Why is power additive in the AC circuit

Forum: Analog electronics and circuit technology Electrical power with series and parallel connection of resistors





Hello, I am totally confused by this solution: http://www.dl9hcg.a36.de/loesungenA/TD115.pdf Why does he use 10V when connected in series? Aren't there 3.33V across each resistor? No matter how I calculate it, I come up with the 3W with the parallel connection, but me with the series connection. Isn't the total power with the series connection P = square (U) / R = 100V / 300Ohm = 0.3W Also the total current with the series connection would be: I = U / R = 10V / 300 Ohm = 0.033 A Resistance: P = square (I) * R = square (0.033 A) * 100 Ohm = 0.1 W So again in total 0.3 W. What am I doing wrong?



Markus wrote:> No matter how I calculate it, I come to the 3W with the parallel connection,> but me with the series connection. Not with the same Tension. But it is, if every resistance is up to its own Load limit is maxed out by 1W. Because the question is like this:
1What load capacity can the interconnection of three equal
2Achieve resistances with an individual load capacity of 1 W each,
3if all 3 resistors are connected either in parallel or in series?
You don't have to do anything Arithmetic, but only Think : with 3 resistors with a maximum of 1W each one can burn a maximum of 3W. Regardless of any interconnections ...

: Edited by the moderator

The doc is stupid. Very simple: 3x100 Ohm total resistance at 10V -> P = U² / R -> 10² / 300 = 0.33W Edit: Arrgh, not even read. :-(

: Edited by user
by Esmeralda P. (Company: private) (max707)


Lothar M. wrote:> You don't have to calculate anything, just think ... Because if you are already calculating, then you have to calculate correctly.


Esmeralda P. wrote:> Lothar M. wrote: >> You don't have to calculate anything, just think ... >> Because if you are already calculating, then you have to calculate correctly. Right, the math in this doc sucks.



Andreas B. wrote:> The doc is dumb-talk. The good guy somehow bent the numbers so that he came up with a plausible result. And this solution is somehow being sold now ...



Markus wrote:> I am totally confused by this solution: >> http://www.dl9hcg.a36.de/loesungenA/TD115.pdf That is not surprising, because there is also Quark. > Why does he use 10V when connected in series? Yeah, that's wrong. It should be 30V. > Are there not 3.33V at every resistor? Correct. > [...]> What am I doing wrong? Nothing. The core statement in the linked PDF that the power is additive is correct - but if the partial resistances are the same (e.g. 100 ohms), then the total impedance is of course different, depending on whether you connect in series or in parallel. When connected in parallel, the same voltage is applied to all partial resistors and the current is divided; When connected in series, the same current flows through the partial resistors and the voltage is divided. Parallel connection gives 10V * 0.3A = 3W total power; Series connection gives 30V * 0.1A = 3W total power. HTH


Lothar M. wrote:> You don't have to calculate anything, just think: with 3> resistors with a maximum of 1W you can burn a maximum of 3W.> Regardless of any interconnections ... This is probably the stupidest and most confusing example I've seen have! : DA eg. 1W resistor can always withstand a maximum of 1W, regardless of whether there are more consumers (resistors) attached to the power source ... We were never explicitly told that ... why?

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