Question

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.

Decrease, decrease
Decrease, increase
Increase, increase
Increase, decrease

Answer 1

Increase, increase

Step-by-step explanation:

For a light bulb, voltage, resistance and current follows the Ohm's law:

`V=RI`

where

V is the voltage

R is the resistance

I is the current

In this problem, the voltage is fixed: V=120 V. We can rewrite the equation as

`I=(V)/(R)` (1)

we see that the current is inversely proportional to the resistance: so, if we decrease the resistance of the filament, the current will increase.

Moreover, the power dissipated by the bulb is

`P=I^2 R`

which can be rewritten using (1) as

`P=(V^2)/(R)`

We see that the power dissipated is inversely proportional to the resistance: therefore, if we decrease the resistance of the filament, the power dissipated will increase.