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A lightbulb 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.

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thisisbenmanley

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).