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irina [24]
3 years ago
14

A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V house

hold 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
Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

Increase, increase

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=\frac{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=\frac{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.

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Answer:

a = 0.009 J

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Explanation:

Given that

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a

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Potential Energy, P = 1/2 * k * x²

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