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Serhud [2]
3 years ago
10

An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto

r is the diameter of the filament reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
Physics
2 answers:
HACTEHA [7]3 years ago
8 0

Answer:

Explanation:

Po = 60 W

P = 54.3 W

Let the initial diameter of the filament is do and the final diameter of the filament is d.

Let the voltage is V and the initial resistance is Ro and the final resistance is R.

The formula for power is given by

P = V²/R

The resistance of the filament is inversely proportional to the square of the diameter of the filament. As voltage is constant so the power is

Power α diameter²

So, initial power is

Po α do²     ..... (1)

Final power is

P α d²         ..... (2)

Divide equation (2) by equation (1), we get

P / Po = d² / do²

54.3 / 60 = d² / do²

d² / do² = 0.905

d = 0.95 d

Thus, the diameter of the filament is reduced to a factor of 0.95 .  

Lemur [1.5K]3 years ago
7 0

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

R\propto\dfrac{1}{D}

The power is inversely proportional to the resistance.

P\propto\dfrac{1}{R}

P\propto D^2

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}

Put the value into the formula

\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}

\dfrac{D_{i}}{D_{f}}=0.95

D_{i}=0.95 D_{f}

Hence, The factor of the diameter is 0.95.

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