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A blow-dryer and a vacuum cleaner each operate with a voltage of \(120 \mathrm{V} .\) The current rating of the blow-dryer is \(11 \mathrm{A}\), and that of the vacuum cleaner is 4.0 A. Determine the power consumed by (a) the blowdryer and (b) the vacuum cleaner. (c) Determine the ratio of the energy used by the blow-dryer in 15 minutes to the energy used by the vacuum cleaner in one-half hour.

Short Answer

Expert verified
(a) 1320 W, (b) 480 W, (c) \(\frac{11}{8}\)

Step by step solution

01

Formula for Electrical Power

To find the power consumed by an electrical device, use the formula: \( P = V \times I \), where \(P\) is power in watts, \(V\) is voltage in volts, and \(I\) is current in amperes.
02

Calculate Power of the Blow-dryer

Given: \(V = 120 \, V\) and \(I = 11 \, A\) for the blow-dryer. Using the formula: \[ P_{blow-dryer} = 120 \, V \times 11 \, A = 1320 \, W \]
03

Calculate Power of the Vacuum Cleaner

Given: \(V = 120 \, V\) and \(I = 4.0 \, A\) for the vacuum cleaner. Using the formula: \[ P_{vacuum} = 120 \, V \times 4.0 \, A = 480 \, W \]
04

Calculate Energy Used by Each Device

Energy used is given by \( E = P \times t \), where \(P\) is power in watts and \(t\) is time in seconds. - For the blow-dryer, \( t = 15 \times 60 = 900 \, s \): \[ E_{blow-dryer} = 1320 \, W \times 900 \, s = 1,188,000 \, J \] - For the vacuum cleaner, \( t = 30 \times 60 = 1800 \, s \): \[ E_{vacuum} = 480 \, W \times 1800 \, s = 864,000 \, J \]
05

Calculate the Ratio of Energies

The ratio of energy used by the blow-dryer to energy used by the vacuum cleaner is: \[ \text{Ratio} = \frac{E_{blow-dryer}}{E_{vacuum}} = \frac{1,188,000 \, J}{864,000 \, J} = \frac{33}{24} = \frac{11}{8} \]

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Key Concepts

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

Ohm's Law
Understanding how electricity works and is measured is a foundational concept in physics. Ohm's Law is one of the fundamental principles used to analyze electric circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. The formula is given by:\[ V = I \times R \]where:
  • \( V \) is the voltage in volts (V),
  • \( I \) is the current in amperes (A),
  • \( R \) is the resistance in ohms (Ω).
Ohm’s Law helps in predicting how much current will flow in a circuit for any given voltage and resistance. It shows that voltage is a product of the current flowing through a conductor and the resistance it encounters. This relationship is critical when designing and troubleshooting circuits to ensure they operate safely and efficiently.
Energy Consumption
Energy consumption refers to the amount of energy used by electrical devices over time. This is especially crucial for understanding the efficiency of electrical appliances and managing electricity costs. The energy used by a device can be calculated by:\[ E = P \times t \]where:
  • \( E \) is the energy in joules (J),
  • \( P \) is the power in watts (W),
  • \( t \) is the time in seconds (s).
In practical terms, energy consumption gives us insights into how much electricity our devices use over a specific period. For example, knowing how much energy a blow-dryer uses in 15 minutes can help estimate its contribution to the electricity bill. By calculating the energy consumption, one can make informed decisions about using electrical devices more efficiently and potentially reduce costs.
Power Formula
Electrical power is the rate at which energy is consumed. It is expressed in watts (W) and can be determined using the power formula:\[ P = V \times I \]This indicates that the power ( \( P \) ) consumed by a device is the product of the voltage ( \( V \) ) it operates at and the current ( \( I \) ) flowing through it.

Example Calculation

To find the power consumed by a blow-dryer with a voltage of 120 V and a current of 11 A:\[ P = 120 \, V \times 11 \, A = 1320 \, W \]This means the blow-dryer consumes 1320 watts of power.
Understanding this formula is crucial for calculating the energy consumption of any electrical device and planning their use. Devices with higher power ratings consume more electricity and thus are typically more expensive to operate.
Electrical Devices
Electrical devices are everyday gadgets that use electricity to function. Their operation can be analyzed and optimized using principles of power and energy. Devices like blow-dryers and vacuum cleaners have specific voltage and current ratings which determine their energy consumption levels. In turn, this affects both their performance and operating costs.
  • **Blow-Dryers:** These are typically high-power devices due to their heating elements, leading to significant energy use. A blow-dryer consuming 1320 W of power will use more energy and result in higher costs if used frequently.
  • **Vacuum Cleaners:** Often used longer than other devices, vacuum cleaners must be efficient to avoid excessive energy consumption. A model using 480 W is relatively efficient.
Understanding the power and energy metrics of these devices helps users make informed choices, such as opting for energy-efficient models. This not only benefits personal finances but also contributes towards reducing one's carbon footprint.

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Most popular questions from this chapter

You have three capacitors: \(C_{1}=67 \mu \mathrm{F}, C_{2}=45 \mu \mathrm{F},\) and \(C_{3}=33 \mu \mathrm{F}\). Determine the maximum equivalent capacitance you can obtain by connecting two of the capacitors in parallel and then connecting the parallel combination in series with the remaining capacitor.

The current in a \(47-\Omega\) resistor is 0.12 A. This resistor is in series with a \(28-\Omega\) resistor, and the series combination is connected across a battery. What is the battery voltage?

The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 52-gal \(\left(1.00\right.\) gal \(\left.=3.79 \times 10^{-3} \mathrm{m}^{3}\right)\) unit starts with cold water at \(11^{\circ} \mathrm{C}\) and delivers hot water at \(53^{\circ} \mathrm{C}\). The unit is electric and utilizes a resistance heater \((120 \mathrm{V}\) ac, \(3.0 \Omega\) ) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.

Suppose that two capacitors \(\left(C_{1}\right.\) and \(\left.C_{2}\right)\) are connected in series. Show that the sum of the energies stored in these capacitors is equal to the energy stored in the equivalent capacitor. [ Hint: The energy stored in a capacitor can be expressed as \(\left.q^{2} /(2 C) .\right]\)

The drawing shows two capacitors that are fully charged \(\left(C_{1}=2.00 \mu \mathrm{F}, q_{1}=6.00 \mu \mathrm{C} ; C_{2}=\right.\) \(\left.8.00 \mu F, q_{2}=12.0 \mu C\right) .\) The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor.

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