Answer:d
Explanation:
Suppose V is the voltage of battery and R is the resistance of bulb
so Power drop for initial stage
[tex]P_1=\frac{V^2}{R}[/tex]
When another identical bulb of same resistance is applied in parallel so voltage Drop across both the resistor will be same i.e. V
so Power consumed [tex]P_2=\frac{V^2}{R}[/tex]
so there is no change in power and hence no dip in brightness