is 32V. The 4Ω is dissipating 16V x 4A = 64W. The 8Ω is dissipating 32V x 4A = 128W. This means that voltage or current must be reduced if the 8Ω is to be within its power rating. If the current remains at 4A, the voltage across the 8Ω must be reduced to 25V (4A x 25V = 100W), so the input voltage must be reduced to 37.5V. The maximum permitted dissipation of the pair of resistors in series is 37.5V x 4A = 150W.

I thought about it some more and came up with:
If 1A were flowing in R1, a 4Ω 25W resistor and R2, an 8Ω 25W resistor connected in series, then a voltage of 25V across R2 would result in its maximum permitted dissipation. The total resistance is 8 + 4 = 12Ω. The voltage across the two resistors would be 25V x 12 / 8 = 37.5V. 37.5V x 1A results in a total power rating of 37.5W. The power rating of two unequal value resistors with the same power rating where R2 is the greater resistance is given by:
P(total) = P(R2) x (R1 + R2) / R2

[Dave H]

If a pair of different value resistors of the same power rating are connected in series, the power handing of the pair is determined by the power rating of the larger value resistor. Consider a 4Ω 100W resistor in series with an 8Ω 100W resistor. Assume an input voltage of 48V and a current of 4A, i.e. 192W. The voltage across the 4Ω is 16V and the voltage across the 8Ω is 32V. The 4Ω is dissipating 16V x 4A = 64W. The 8Ω is dissipating 32V x 4A = 128W. This means that voltage or current must be reduced if the 8Ω is to be within its power rating. If the current remains at 4A, the voltage across the 8Ω must be reduced to 25V (4A x 25V = 100W), so the input voltage must be reduced to 37.5V. The maximum permitted dissipation of the pair of resistors in series is 37.5V x 4A = 150W.

To simplify it you could say the current is the same for both resistors therefore power is proportional to resistor value so when the 8R resistor is dissipating 100W the 4R resistor must be dissipation 50W therefore max permitted dissipation for the pair is 150W.

[Dave H]

You could also do something similar for two 100W, 8 and 4 ohm resistors in parallel. Parallel resistors have the same voltage therefore power is inversely proportional to resistor value. When the 4R resistor is dissipating 100W the 8R resistor must be dissipating 50W so the combined max power is again 150W.

[nevetslab]

Advanced "trick" question:
Twelve 1-ohm resistors are configured so that each resistor is the edge of a cube.
What is the resistance between diagonally opposite corners, i.e. corners that are points on a line that passes through the cube's center?

Trick: Inject an imaginary 1 amp current at one corner and calculate the voltage drops across each resistor- noting that due to symmetry, the current will split equally at each node.

-rb

Every time I re-start the thinking process on this, knowing I didn't go far enough in the math/theory classes, I don't have the mathematical tools to work this one out. Taking two of the 1 ohm 'vertical' members out of the cube, it's easy, and you'd get 1 ohm from the opposite diagonal corners. But, WITH those two members back in place, my head starts hurting! I notice nobody else has yet replied to this advanced 'trick' question. It's a bit beyond my skills still.

[rjb]

Every time I re-start the thinking process on this, knowing I didn't go far enough in the math/theory classes, I don't have the mathematical tools to work this one out. Taking two of the 1 ohm 'vertical' members out of the cube, it's easy, and you'd get 1 ohm from the opposite diagonal corners. But, WITH those two members back in place, my head starts hurting! I notice nobody else has yet replied to this advanced 'trick' question. It's a bit beyond my skills still.

I guess this is the piece of theory you need:
https://en.wikipedia.org/wiki/Kirchh...s_circuit_laws
At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. The algebraic sum of currents in a network of conductors meeting at a point is zero.

So:
Inject 1A into one corner. Make an educated guess that due to the symmetry of the configuration, the current leaving that node is going to split evenly in every direction. So (1/3)A travels through each of the 3 resistors touching that corner. The voltage drop across each of those resistors is V=I*R=(1/3)A*1ohm=(1/3)V. Then each of the 3 current paths makes a 2-way split, so the voltage drop across each of the next resistors in line is (1/6)V. Then pairs of current paths join, and the voltage drop across each of the 3 resistors touching the "opposite" corner is (1/3)V.

That sure could've been said a lot more clearly- but not by me. It'd be a lot easier to show with a diagram.

Anyways:
The voltage drop across any path between diagonally opposite corners is (1/3)V+(1/6)V+(1/3)V=(5/6)V
The current through the configuration (entering one corner and leaving the other) is 1A
So the resistance of the configuration is R=V/I=(5/6)V/1A=(5/6)ohm
I think.

-rb

PS: This really is a "trick" question that has no practical application I'm aware of.

[Enzo]

A lot of trick questions have no practical value, but they become a special case that forces you to use the correct reasoning to get the answer.

[SoulFetish]

It was a dead giveaway!

The "H" point impedance can be anything between a short circuit and an open circuit with no effect whatsoever on the other elements. The logic behind this is a special case of the logic for all bridge-type sensing circuits.

Exactly! Makes perfect sense.
.... but I’m still glad Dave H got to it before I did