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Chapter 18: Problem 113

(a) What is the resistance of the heater element in a \(1500-\mathrm{W}\) hair dryer that plugs into a \(120-\mathrm{V}\) outlet? (b) What is the current through the hair dryer when it is turned on? (c) At a cost of \(\$ 0.10\) per kW .h, how much does it cost to run the hair dryer for 5.00 min? (d) If you were to take the hair dryer to Europe where the voltage is \(240 \mathrm{V},\) how much power would your hair dryer be using in the brief time before it is ruined? (e) What current would be flowing through the hair dryer during this time?

Short Answer

Expert verified
Based on the calculations above, the hair dryer has the following properties: (a) The resistance of the heater element is 9.60 ohms. (b) The current through the hair dryer is 12.5 A when used in the US. (c) The cost to run the hair dryer for 5.00 minutes is $0.0125. (d) The power the hair dryer would be using in Europe is 6000 W, a brief time before getting ruined. (e) The current flowing through the hair dryer in Europe is 25 A.

Step by step solution

01

(a) Calculate the resistance of the heater element

To calculate the resistance of the heater element, we can use the formula: \(P = \frac{V^2}{R}\) Where \(P\) is the power (\(1500\ \mathrm{W}\)), \(V\) is the voltage (\(120\ \mathrm{V}\)), and \(R\) is the resistance. Rearranging the formula to solve for \(R\), we get: \(R = \frac{V^2}{P}\) Now, plug in the values and solve for \(R\): \(R = \frac{(120\ \mathrm{V})^2}{1500\ \mathrm{W}} = 9.60\ \Omega\)
02

(b) Calculate the current through the hair dryer

To calculate the current through the hair dryer, we can use Ohm's law: \(I = \frac{V}{R}\) Where \(I\) is the current, \(V\) is the voltage (\(120\ \mathrm{V}\)), and \(R\) is the resistance (calculated in part (a)). Now, plug in the values and solve for \(I\): \(I = \frac{120\ \mathrm{V}}{9.60\ \Omega} = 12.5\ \mathrm{A}\)
03

(c) Calculate the cost to run the hair dryer for 5.00 minutes

First, we need to determine how much energy the hair dryer uses in 5.00 minutes. We can use the formula: \(E = P \times t\) Where \(E\) is the energy, \(P\) is the power, and \(t\) is the time. Since the hair dryer uses \(1500\ \mathrm{W}\) of power, we need to convert it to kilowatts: \(P = 1.5\ \mathrm{kW}\) Also, convert the time to hours: \(t = \frac{5.00\ \mathrm{min}}{60\ \mathrm{min/h}} = 0.08333\ \mathrm{h}\) Now, plug in the values and solve for \(E\): \(E = 1.5\ \mathrm{kW} \times 0.08333\ \mathrm{h} = 0.125\ \mathrm{kWh}\) The cost of running the hair dryer for 5.00 minutes is: \(Cost = 0.125\ \mathrm{kWh} \times \$0.10\ \mathrm{per\ kWh} = \$0.0125\)
04

(d) Calculate the power the hair dryer would be using in Europe

To calculate the power the hair dryer would be using in the brief time before it gets ruined in Europe, we can use the formula: \(P = \frac{V^2}{R}\) Where \(P\) is the power, \(V\) is the voltage in Europe (\(240\ \mathrm{V}\)), and \(R\) is the resistance (calculated in part (a)). Now, plug in the values and solve for \(P\): \(P = \frac{(240\ \mathrm{V})^2}{9.60\ \Omega} = 6000\ \mathrm{W}\)
05

(e) Calculate the current flowing through the hair dryer in Europe

To calculate the current flowing through the hair dryer in Europe, we can use Ohm's law: \(I = \frac{V}{R}\) Where \(I\) is the current, \(V\) is the voltage in Europe (\(240\ \mathrm{V}\)), and \(R\) is the resistance (calculated in part (a)). Now, plug in the values and solve for \(I\): \(I = \frac{240\ \mathrm{V}}{9.60\ \Omega} = 25\ \mathrm{A}\)

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Most popular questions from this chapter

We can model some of the electrical properties of an unmyelinated axon as an electric cable covered with defective insulation so that current leaks out of the axon to the surrounding fluid. We assume the axon consists of a cylindrical membrane filled with conducting fluid. A current of ions can travel along the axon in this fluid and can also leak out through the membrane. The inner radius of the cylinder is \(5.0 \mu \mathrm{m} ;\) the membrane thickness is \(8.0 \mathrm{nm} .\) (a) If the resistivity of the axon fluid is \(2.0 \Omega \cdot \mathrm{m},\) calculate the resistance of a 1.0 -cm length of axon to current flow along its length. (b) If the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\).m, calculate the resistance of the wall of a 1.0 -cm length of axon to current flow across the membrane. (c) Find the length of axon for which the two resistances are equal. This length is a rough measure of the distance a signal can travel without amplification.
If a \(93.5-\mathrm{V}\) emf is connected to the terminals \(A\) and \(B\) and the current in the \(4.0-\Omega\) resistor is \(17 \mathrm{A},\) what is the value of the unknown resistor \(R ?\)
A 20 - \(\mu\) F capacitor is discharged through a 5 -k\Omega resistor. The initial charge on the capacitor is \(200 \mu \mathrm{C}\) (a) Sketch a graph of the current through the resistor as a function of time. Label both axes with numbers and units. (b) What is the initial power dissipated in the resistor? (c) What is the total energy dissipated?
A \(1.5-\mathrm{V}\) flashlight battery can maintain a current of $0.30 \mathrm{A}\( for \)4.0 \mathrm{h}$ before it is exhausted. How much chemical energy is converted to electrical energy in this process? (Assume zero internal resistance of the battery.)
An aluminum wire of diameter 2.6 mm carries a current of 12 A. How long on average does it take an electron to move \(12 \mathrm{m}\) along the wire? Assume 3.5 conduction electrons per aluminum atom. The mass density of aluminum is \(2.7 \mathrm{g} / \mathrm{cm}^{3}\) and its atomic mass is $27 \mathrm{g} / \mathrm{mol}$.
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