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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 \(1.00\) mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving lowresistance devices? Explain your reasoning.

Short Answer

Expert verified
After calculating the steps: (b) The current through the body is 1.82 mA. (c) The power dissipated in the body is 33 W. (d) The internal resistance should be increased to at least 18 M Ohms to limit the current to 1 mA. (e) This modification will reduce the effectiveness of the power supply for driving low-resistance devices, as it would drop more voltage across its internal resistance, leaving a lower voltage across the device.

Step by step solution

01

- Circuit Diagram

Create a circuit diagram involving a voltage source (the 20 kV power supply), which has an internal resistance of 2000 ohms in series with the person's body resistance of 10 k Ohms.
02

- Calculate Current Through the Body

Apply Ohm's Law, which states that current (I) is equal to voltage (V) over resistance (R). Calculate the combined resistance of person's body and internal resistance. Then, calculate the current as I = V/(R_internal + R_body).
03

- Calculate Power Dissipated in the Body

Use the power formula, P = I^2 * R, where I is the current from Step 2 and R is the resistance of the person's body, to calculate the power dissipated in the person's body.
04

- Determine New Internal Resistance for Safety

To ensure the current does not exceed 1.00 mA, set I = 1.00 mA in Ohm's Law and solve for the new internal resistance (R_safety) while keeping the body resistance and voltage the same, using the equation V = I * (R_safety + R_body).
05

- Analyze Effectiveness with New Internal Resistance

Discuss the requirements for the power supply to drive low-resistance devices and how the increased internal resistance could affect its efficiency and output voltage.

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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 principle in physics and electrical engineering that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance of the conductor.

The formula for Ohm's Law is expressed as: \[ I = \frac{V}{R} \] where \(I\) is the current in amperes, \(V\) is the voltage in volts, and \(R\) is the resistance in ohms. This law is critical in analyzing and understanding electrical circuits, especially when determining how much current flows through a component, like the unfortunate person in our exercise who grasped the terminals of a power supply.

Applying Ohm's Law, by knowing the voltage and total resistance in the circuit, we can calculate the dangerous current that would flow through the person's body. This could be the stepping stone for discussing safety measures and the potential dangers associated with high-voltage circuits.
Circuit Safety Precautions
Circuit safety is paramount when dealing with electrical devices and power supplies. To prevent electric shocks, circuits are designed with various safety features like fuses, circuit breakers, and proper grounding.

In our example involving a person and a high-voltage power supply, a couple of safety measures could be discussed. For instance, the internal resistance of the power supply can be increased to reduce the current flow, which would minimize the risk of harmful electric shocks. Using insulation materials on terminals or employing barriers can prevent accidental contact altogether.

Moreover, safety equipment such as rubber gloves, mats, and boots can protect against electric shocks. It is also essential to follow safety protocols, such as not working alone on high-voltage circuits and always testing circuits to ensure they are de-energized before interfacing with them. Learning about circuit safety helps in making informed decisions when designing or modifying electrical systems to ensure they are both safe and functional.
The Role of Power Dissipation in Circuits
Power dissipation in electrical circuits is the process of converting electrical energy into heat. This phenomenon is observed when an electric current flows through a resistive component, releasing energy in the form of heat as a byproduct. The power dissipated can be calculated using the formula: \[ P = I^2 \times R \] where \(P\) represents the power in watts, \(I\) is the current in amperes, and \(R\) is the resistance in ohms.

In the context of the exercise, the power dissipated in the person's body would be indicative of the energy converted to heat due to the current flow, potentially damaging tissues and organs. Understanding power dissipation is essential for designing circuits that are not only efficient but also safe. Incorporating components like heat sinks or using materials with higher thermal tolerance can manage and mitigate the effects of power dissipation. Furthermore, when the power supply's internal resistance is modified to be safer, it could also change the level of power dissipation in the system, which must be taken into account to maintain circuit integrity and performance.

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Most popular questions from this chapter

Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 39 remaining identical bulbs, what is then the operating voltage of each?

(a) What is the internal resistance of a 1.54-V dry cell that supplies \(1.00 \mathrm{~W}\) of power to a \(15.0-\Omega\) bulb? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

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Using the exact exponential treatment, find how much time is required to charge an initially uncharged \(100-\mathrm{pF}\) capacitor through a \(75.0-\mathrm{M} \Omega\) resistor to \(90.0 \%\) of its final voltage.

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