Question

A North American tourist takes his \(25.0 - W\), \(120 - V\) AC razor to Europe, finds a special adapter, and plugs it into \(240{\rm{ }}V\) AC. Assuming constant resistance, what power does the razor consume as it is ruined?

Short answer

Value of the power consumed by the razor is \({P'} = 100\;W\).

Step-by-step solution

Relation between the potential difference and power.

The power, or rate at which energy is delivered to a circuit element, is\(P = I\Delta V\)if a potential difference\(\Delta V\)is maintained across the element.

We can represent the power applied to a resistor as\(\begin{align}{}P = {I^2}R\\P = \frac{{{{(\Delta V)}^2}}}{R}\end{align}\)since the potential difference across a resistor is given by\(\Delta V = IR\).

A resistor's inherent energy manifests as the energy transmitted to it through electrical transmission.

Given data

The standard voltage in North America is:

\(\Delta V = 120\;V\)

The power consumption of the razor at\(\Delta V\)is:\(P = 25.0\;W\)

The voltage in Europe is: \(\Delta {V'} = 240\;V\)

Calculation for the power consumption of the razor.

The power consumption of the razor at voltage\(\Delta V\)is found from Equation:

\(P = \frac{{{{(\Delta V)}^2}}}{R}\)

Solving for\(R\), we obtain:

\(R = \frac{{{{(\Delta V)}^2}}}{P}\)

Entering the values for\(\Delta V\)and\(P\), we obtain:

\(\begin{align}{}R = \frac{{{{(120\;V)}^2}}}{{25.0\;W}}\\R = 576{\rm{ }}\Omega \end{align}\)

The power consumption of the razor at voltage\(\Delta {V'}\)is found from equation:

\({P'} = \frac{{{{\left( {\Delta {V'}} \right)}^2}}}{R}\)

Entering the values for\(\Delta {V'}\)and\(R\), we get:

\(\begin{align}{}{P'} = \frac{{{{(240\;V)}^2}}}{{576{\rm{ }}\Omega }}\\{P'} = 100\;W\end{align}\)

Therefore, the power value is obtained as \(100\;W\).