Why is it that sometimes current is not inversely proportional to resistance? 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?
 A: 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).  

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? 
