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Chapter 25: Problem 70

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) 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 intermal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

Short Answer

Expert verified
(a) 1.17 A; (b) 13,690 W; (c) 13.99 M惟.

Step by step solution

01

Identify given values

Begin by listing all known values from the problem:- Resistance of the person's body, \( R_{b} = 10,000 \, \Omega \).- Internal resistance of the power supply, \( R_{i} = 2,000 \, \Omega \).- Potential difference (voltage) of the power supply, \( V = 14,000 \, \text{V} \). - Desired maximum current, \( I_{max} = 0.001 \, \text{A} \).
02

Calculate the current through the body

Use Ohm's law to calculate the current, which states \( V = I \cdot R \). Here, the total resistance \( R_{t} \) is the sum of the internal and body resistance:\[ R_{t} = R_{b} + R_{i} = 10,000 \, \Omega + 2,000 \, \Omega = 12,000 \, \Omega \]Now, calculate the current \( I \):\[ I = \frac{V}{R_{t}} = \frac{14,000 \, \text{V}}{12,000 \, \Omega} = 1.17 \, \text{A} \]
03

Calculate the power dissipated in the body

The power dissipated \( P \) is calculated using the equation \( P = I^2 \cdot R \), where \( R \) is the body resistance:\[ P = (1.17 \, \text{A})^2 \times 10,000 \, \Omega = 13,690 \, \text{W} \]
04

Determine new internal resistance for safety

To ensure the current is 1.00 mA or less, calculate the total resistance needed with the desired current:\[ I_{max} = 0.001 \, \text{A} \]The total required resistance is:\[ R_{t} = \frac{V}{I_{max}} = \frac{14,000 \, \text{V}}{0.001 \, \text{A}} = 14,000,000 \, \Omega \]To find the necessary internal resistance \( R_{i}^* \), subtract the body resistance:\[ R_{i}^* = R_{t} - R_{b} = 14,000,000 \, \Omega - 10,000 \, \Omega = 13,990,000 \, \Omega \]

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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. It is an essential concept in understanding how electricity works. In simple terms, imagine it as water flowing through a pipe. The amount of flow is similar to the current in a circuit. It is measured in amperes, usually abbreviated as "A." To calculate the electric current, Ohm's Law is a powerful tool. It establishes a relationship between voltage, current, and resistance.

  • Voltage (V) is the electrical potential difference that pushes the charge through the circuit.
  • Resistance (R) measures how much the conductor resists the flow of current.
  • Current (I) is calculated using the formula: \( I = \frac{V}{R} \).
Ohm's Law helps us understand the relationship between these three components. In the given exercise, it is used to find the current flowing through a person鈥檚 body when exposed to high voltage. Here, by summing up resistances of the power supply and the person's body, the total resistance can be used to find the current. This demonstrates the practical application of Ohm鈥檚 Law in safety scenarios.
Power Dissipation
Power dissipation in a circuit refers to the process by which electrical energy is converted into heat energy in the resistance of the circuit. This happens when electric current flows through a resistor. Essentially, power dissipated is the amount of energy lost as heat within a system.

The calculation of power dissipation is crucial in design and safety. It lets us calculate how much heat energy is produced, helping to prevent overheating in electrical circuits.

  • Power (P) is calculated using the formula: \( P = I^2 \cdot R \), where \( I \) is the current and \( R \) is the resistance.
  • This formula shows that power dissipated increases with the square of the current or directly with resistance.
  • In the exercise, the power dissipated through the person鈥檚 body is substantial, which illustrates the importance of managing power levels to ensure safety.
Understanding these calculations helps ensure that circuits operate safely and efficiently, preventing damage or injury.
Electrical Resistance
Electrical resistance is a measure of the difficulty faced by the current as it flows through a conductor. High resistance means less current can pass for a given voltage, while low resistance allows more current to flow.

Resistance is expressed in ohms (惟) and is influenced by several factors:

  • Material: Different conductor materials have different inherent resistances. For example, copper has low resistance, making it ideal for wires.
  • Length and Cross-sectional Area: Longer conductors and those with smaller cross-sectional areas have higher resistance.
In practice, resistance plays a crucial role in controlling the amount of current that can flow in an electrical circuit. In safety terms, like in the provided problem scenario, controlling resistance helps limit electric current to safe levels. In the problem's context, adjusting internal resistance was necessary to lower current passing through the person鈥檚 body, ensuring it remains at a non-lethal level. This demonstrates how manipulating resistance values is key in designing safe electrical systems.

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Most popular questions from this chapter

A silver wire 2.6 \(\mathrm{mm}\) in diameter transfers a charge of 420 \(\mathrm{C}\) in 80 \(\mathrm{mm}\) . Silver contains \(5.8 \times 10^{23}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

A \(25.0-\Omega\) bulb is connected across the terminals of a \(12.0-\mathrm{V}\) battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

A typical small flashlight contains two batteries, each having an emf of 1.5 \(\mathrm{V}\) , connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for 5.0 \(\mathrm{h}\) , what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

You apply a potential difference of 4.50 \(\mathrm{V}\) between the ends of a wire that is 2.50 \(\mathrm{m}\) in length and 0.654 \(\mathrm{mm}\) in radius. The resulting current through the wire is 17.6 \(\mathrm{A}\) . What is the resistivity of the wire?

A \(5.00-\mathrm{A}\) current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} )\) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?
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