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enter image description here

In this picture, as we know power due to heating of the resistance is $V^2/R$ .

Now, if we add a parallel resistance to the circuit, the power dissipated through this resistor is also $V^2/R$ because the voltage applied at the end of each resistor is the same. So, if we proceed further, the total power dissipation becomes very large if a large number of parallel resistance is added, which violates the power rating of the source. How can we explain this?

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closed as unclear what you're asking by Aaron Stevens, ZeroTheHero, Jon Custer, Kyle Kanos, Chair Feb 8 at 13:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Eventually the power source will not be able to provide the needed current. Can you explain what you are having trouble understanding? $\endgroup$ – garyp Feb 5 at 14:48
  • $\begingroup$ @garyp Based on comments in my answer, it looks like the OP meant to ask a different question $\endgroup$ – Aaron Stevens Feb 5 at 15:14
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Considering the comments/chats you have with @Aaron_Stevens I interpret your question as to what limits the current of a voltage source, or what limits the voltage of a current source. There are no real voltage or current sources. These concepts are idealizations (models) of real electromagnetic energy supplies that are easy to apply mathematically when the laws of Kirchhoff are applied.

A real battery, for example, is neither an ideal voltage nor an ideal current source. It is a monster electrical, chemical, thermal, mechanical, etc., system for which we ignore almost everything and assume that while we are using it the voltage at its outside accessible terminals and the current through them can be modeled if it was as an ideal voltage source in series with an ideal linear resistor. In the usual model, this ideal voltage source has a voltage drop between its internal terminals (these are inaccessible to us for it is inside the battery) is a known value and it is independent of any thing we do when we connect the outside (accessible) terminals to our circuit. We also usually assume that this internal voltage drop is known at any instant of time and usually taken as a constant, but that is not very important.

The only thing you are not allowed to do with an ideal voltage source is to connect its terminals with a zero resistance wire, otherwise everything is allowed and it then can supply arbitrary amount of currents. To prevent inadmissible connections and getting into contradictions with Kirchhoff's laws, e.g., infinite current, we must have some positive resistance attached to it and the real internal resistance of a real battery will do just fine for this purpose. In fact, the maximum current through the model will be then the ratio of the open circuit voltage drop and the internal resistance. Anything else you hang on it will produce lower current.

A 1Ah battery, say, will deliver into the circuit to which it is connected over an hour 3600Cb, an enormous amount of charge if it was freely available. These charges are not stored statically in the battery but instead are produced continuously by the internal electro-chemical process. When we are using Kirchhoff's laws to calculate what is going on within the circuit connected to the battery we completely ignore all that electro-chemistry and just assume that the voltage is constant as long as the current draw does not exceed some limit.

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Your issue makes complete sense. You can look at it from multiple perspectives, although they are all related.

1) If you are familiar with how resistance adds in parallel, then (using the simpler case of adding resistors with the same resistance each time) we have for the total (or effective) resistance of $n$ resistors in parallel $$R_T=\frac Rn$$

So, as $n$ becomes larger, the effective resistance drops, and therefore if we assume a constant voltage $V$ the circuit will be using more current, and thus a larger power dissipation

2) This is essentially where the above parallel resistance equation comes from, but the more resistors you have at the same potential difference, the more current you will be pulling through the circuit. For example, since each resistor will have a current $I=\frac VR$ going through it, then the total amount of current being "pulled from the battery" is going to be $$I_T=nI=n\frac VR$$

Essentially by adding more resistors in parallel you are trying to get more from the power source. This is a great example as to why you don't want to plug so many things into outlets, since you are essentially adding more and more "resistors" in parallel with the power supply.

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  • $\begingroup$ Sir if we use this power or stored this power any way by our advanced technology then there is a big source, we do not need a power plant and also what do we mean by watt ratting of a source. it shows that power absorbed from a source depend on my adding resistance.there is no maximum power rating of a source. $\endgroup$ – abhijit halder Feb 5 at 14:26
  • $\begingroup$ @abhijithalder Yes, power absorbed by a source definitely depends on what system of resistors is hooked up to it. For example, if I have no resistors hooked up to my battery, then my battery is not outputting any power at all. The issue is that in the real world you can't pull an unlimited amount of energy at an very high rate from power supplies. There are limits. If you ask too much of your power supply, it will break down, or at the least run out of energy to give. $\endgroup$ – Aaron Stevens Feb 5 at 14:29
  • $\begingroup$ but why there is this type of limit. Is there any physical reason behind it. $\endgroup$ – abhijit halder Feb 5 at 14:33
  • $\begingroup$ @abhijithalder The comments of an answer is not the place to ask new questions and to answer new questions. If you have a new question, please ask a new, separate question. The comments on this answer are to have the purpose of clarifying anything in my answer that pertains to the original question above. $\endgroup$ – Aaron Stevens Feb 5 at 14:35
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    $\begingroup$ @abhijithalder I am not going to do that here. Like I said that is a separate question along the lines of "What causes the watt rating of a power supply". If you want to know the answer to this question, just please ask a new question. $\endgroup$ – Aaron Stevens Feb 5 at 15:07

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