/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 A hair dryer consumes \(1600 .\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A hair dryer consumes \(1600 .\) W of power and operates at \(110 .\) V. (Assume that the current is \(D C .\) In fact, these are root-mean-square values of AC quantities, but the calculation is not affected. Chapter 30 covers AC circuits in detail.) a) Will the hair dryer trip a circuit breaker designed to interrupt the circuit if the current exceeds \(15.0 \mathrm{~A} ?\) b) What is the resistance of the hair dryer when it is operating?

Short Answer

Expert verified
Answer: No, the hair dryer will not trip the circuit breaker as it has a current of 14.55 A, which is less than the limit of 15 A. The resistance of the hair dryer is approximately 7.56 Ω.

Step by step solution

01

Determine the current

Use the power formula: \(P = IV\). We have the power (\(P = 1600\) W) and voltage (\(V = 110\) V) and need to solve for the current \(I\): $$ I = \frac{P}{V} $$ Plug in the given values: $$ I = \frac{1600}{110} \approx 14.55\,\text{A} $$
02

Check if the hair dryer will trip the circuit breaker

The hair dryer has a current of 14.55 A. The circuit breaker is designed to interrupt the circuit if the current exceeds 15.0 A. Since 14.55 A is less than 15.0 A, the hair dryer will not trip the circuit breaker.
03

Calculate the resistance of the hair dryer

Now, we'll use Ohm's law to calculate the resistance \(R\), which states: \(V = IR\). We need to solve for \(R\): $$ R = \frac{V}{I} $$ Plug in the given values for voltage \(V\) and calculated current \(I\): $$ R = \frac{110}{14.55} \approx 7.56\,\text{Ω} $$ So the resistance of the hair dryer when operating is approximately \(7.56\,\text{Ω}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle in the study of electric circuits, serving as a cornerstone for understanding how voltage, current, and resistance interact with each other. It is succinctly expressed by the formula:

\[ V = IR \]
where:
  • \(V\) represents the voltage across the component in volts (V),
  • \(I\) is the current flowing through the component in amperes (A), and
  • \(R\) is the resistance of the component in ohms (\(\Omega\)).
By rearranging the formula, you can calculate any one of these three values if the other two are known. In the case of our hair dryer example, once the current is found using the electrical power formula, Ohm's Law is used to compute the resistance. Understanding this relationship helps troubleshoot and design electrical systems by predicting how they react to different currents and voltages.
Electrical Power

Power in Electric Circuits

Electrical power is a measure of how much work can be done by an electric current over a certain period of time. It's typically measured in watts (W), and the formula linking power to current and voltage is:

\[ P = IV \]
where:
  • \(P\) is the power in watts,
  • \(I\) is the current in amperes, and
  • \(V\) is the voltage in volts.
This equation indicates that for a constant voltage, as the current increases, the power consumption also grows. Taking our hair dryer, rated at \(1600\) W for its wattage, we found how much current it draws using this power relation, key to determining whether it would exceed the limit of a \(15\) A circuit breaker.
AC DC Current Difference

Alternating Current (AC) versus Direct Current (DC)

The difference between AC and DC current is in the direction of flow. DC current, which comes from sources like batteries, flows in a single constant direction. It's the simple, stable current that you'd calculate when using Ohm's Law in basic circuit analysis, as shown with our hair dryer's assumed DC current.

AC, on the other hand, is the type of current delivered to our homes and used by the majority of appliances. It periodically reverses direction. The root-mean-square (RMS) values for AC quantities such as voltage and current offer a 'DC equivalent' value—useful for calculations like the ones in our exercise. Understanding these differences is crucial when working with real-world electronics and when considering the safety, efficiency, and function of the devices we use every day.

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Most popular questions from this chapter

Two resistors with resistances \(200 . \Omega\) and \(400 . \Omega\) are connected (a) in series and (b) in parallel with an ideal 9.00-V battery. Compare the power delivered to the \(200 .-\Omega\) resistor.

When a battery is connected to a \(100 .-\Omega\) resistor, the current is \(4.00 \mathrm{~A}\). When the same battery is connected to a \(400 .-\Omega\) resistor, the current is 1.01 A. Find the emf supplied by the battery and the internal resistance of the battery.

A copper wire has radius \(r=0.0250 \mathrm{~cm},\) is \(3.00 \mathrm{~m}\) long, has resistivity \(\rho=1.72 \cdot 10^{-8} \Omega \mathrm{m},\) and carries a current of \(0.400 \mathrm{~A}\). The wire has density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\) a) What is the resistance, \(R,\) of the wire? b) What is the electric potential difference, \(\Delta V\), across the wire? c) What is the electric field, \(E\), in the wire?

A voltage spike causes the line voltage in a home to jump rapidly from \(110 . \mathrm{V}\) to \(150 . \mathrm{V}\). What is the percentage increase in the power output of a 100.-W tungsten-filament incandescent light bulb during this spike, assuming that the bulb's resistance remains constant?

A 34 -gauge copper wire, with a constant potential difference of \(0.10 \mathrm{~V}\) applied across its \(1.0 \mathrm{~m}\) length at room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right),\) is cooled to liquid nitrogen temperature \(\left(77 \mathrm{~K}=-196^{\circ} \mathrm{C}\right)\) a) Determine the percentage change in the wire's resistance during the drop in temperature. b) Determine the percentage change in current flowing in the wire. c) Compare the drift speeds of the electrons at the two temperatures.

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