Problem 23 - Physics

Question

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 \(4.0 \mathrm{~A} .\) Determine the power consumed by (a) the blow-dryer 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.

Step-by-Step Solution

Answer

a) 1320 W; b) 480 W; c) 1.375.

Step 1: Identify Given Values

For the blow-dryer: Voltage, \( V = 120 \mathrm{~V} \) and Current, \( I = 11 \mathrm{~A} \).For the vacuum cleaner: Voltage, \( V = 120 \mathrm{~V} \) and Current, \( I = 4.0 \mathrm{~A} \).

Step 2: Calculate Power for Blow-dryer

Use the power formula, \( P = VI \).\[ P_{\text{blow-dryer}} = 120 \mathrm{~V} \times 11 \mathrm{~A} = 1320 \mathrm{~W} \]

Step 3: Calculate Power for Vacuum Cleaner

Use the same power formula, \( P = VI \).\[ P_{\text{vacuum cleaner}} = 120 \mathrm{~V} \times 4.0 \mathrm{~A} = 480 \mathrm{~W} \]

Step 4: Determine Energy Usage Over Time

First, convert the time into hours: 15 minutes = 0.25 hours and 30 minutes = 0.5 hours.Next, calculate the energy using \( E = P \times t \).For the blow-dryer: \[ E_{\text{blow-dryer}} = 1320 \mathrm{~W} \times 0.25 \mathrm{~h} = 330 \mathrm{~Wh} \]For the vacuum cleaner:\[ E_{\text{vacuum cleaner}} = 480 \mathrm{~W} \times 0.5 \mathrm{~h} = 240 \mathrm{~Wh} \]

Step 5: Calculate Energy Ratio

The ratio of energy used by the blow-dryer to the vacuum cleaner is:\[ \text{Ratio} = \frac{E_{\text{blow-dryer}}}{E_{\text{vacuum cleaner}}} = \frac{330 \mathrm{~Wh}}{240 \mathrm{~Wh}} = 1.375 \]

Key Concepts

Current and Voltage Relationship

Understanding the relationship between current and voltage is crucial when analyzing electrical devices. Voltage is the electric potential difference; essentially, it's what pushes electric charges through a circuit. Think of it as the force that propels electrons to move. Current, on the other hand, is the flow of these charges. It is measured in amperes (A) and indicates how many electrons pass a point in the circuit per second.

In our example, both the blow-dryer and the vacuum cleaner operate at a voltage of 120 V. However, they have different current ratings: 11 A for the blow-dryer and 4 A for the vacuum cleaner. This difference in current explains how much electrical charge each device is drawing from the same voltage source.

The varying current affects the power consumption of these devices, as power depends on both the voltage applied and the current flowing through the device.

Power Formula in Physics

The power formula is a fundamental aspect of physics when dealing with electrical systems. Power, measured in watts (W), is the rate at which energy is used or converted in a device. The formula to calculate power is simple and can be expressed as \( P = VI \), where \( P \) represents power, \( V \) is voltage, and \( I \) is current.

In our scenario, to find the power consumed by the blow-dryer, we apply the formula: \( P_{\text{blow-dryer}} = 120 \, \mathrm{V} \times 11 \, \mathrm{A} \), resulting in \( 1320 \, \mathrm{W} \). For the vacuum cleaner, the same formula gives us: \( P_{\text{vacuum cleaner}} = 120 \, \mathrm{V} \times 4 \, \mathrm{A} = 480 \, \mathrm{W} \).

This calculation underlines how the same voltage can lead to different power consumption due to the variance in current, illustrating the practical application of the power formula in everyday electronics.

Energy Consumption and Usage Ratio

When examining energy consumption, understanding how much power a device consumes over time is essential. Energy is measured in watt-hours (Wh) and is the product of power and time. In mathematical terms, energy is calculated with \( E = P \times t \).

In our exercise, the blow-dryer's energy consumption for 15 minutes, or 0.25 hours, is \( E_{\text{blow-dryer}} = 1320 \, \mathrm{W} \times 0.25 \, \mathrm{h} = 330 \, \mathrm{Wh} \). The vacuum cleaner consumes \( E_{\text{vacuum cleaner}} = 480 \, \mathrm{W} \times 0.5 \, \mathrm{h} = 240 \, \mathrm{Wh} \) in half an hour.

To find out the usage efficiency, calculate the ratio of energy used by the two devices. By dividing the energy used by the blow-dryer by the energy used by the vacuum cleaner, \( \frac{330 \, \mathrm{Wh}}{240 \, \mathrm{Wh}} \), we obtain a ratio of 1.375. This means, for the times under consideration, the blow-dryer uses 1.375 times more energy than the vacuum cleaner, highlighting the blow-dryer's higher energy demand per unit of time.