The power in a lightbulb is given by the equation p=i^2 r, where i is the current flowing through the lightbulb and r is the resistance of the lightbulb. what is the current in a circuit that has a resistance of 30.0 ω and a power of 55.0 w?

The power DISSIPATED BY a lightbulb is given by the equation P=I² R,
where I is the current flowing through the lightbulb and R is the resistance
of the lightbulb.

If a circuit has a resistance of 30 Ω and it dissipates 55 W of power in the form of heat or light that it never gets back, then ...

          P  =  I² R

        55 W  =  I² · 30 Ω 

            I²  =  55 W / 30 Ω 

            I   =  √(55/30)  =  √1.833   =   1.35 Amperes

Independent variable is something that is changed in a experiment.