Alightbulb is rated by the power that it dissipates when connected to a given voltage. for a lightbulb connected to 120 v household electricity, decreasing the resistance of the filament will the current through the bulb and the power dissipated by the bulb.

Here, we're going to make use of two important equations: V = IR and P = IV.

V = IR shows us that the product of current (I) and resistance (R) must equal voltage (V). If V is held constant at 120 (which it is in the question), then, if the resistance R decreases, the current I must increase in order to make the equation work (if V = IR and R goes down, but V must stay the same, I must go up to balance the change out).

P = IV shows us that the product of current (I) and voltage (V) must equal power (P). If V is held constant at 120 (which it is in the question), then, if the resistance R decreases, the power P must decrease (because P = IV, so if I goes down and V stays the same, P must go down).

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