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5\. Electrocution A human being can be electrocuted if a current as small as \(50 \mathrm{~mA}\) 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 \(2000 \Omega\), what might the fatal voltage be?

Short Answer

Expert verified
100 V

Step by step solution

01

Identify Given Values

Determine the values provided in the problem. The current that can cause electrocution is given as \( I = 50 \text{ mA} = 0.050 \text{ A} \) and the resistance of the person is given as \( R = 2000 \text{ } \Omega \).
02

Use Ohm's Law

Apply Ohm's Law, which states \( V = IR \). This equation relates voltage (V), current (I), and resistance (R).
03

Substitute Values Into Ohm's Law

Substitute the given current (I) and resistance (R) into Ohm's Law to find the voltage (V). \( V = 0.050 \text{ A} \times 2000 \text{ } \Omega \) which simplifies to \( V = 100 \text{ V} \).

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Key Concepts

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

Electrical Current
Electrical current is the flow of electric charge through a conductor, typically measured in amperes (A). Think of it like water flowing through a pipe. The amount of current depends on two key factors: voltage and resistance.
The voltage can be thought of as the pressure pushing the electric charge through the conductor, while the resistance is like the pipe size that restricts the water flow.
Higher voltage increases the current, whereas higher resistance decreases it. Ohm's Law helps us understand this relationship clearly through the formula:
  • Ohm's Law:
    • \( V = IR \) where:
      • V is Voltage (Volts)
      • I is Current (Amperes)
      • R is Resistance (Ohms)
      In our example problem, the electrician experiences a current of \(0.050\) A when holding the conductors with sweaty hands. This current is significant enough to cause electrocution if near the heart. Always handle electrical devices with care!

Human Resistance
Human resistance varies greatly depending on several factors such as skin moisture, thickness, and the part of the body through which the current passes. Typically, dry skin has higher resistance, ranging from \(1,000 \text{ } \text{to } 100,000 \text{ } \text{Ohms}\). However, when the skin is wet (e.g., sweaty hands), the resistance drops significantly.
In the given problem, the electrician's resistance is \(2000 \text{ } \text{Ohms}\) due to his sweaty hands. This lower resistance can allow a more significant current to flow through his body.
Consider some safety measures to avoid electrical hazards:
  • Wear insulated gloves
  • Ensure hands are dry
  • Use tools with non-conductive handles
Understanding how human resistance impacts current flow is crucial in preventing electrical accidents.
Voltage Calculation
To find the voltage that could be fatal in this scenario, we use the given values and apply Ohm's Law. Ohm's Law states that
  • V = IR
  • In our example: \( I = 0.050 \text{ } A \) \( R = 2000 \text{ } \text{Ohms} \)

      Substituting these values into the equation gives us:

      \( V = 0.050 \text{ } \text{A} \times 2000\text{ } \text{Ohms} \) Which simplifies to: \( V = 100 \text{ } \text{V} \)
      This means the voltage that could cause a fatal shock under these conditions is 100 Volts (V).
      It's important to realize that even household voltages can be dangerous. Ordinary outlets provide about \(120 \text{ } V\), which is above the fatal threshold for electrocution. Always disconnect power before handling electrical components and exercise caution to prevent accidental shocks.
      By understanding voltage calculations and practicing safety measures, we can significantly reduce the risk of electrical accidents.

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Most popular questions from this chapter

50\. 1994 Honda Accord Consider a 1994 Honda Accord with a battery that is rated at 52 ampere-hours. This battery is supposed to be able to deliver 1 ampere of current to electrical devices in a car for at least 52 hours or 2 amperes for 26 hours, and so on. Suppose you leave the car lights turned on when you park the car and the car lights draw 20 amperes of current. How long will it be before your battery is dead?

39\. When Applied When \(115 \mathrm{~V}\) is applied across a wire that is \(10 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the current density is \(1.4 \times 10^{4} \mathrm{~A} / \mathrm{m}^{2}\). Find the resistivity of the wire.

36\. Current Density (a) The current density across a cylindrical conductor of radius \(R\) varies in magnitude according to the equation $$ J=J_{0}\left(1-\frac{r}{R}\right) $$ where \(r\) is the distance from the central axis. Thus, the current density has a maximum magnitude of \(J_{0}=\left|\vec{J}_{0}\right|\) at that axis \((r=0)\) and decreases linearly to zero at the surface \((r=R) .\) Calculate the current in terms of \(J_{0}\) and the conductor's cross-sectional area \(A=\) \(\pi R^{2}\). (b) Suppose that, instead, the current density is a maximum \(J_{0}\) at the cylinder's surface and decreases linearly to zero at the axis: \(J=J_{0} r / R .\) Calculate the magnitude of the current. Why is the result different from that in (a)?

30\. Small But Measurable A small but measurable current of \(1.2 \times 10^{-10} \mathrm{~A}\) exists in a copper wire whose diameter is \(2.5 \mathrm{~mm}\). Assuming the current is uniform, calculate (a) the current density and (b) the average electron speed.

32\. The U.S. Electric Code The (United States) National Electric Code, which sets maximum safe currents for insulated copper wires of various diameters, is given (in part) in the table. Plot the safe current density as a function of diameter. Which wire gauge has the maximum safe current density? ("Gauge" is a way of identifying wire diameters, and \(1 \mathrm{mil}=10^{-3}\) in. $$ \begin{array}{lrrrrrrrr} \hline \text { Gauge } & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ \text { Diameter, mils } & 204 & 162 & 129 & 102 & 81 & 64 & 51 & 40 \\ \text { Safe current, A } & 70 & 50 & 35 & 25 & 20 & 15 & 6 & 3 \\ \hline \end{array} $$

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