Chapter 18, Problem 32P

To determine

The brightness of the bulb in Europe.

Answer to Problem 32P

  $\frac{1}{4}$ of the original brightness.

Explanation of Solution

Given info:

Power, $P=75W$

Voltage in Europe, $V_{Europe}=240V$

Voltage in USA, $V_{USA}=120V$

Formula used:

Expression for the power in terms of resistance and voltage:

  $P=\frac{V^2}{R}$

Where, $R$ is resistance, $P$ is power and $V$ is voltage.

Calculation:

It is known that the power is $P=\frac{V^2}{R}$ . The resistance is supposed to be constant in both of the given cases. Therefore, power is directly proportional to $V^2$ . It is also known that the power is directly proportional to the brightness of the lightbulb. Because of this increasing power will increase the brightness of the lightbulb. Also, increasing the voltage will increase the power and increasing the power will increase the brightness. Thus, increasing voltage will increase the brightness of the bulb.

According to this, the brightness of Europe’s lightbulb will be greater than USA’s lightbulb.

Now, find the relation between two brightness’s. For the find the ration of $\frac{P_{Europe}}{P_{USA}}$ .

  $\frac{P_{Europe}}{P_{USA}}=\frac{\frac{V_{Europe}^2}{R}}{\frac{V_{USA}^2}{R}}$

  $\frac{P_{Europe}}{P_{USA}}=\frac{V_{Europe}^2}{V_{USA}^2}$

Substitute the given values,

  $\frac{P_{Europe}}{P_{USA}}=\frac{{\left(240\right)}^2}{{\left(120\right)}^2}=\frac{57600}{14400}=4$

Therefore, $P_{Europe}=4P_{USA}$

Conclusion: The brightness in USA is one fourth of the original brightness.