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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

A constant electric field is maintained inside a semiconductor. As the temperature is lowered, the magnitude of the current density inside the semiconductor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

A rectangular wafer of pure silicon, with resistivity \(\rho=2300 \Omega \mathrm{m},\) measures \(2.00 \mathrm{~cm}\) by \(3.00 \mathrm{~cm}\) by \(0.010 \mathrm{~cm}\) Find the maximum resistance of this rectangular wafer between any two faces.

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 battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

A material is said to be ohmic if an electric field, \(\vec{E}\), in the material gives rise to current density \(\vec{J}=\sigma \vec{E},\) where the conductivity, \(\sigma\), is a constant independent of \(\vec{E}\) or \(\vec{J}\). (This is the precise form of Ohm's Law.) Suppose in some material an electric field, \(\vec{E}\), produces current density, \(\vec{J},\) not necessarily related by Ohm's Law; that is, the material may or may not be ohmic. a) Calculate the rate of energy dissipation (sometimes called ohmic heating or joule heating) per unit volume in this material, in terms of \(\vec{E}\) and \(\vec{J}\). b) Express the result of part (a) in terms of \(\vec{E}\) alone and \(\vec{J}\) alone, for \(\vec{E}\) and \(\vec{J}\) related via Ohm's Law, that is, in an ohmic material with conductivity \(\sigma\) or resistivity \(\rho .\)

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