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A person with body resistance between his hands of \(10.0 \mathrm{k} \Omega\) accidentally grasps the terminals of a \(20.0-\mathrm{kV}\) power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is \(2000 \Omega\), what is the current through his body? (c) What is the power dissipated in his body? (d) 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 this situation to be \(1.00 \mathrm{mA}\) or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

Short Answer

Expert verified
A person accidentally grasps a 20.0 kV power supply with body resistance of 10 kΩ and internal resistance of 2000 Ω. The current through his body is 1.67 A, and the power dissipated in his body is 16,700 W. To make the power supply safe, the internal resistance should be increased to 19,990,000 Ω for a maximum current of 1.00 mA, but this modification may compromise the power supply's effectiveness for driving low-resistance devices.

Step by step solution

01

Draw a circuit diagram

Illustrate the circuit diagram with a 20.0 kV power supply represented by a battery, connected in series with an internal resistance (2000 Ω) and the person's body resistance (10.0 kΩ).
02

Calculate the total resistance

Add the internal resistance (2000 Ω) and the person's body resistance (10.0 kΩ) to find the total resistance in the circuit. Total resistance, \(R_T = R_{internal}+R_{person} = 2000 \Omega + 10,000 \Omega = 12,000 \Omega\)
03

Calculate the current through the person's body

Use Ohm's Law to find the current through the person's body, knowing the voltage (20.0 kV) and total resistance (12,000 Ω). \(I = \frac{V}{R_T} = \frac{20,000 \mathrm{V}}{12,000 \Omega} = 1.67 \mathrm{A}\)
04

Calculate the power dissipated in the person's body

Use the power formula to find the power dissipated in the person's body, knowing the current (1.67 A) and the person's resistance (10.0 kΩ). \(P = IV = (1.67 A)(10,000 \Omega) = 16,700 \mathrm{W}\)
05

Calculate the internal resistance required for a maximum current of 1.00 mA

Use Ohm's Law to determine the new total resistance needed to achieve the desired maximum current of 1.00 mA. Then, subtract the person's resistance to find the required internal resistance. \(R_T = \frac{V}{I_{max}} = \frac{20,000 V}{0.001 A} = 20,000,000 \Omega\) New internal resistance: \(R'_{internal} = R_T - R_{person} = 20,000,000 \Omega - 10,000 \Omega = 19,990,000 \Omega\)
06

Discuss the effectiveness of the modified power supply

The modification will make the power supply safer for humans; however, it will significantly reduce the power supply's ability to deliver power to low-resistance devices. This is because the increased internal resistance will result in lower current and, consequently, lower power available for external devices. This may compromise the power supply's effectiveness for driving low-resistance devices.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the field of electricity which relates the current flowing through a conductor to the voltage across it and its resistance. It can be expressed in the mathematical formula:
\( I = \frac{V}{R} \)
where \( I \) is the current in amperes (A), \( V \) is the voltage in volts (V), and \( R \) is the resistance in ohms (Ω).

In our exercise, Ohm's Law helps us to determine the current flowing through a person's body when they come into contact with a voltage source. Knowing the body's resistance and the voltage, we can calculate the dangerous levels of current that would pass through the body. This calculation is critical for understanding electrical safety and designing systems to minimize the risk of electric shock.
Electrical Resistance
Electrical resistance is the measure of the opposition that a material offers to the flow of electric current. It’s often compared to water flowing through a pipe: the thinner the pipe or the more obstructions it has, the harder it is for water to flow. Similarly, with higher resistance in a circuit, less current will flow for a given voltage.

In the context of our exercise, the person's body and the power supply's internal components both have resistance. These resistances add up to determine the total resistance in the circuit. When designing electrical systems, understanding the resistance within components, such as the human body, is important to ensure safety and proper functioning of the system.
Circuit Diagrams
Circuit diagrams are visual representations of electrical circuits. They use symbols to depict different components like resistors, batteries, and connectors, simplifying the understanding of complex circuits. In safety scenarios, such diagrams can illustrate potential hazards and clarify how various resistances are connected.

In our exercise, the circuit diagram helped us visualize the connection between the power supply, its internal resistance, and the human body. By drawing the circuit, we can better assess the impact of modifications to the circuit, such as increasing the power supply’s internal resistance, to reduce the risk of electric shock.
Power Dissipation
Power dissipation refers to the conversion of electrical energy into heat energy in a resistive component. The formula to calculate power dissipated is given by:
\( P = IV \)
where \( P \) is the power in watts (W), \( I \) is the current in amperes (A), and \( V \) is the voltage across the component in volts (V).

For our exercise, this concept is crucial because power dissipation within a human body can lead to severe injuries or even death. Understanding the amount of power dissipated helps evaluate the hazards and implement safety measures, such as increasing the internal resistance of the power supply, to reduce the risk to an acceptable level.

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Most popular questions from this chapter

When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the \(R C\) constant of the circuit-it is not possible to measure time variations shorter than \(R C .\) How would you manipulate \(R\) and \(C\) in the circuit to allow the necessary measurements?

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