Question

In Europe the standard voltage in homes is 220 V instead of the 120 V used in the United States. Therefore "100-W" European bulb would be intended for use with a 220-V potential difference. (a) If you bring a "100-W" European bulb home to the untied States, what should be its US power rating? (b) How much current will the 100-W European bulb draw in normal use in the United States? [Answers: 29.8 W, 0.248 A]

The answers are provided. I need the work showing how to get to the answers.

Answer 1

Concepts and reason

The concept required to solve the given problem is electric power.

Firstly, calculate the resistance of the electric bulb using the formula of electric power in terms of resistance and voltage.

Then, calculate the US power rating of the 100 W European bulb.

Lastly, calculate the current rating of the bulb in US with the help of formula for electric power in terms of voltage and current.

Fundamentals

The electric power is given by,

$P = VI$

Here, $V$ is the voltage and $I$ is the current.

Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference across its ends. Mathematically the law may be written as,

$\begin{array}{c}\\I \propto V\\\\V \propto I\\\\R = \frac{V}{I}\\\end{array}$

Here, $V$ is the potential difference, $R$ is the resistance, and $I$ is the current.

The electric power may also be written as,

$P = \frac{{{V^2}}}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

(a)

The standard voltage in homes in Europe is,

${V_{{\rm{Europe}}}} = 220{\rm{ V}}$

The standard voltage in homes in US is,

${V_{{\rm{US}}}} = 120{\rm{ V}}$

Power of the bulb is,

$P = 100{\rm{ W}}$

The electric power is given by,

$P = \frac{{{V^2}}}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

The resistance of the bulb remains same irrespective of the places where it would be used.

For the bulb being used in Europe, the resistance of the bulb will be given by,

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

Rearrange the above equation.

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

Substitute $100{\rm{ W}}$ for ${P_{{\rm{Europe}}}}$ and $220{\rm{ V}}$ for ${V_{{\rm{Europe}}}}$ in the above equation.

$\begin{array}{c}\\R = \frac{{{{\left( {220{\rm{ V}}} \right)}^2}}}{{100{\rm{ W}}}}\\\\ = 484{\rm{ }}\Omega \\\end{array}$

The power rating of the bulb in US should be,

${P_{{\rm{US}}}} = \frac{{{V_{{\rm{US}}}}^2}}{R}$

Substitute $120{\rm{ V}}$ for ${V_{{\rm{US}}}}$ and $484{\rm{ }}\Omega$ for $R$ in the above equation.

$\begin{array}{c}\\{P_{{\rm{US}}}} = \frac{{{{\left( {120{\rm{ V}}} \right)}^2}}}{{484{\rm{ }}\Omega }}\\\\ = 29.8{\rm{ W}}\\\end{array}$

(b)

The current drawn by a $100{\rm{ W}}$ European bulb in US can be found by the equation,

${P_{{\rm{US}}}} = {V_{{\rm{US}}}}I$

Rearrange the above equation.

$I = \frac{{{P_{{\rm{US}}}}}}{{{V_{{\rm{US}}}}}}$

Substitute $29.8{\rm{ W}}$ for ${P_{{\rm{US}}}}$ and $120{\rm{ V}}$ for ${V_{{\rm{US}}}}$ in the above equation.

$\begin{array}{c}\\I = \frac{{29.8{\rm{ W}}}}{{120{\rm{ V}}}}\\\\ = 0.248{\rm{ A}}\\\end{array}$

Ans: Part a

The US power rating of the 100 W European bulb will be $29.8{\rm{ W}}$ .

Part b

The current drawn by a 100 W European bulb is $0.248{\rm{ A}}$ .