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Suppose we add another cell and connect it to the same bulb shown(Two cells connected to the same bulb).

From the second cell perspective, it sees a resistor and the same chemical reaction should take place with in the cell, and it should try to push the same amount of charge per second ? If so, the current through the bulb should double ? This should violate the ohm's law (I=E/R) ?

I know that ohm's law cannot be violated. So bulb doesn't get any brighter. But how does the second cell know about the presence of the first cell in order to slow down its reaction rate ?

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    $\begingroup$ The second cell notices that pushing electrons through the wire suddenly became twice as hard. $\endgroup$ Jun 1, 2017 at 7:59
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    $\begingroup$ May I know how it notices, as it sees the same resistor ? $\endgroup$
    – Hiiii
    Jun 1, 2017 at 8:10
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    $\begingroup$ True, but there is also a bunch of extra electrons trying to get through. Or let's put it this way: the resistor is the same, so is the voltage, hence so is the current, but somebody (the second cell) is stealing half of the current, so it doesn't reach us (the first cell). $\endgroup$ Jun 1, 2017 at 8:26

3 Answers 3


It doesn't push charge. You are visualizing it wrong.

It sets up an electric field with a certain voltage. If you add another cell in parallell, it does nothing for the bulb, it already was under the influence of an electric field of similar strength. The original cell will "slow down", and your new cell will "speed up" since there are miniscule differences in the cell-internal resistance which will balance out when they are sharing the load (literally, and figuratively. I love when sentences turn out like that)

And Ohms law can most certainly be violated, it is violated every single day by any equipment using alternating current (AC).

  • $\begingroup$ it is violated every single day by any equipment using alternating current I have to disagree. Ohms law, in it's purest state is: $$j=\sigma E$$ where $j$ is current density and $E$ is electric field. $\sigma$ is conductivity. In a more simplified way, it would be: $$V=iR$$ In an alternating current, if you consider rms values, it seems to be violated, but it's followed at every single instant where ohmic resistors are present. Rms values are just averages, you cannot use them to generalise the law. Ohm's law is only violated when non-ohmic devices are used. $\endgroup$ Jun 3, 2017 at 4:04
  • $\begingroup$ Yes, and everywhere AC is being conducted, you have non ohmic behaviour... Because no devices are purely ohmic. $\endgroup$
    – Stian
    Jun 3, 2017 at 12:19
  • $\begingroup$ Because no devices are purely ohmic are you attempting to conclude that Ohm's law is not at all valid? Your argument is not just for AC, but also for DC. $\endgroup$ Jun 3, 2017 at 15:43
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    $\begingroup$ Yes, that is basically what is the conclusion of that. You can of course, construct a complex version of Ohm's law with impedance (and a phase relationship variable for AC) which would be much more correct. It is fairly important to realize the limitations of Ohm's law, and that is why I state that it is broken every single day. You can improve by translating to phasors and whatnot, but the point remains, Ohm's law is a simplification and should not be taken as any sort of truth about the natural world. It $\endgroup$
    – Stian
    Jun 5, 2017 at 5:56

I don't believe it is correct to state that another cell "does nothing for the bulb". Before I explain, I need to admit that this (simple DC circuits, batteries) is not an area I'm particularly competent in - I'm relying on my common sense and my decades and decades and decades old University studies. If it were true, then connecting 1000 identical cells would "do nothing", but I believe the filament would burn out in an instant. (I could be wrong here!) The place where you are clearly mistaken is your assumption that the resistance is constant. For a metal conductor, resistance is an increasing function of temperature - higher temperature, more resistance. This means that doubling the current flow will heat up the filament and increase resistance. So, one question you need to ask yourself is: is light intensity (brightness) proportional to work done? And are your eyes able to discern a increase in intensity? I don't know, but I suggest that since our eyes are can sense a single photon (under optimum conditions) and can also function under lighting conditions far more intense than the noon desert Sun, that its more of a log(I) measurement device than an I measurement device (I=intensity). My experience is that as you increase the current to a filament (starting from zero) at a fixed voltage the bulb will brighten until it reaches a "plateau" and little increase in brightness is noted (some of this is almost certainly my eyes adjusting to the brightness increase). I note some increase but not as dramatic as its ramp up to "full" brightness. If I continue to increase current, not much happens until pop! the filament fails (possibly catastrophically). So, evaluate this answer and see what others say about it. If it gets a lot of down votes, well I warned you. You know what they say about touching a live electrical wire, right? It's not the voltage that kills you, it's the current. Same idea with the filament. Power = IE. Work is power times duration. (I should also note that the relationship between resistance and temperature is more complicated that I care to discuss here).


Ideal case (internal resistance of the cell is zero):

The E.M.F across the cell does not change according to the external circuit. If you assume that the cell has no internal resistance, the potential drop across the cell is equal to its E.M.F in a closed circuit.

The potential drop across the bulb in the first case (one bulb connected to the cell) is equal to the potential drops across the bulb in the second case (two cells in parallel). This is because the bulbs are connected to the same ends of the cell and the potential difference across the cell does not change.

The power consumption of any bulb is given by:

$$P = VI$$

If the device obeys Ohm's law, the above equation can be written in terms of $V$ and $R$ as follows: $$P = \frac{V^2}{R}$$

As the resistance of the bulb is a property of the bulb and does not change in the two cases, the power consumed by each of the two bulbs is the same.

Therefore, you shouldn't expect any appreciable increase in brightness.

Non-ideal case:

If internal resistance of the cell was taken into account, the potential drop acros the two bulbs would decrease (net resistance in the circuit decreased as the two internal resistances are now in parallel, current increased, net potential drop across the internal resitance increased and therefore, the potential drop across the cell decreased) and hence, they would glow less brightly than the bulb in the first case.

  • $\begingroup$ Unless I'm missing something, the question is about multiple batteries with a single bulb, but your answer addresses the case of a single battery with multiple bulbs. $\endgroup$
    – airhuff
    Jun 1, 2017 at 19:22
  • $\begingroup$ I edited the answer. Thanks for pointing out. $\endgroup$
    – Yashas
    Jun 2, 2017 at 3:22

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