Answer:
The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
Step-by-step explanation:
Brightness of the bulb is given as
![P= (V^2)/(R)](https://img.qammunity.org/2020/formulas/physics/high-school/yndcfv1vs4999iejoxfsieex4wdpdqp3if.png)
since all bulbs are connected in parallel so here voltage across each bulb will remain same and resistance of each bulb is "R"
So here power across each bulb will remain the same always.
So there will be no effect on the power or brightness of bulb.
Now we also know that equivalent resistance is given as
![(1)/(R_(eq)) = (1)/(R_1) + (1)/(R_2) + (1)/(R_3)............](https://img.qammunity.org/2020/formulas/physics/high-school/26x9d2xo6pxrmbek9xdd617xdaw1larn72.png)
![R_(eq) = (R)/(n)](https://img.qammunity.org/2020/formulas/physics/high-school/ymhpqzw8x36sgs6fodgfkyjcioco6t4ecm.png)
so here equivalent resistance will decrease on adding more resistance in parallel.
so correct answer will be
The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.