/*! 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 38 Electrical safety. This procedur... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 38

Electrical safety. This procedure is not recommended! You'll see why after you work the problem. You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.0 \(\mathrm{k} \Omega\) . (a) If you accidentally touch the "hot" wire connected to the 120 \(\mathrm{V}\) line, how much current will pass through your body? Is this amount enough to be dangerous? (The maximum safe current is about 5 \(\mathrm{mA}\). (b) How much electrical power is delivered to your body?

Short Answer

Expert verified
(a) 120 mA, dangerous. (b) 14.4 W power delivered.

Step by step solution

01

Understanding Ohm's Law

To calculate the current through the body, we use Ohm's Law: \( V = IR \). Where \( V \) is the voltage across the body, \( I \) is the current through the body, and \( R \) is the resistance of the body. We know \( V = 120 \, \text{V} \) and \( R = 1.0 \, \text{k}\Omega = 1000 \, \Omega \).
02

Calculating Current

Using Ohm's Law, we rearrange it to find the current: \( I = \frac{V}{R} = \frac{120}{1000} \). Calculate this to determine the current through the body.
03

Result for Current

Calculate \( I = 0.12 \, \text{A} \) or \( 120 \, \text{mA} \). This is significantly higher than the maximum safe current of 5 \( \text{mA} \), making it extremely dangerous.
04

Calculating Power Delivered

The power delivered to the body can be calculated using the formula \( P = IV \). Substitute the values: \( P = 0.12 \, \text{A} \times 120 \, \text{V} \).
05

Result for Power Delivered

Calculate \( P = 14.4 \, \text{W} \). This is the amount of power delivered to the body, which is quite substantial given that safe levels are typically in milliwatts.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Ohm's Law
Ohm's Law is a fundamental concept in electrical engineering and physics. It describes the relationship between voltage, current, and resistance in a circuit. The law is expressed by the formula \( V = IR \), where:
  • \( V \) stands for voltage, measured in volts (V).
  • \( I \) represents the current, measured in amperes (A).
  • \( R \) is the resistance, measured in ohms (惟).
Ohm's Law implies that the current through a conductor between two points is directly proportional to the voltage across the two points. Simultaneously, it is inversely proportional to the resistance between them. Understanding how to manipulate this formula is crucial for solving a variety of electrical problems. For instance, if we know the voltage and resistance, we can easily calculate the current.
Current Calculation
The current calculation is vital to determine how safe an electrical connection is, especially under potentially hazardous conditions. To compute the current through the body when exposed to a voltage of 120 V with a resistance of 1.0 k惟 (or 1000 惟), we use Ohm鈥檚 Law rearranged as \( I = \frac{V}{R} \).
By substituting the known values into the equation, we have:
  • \( I = \frac{120}{1000} \)
This results in a current of \( 0.12 \, \text{A} \) or \( 120 \, \text{mA} \), which is far above the safety threshold of 5 mA. Such a high current is dangerous as it exceeds safe levels, and can lead to severe injury or even fatality under certain circumstances.
Power Calculation
Power calculation in electrical contexts allows us to understand how much energy is being consumed or transferred over time. The relevant equation for power is \( P = IV \), where:
  • \( P \) is power, measured in watts (W).
  • \( I \) is the current, in amperes (A).
  • \( V \) is the voltage, in volts (V).
In our example, the power experienced by the body can be calculated using the previously determined current (0.12 A) and the voltage (120 V):
\( P = 0.12 \, \text{A} \times 120 \, \text{V} = 14.4 \, \text{W} \).
This amount of power is considerable when applied to the human body, and significantly above the safe power level under normal conditions. Such an exposure could result in serious injury.
Electrical Resistance
Electrical resistance is a measure of how much a material opposes the flow of electric current. It is determined by factors such as the material's composition, length, cross-sectional area, and temperature. Resistance is measured in ohms (惟).
Key points about resistance:
  • Higher resistance means less current for a given voltage.
  • Conductors like metals have low resistance, while insulators like rubber have high resistance.
  • Resistance increases with temperature in most conductive materials.
In the scenario provided, the human body has a resistance of 1.0 k惟. This resistance, although significant, does not protect against hazardous electrical currents due to the human body's sensitivity to current. Understanding resistance helps in designing safe electrical systems and preventing dangerous scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Electricity through the body, I. A person with a body resistance of 10 \(\mathrm{k} \Omega\) between his hands accidentally grasps the terminals of a 14 \(\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the situation just described to be 1.00 \(\mathrm{mA}\) or less?

Struck by lightning. Lightning strikes can involve currents as high as \(25,000\) A that last for about 40\(\mu\) s. If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is 75 kg, that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega,\) and that his body is all water (which is reasonable for a rough, but plausible, approximation).(a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C}\) , would the person's temperature actually increase that much? Why not? What would happen first?

Typical household currents are on the order of a few amperes. If a 1.50 A current flows through the leads of an electrical appliance, (a) how many electrons per second pass through it, (b) how many coulombs pass through it in 5.0 min, and (c) how long does it take for 7.50 \(\mathrm{C}\) of charge to pass through?

Power rating of a resistor. The power rating of a resistor is the maximum power it can safely dissipate without being damaged by overheating. (a) If the power rating of a certain 15 \(\mathrm{k} \Omega\) resistor is \(5.0 \mathrm{W},\) what is the maximum current it can carry without damage? What is the greatest allowable potential difference across the terminals of this resistor? (b) If a 9.0 \(\mathrm{k} \Omega\) resistor is to be connected across a 120 \(\mathrm{V}\) potential difference, what power rating is required for that resistor?

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Solid State Physics

Read Explanation

Fields in Physics

Read Explanation

Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.