/*! 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 36 A human being can be electrocute... [FREE SOLUTION] | 91Ó°ÊÓ

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A human being can be electrocuted if a current as small as \(50 \mathrm{~m}\) A passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding, one in each hand. If his resistance is \(2100 \Omega\), what might the fatal voltage be?

Short Answer

Expert verified
The fatal voltage is 105 volts.

Step by step solution

01

Understand the Problem

The problem involves calculating the voltage that could be fatal for a human with a given current and resistance. We are given:- Current, \( I = 50 \text{ mA} = 0.050 \text{ A} \)- Resistance, \( R = 2100 \Omega \). We need to find the voltage \( V \) that causes this current to flow through the given resistance.
02

Use Ohm’s Law

Ohm's Law states that \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. We will use this formula to find the voltage \( V \) that corresponds to the given current and resistance.
03

Substitute Values into Formula

We substitute the values for \( I \) and \( R \) into the formula from Step 2:\[ V = 0.050 \text{ A} \times 2100 \Omega \]
04

Calculate the Voltage

Calculate the product of the current and resistance:\[ V = 0.050 \times 2100 = 105 \text{ volts} \]
05

Conclusion

The fatal voltage that might cause electrocution, given the conditions described, is 105 volts.

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Key Concepts

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

Electric Current
Electric current is the flow of electric charge through a conductor, such as a wire. It is generally measured in amperes (A), which represents the flow of electric charge per unit time. An electric current can be either direct (DC), where the flow of charge is in one constant direction, or alternating (AC), where the flow of charge periodically reverses direction. Understanding electric current is crucial for comprehending how electrical circuits operate and how electricity powers devices.

Here are some key points to keep in mind about electric current:
  • The unit of electric current is the ampere (A), which defines how much electric charge flows through a conductor.
  • Even a small electric current, like 50 mA or 0.050 A, can be potentially hazardous depending on where it flows in the human body.
In the given exercise, we deal with an electric current of 50 milliamperes (mA). When this current flows near the heart, it can prove fatal, showcasing the importance of understanding electric currents and their potential impacts.
Electrical Resistance
Electrical resistance is a property of a material that resists the flow of electric current. It is measured in ohms (Ω). The higher the resistance, the more difficult it is for electricity to flow through the material. Resistance plays a crucial role in determining how energy is used and distributed in an electric circuit.

Several factors can affect electrical resistance:
  • Material: Different materials have different levels of resistance. Conductors like copper have low resistance, while insulators like rubber have high resistance.
  • Temperature: Usually, a higher temperature increases the resistance of a material.
  • Cross-sectional area: A larger cross-sectional area reduces resistance, allowing more current to flow through.
  • Length: The longer the material, the greater the resistance.
In the exercise, the electrician's body resistance is expressed as 2100 Ω. This resistance, when combined with a current, helps determine the voltage needed to reach such a current flow, emphasizing the important role electrical resistance plays in circuit calculations.
Voltage Calculation
Voltage, often referred to as electric potential difference, is the driving force that pushes electric current through a circuit. It is akin to the pressure that pushes water through a pipe and is measured in volts (V). Understanding how to calculate voltage is essential for designing and analyzing electrical systems.

Ohm's Law is fundamental when it comes to voltage calculations. The law states that voltage (V) is equal to the product of the current (I) and resistance (R), mathematically expressed as:\[ V = I \times R \]This formula enables us to determine the voltage required to produce a certain current through a known resistance.

In the given problem, we used Ohm's Law to calculate the potentially fatal voltage:
  • Current ( I ) was given as 50 mA, converted to 0.050 A for calculation purposes.
  • Resistance ( R ) was provided as 2100 Ω.
  • Substituting these values into Ohm's formula results in a voltage ( V ) of 105 volts: \[ V = 0.050 \times 2100 = 105 \text{ volts} \]
This calculation highlights the practical application of Ohm's Law in everyday situations, underscoring the importance of voltage calculation in ensuring safety and functionality in electrical circuits.

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Most popular questions from this chapter

Exploding shoes. The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden conversion of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) and requires \(2256 \mathrm{~kJ} / \mathrm{kg}\) to be vaporized. If horizontal current lasts \(2.00 \mathrm{~ms}\) and encounters water with resistivity \(150 \Omega \cdot \mathrm{m}\), length \(12.0 \mathrm{~cm}\), and vertical cross-sectional area \(5.0 \times 10^{-5} \mathrm{~m}^{2}\), what average current is required to vaporize the water?

Figure \(26-18\) shows wire section 1 of diameter \(D_{1}=4.00 R\) and wire section 2 of diameter \(D_{2}=1.75 R\), connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's width. The electric potential change \(V\) along the length \(L=2.00 \mathrm{~m}\) shown in section 2 is \(10.0 \mu \mathrm{V}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). What is the drift speed of the conduction electrons in section \(1 ?\)

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A certain wire has a resistance \(R\). What is the resistance of a second wire, made of the same material, that is twice as long and has twice the diameter?

A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge (electrons) and a sphere. The belt carries charge into the sphere at a rate corresponding to \(76 \mu \mathrm{A}\). (a) Compute the surface charge density on the belt. (b) What is the number density (number per unit area) of the electrons on the belt?

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