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I conducted a simple experiment today in which I tried to determine the power dissipated by each individual bulb when another bulb is connected in parallel. But, I noticed something unusual which goes against what I learned about Ohm's law. For this experiment, I used two 6v 0.1 DC filament light bulbs.

When I added a second identical bulb in parallel to the first one, the total resistance of the circuit halves. So, I expected the total current to double as $I = V/R$. The voltage decreased slightly as expected. However, the current more than quadrupled, which was not what I expected. The result was that the power dissipated by the original bulb increased when another identical bulb was connected in parallel to it. Can someone explain why this is?

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  • $\begingroup$ That's two identical bulbs? $\endgroup$
    – user1583209
    Feb 2, 2017 at 16:37
  • $\begingroup$ Yes, I should have added that. $\endgroup$
    – s.xw
    Feb 2, 2017 at 16:39
  • $\begingroup$ AC or DC? What kind of bulb, LED, traditional filament etc. $\endgroup$
    – JMLCarter
    Feb 2, 2017 at 17:08
  • $\begingroup$ @JMLCarter 6v 0.1A DC filament light bulb. $\endgroup$
    – s.xw
    Feb 2, 2017 at 17:11
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    $\begingroup$ Yes. On the 10 A range, it might not be very accurate measuring currents below 1 A. Try setting it on the lowest range that fits all the measurements (so 0.2 or 0.5 or 1 A full scale) and repeat the measurements. $\endgroup$
    – The Photon
    Feb 2, 2017 at 22:54

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Without more numerical information it is only possible to suggest what is happening.

The resistance of a working $60\, \rm W,\, 230 \, V$ light bulb is approximately $880 \, \rm \Omega$ whilst when cold it is about $70 \, \Omega$.
So the resistance of a light bulb does change a lot with the voltage across (current though) it.

The voltage decreased slightly as expected. Probably indicates the the power supply, voltage $V$, has an internal resistance $X$.

Without reasonable accurate figures the next bit is speculation but it does show that the current through the bulbs can increase by an "unexpected" amount.

Given that the current through one bulb was $0.04\, \rm A$ which is a long way from the rated current of $0.1\, \rm A$ I will assume/guess that the voltage across it was much less than its rating of $6 \, \rm V$.

For one bulb the current through it was $0.04 \, \rm A$ and assume that the voltage across it was $3\, \rm V$ (a guess) and when two bulbs were in parallel the current through one bulb was $0.09 \, \rm A$ and assume that the voltage across them was $1\, \rm V$ (another guess to make the number work).

enter image description here

Using these values one finds that the resistance of one light bulb when the current was $0.04\, \rm A$ was $75\,\rm \Omega$ and the resistance of one light bulb in parallel with another when the current was $0.09\, \rm A$ was $11\,\rm \Omega$.
The voltage of the supply was $3.6 \, \rm V$ and its internal resistance was $14 \,\Omega$.

Importantly, the power dissipated in one bulb when alone was $0.12 \,\rm W$ and when in parallel with another was $0.09 \,\rm W$.
This lowering of the power dissipated in the bulb with another one in parallel with it would result in a lower filament temperature and hence a lower resistance.

So perhaps there is scope for further analysis of the data already obtained but also further experimentation to include more than two bulbs in parallel?
However after reading the comments it sounds as though some more sensitive meters should be used?

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  • $\begingroup$ Hi, thanks for the answer. "This lowering of the power dissipated in the bulb with another one in parallel with it would result in a lower filament temperature and hence a lower resistance." Are you saying that, because there is a lower filament temperature and hence a lower resistance, the current increases? But if current increases, then surely the power must increase as well? I am sorry I did not provide figures; I thought that there was a common explanation to this. $\endgroup$
    – s.xw
    Feb 4, 2017 at 16:28
  • $\begingroup$ I don't have exact figures with me, but the voltage recorded on the voltmeter with one bulb was around 3.6v (I did set it to 6v on the variable power supply). When another bulb was added, voltage dropped to around 3.5v but as I said total current increased from 0.04A to 0.18A and the current of each bulb increased from 0.04A to 0.09A. Using V = IR, it seems as if the resistance of the original bulb decreased by about three quarters. Does this match what you are saying? $\endgroup$
    – s.xw
    Feb 4, 2017 at 16:34
  • $\begingroup$ Also, when I added more bulbs in parallel, the voltage dropped by about 0.1v each time and the current reached a point where, after adding another bulb in parallel, it started to decrease and go below 0.04A for each bulb when I divided total current by the number of bulbs. $\endgroup$
    – s.xw
    Feb 4, 2017 at 16:39
  • $\begingroup$ Your data is at variance with what I wrote because the power dissipated in each of the bulbs which are in parallel is greater than for one bulb alone. So the contradiction is that with a higher power dissipation the resistance has dropped. How were the ammeter and voltmeter arranged? Was the voltmeter across the bulbs of bulbs and ammeter? $\endgroup$
    – Farcher
    Feb 4, 2017 at 16:41
  • $\begingroup$ Are you sure the bulbs were similar? Did you check that each of them had the same one bulb characteristic? $\endgroup$
    – Farcher
    Feb 4, 2017 at 16:43

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