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Chapter 25: Problem 64

BIO 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 internal 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) The current through the person's body is approximately 1.1667 A. b) The power dissipated in his body is 13.61 kW. c) The power supply's internal resistance should be increased to approximately 13.99M惟 in order to be safe.

Step by step solution

01

Calculate the current through the person's body

The total resistance of the circuit is the sum of the internal resistance of the power supply and the person's body resistance i.e. \(2000 \Omega + 10,000 \Omega = 12,000 \Omega = 12 k\Omega\). The current flowing through the person's body can be calculated using Ohm's law: \(I = V/R\), where \(I\) is the current, \(V\) is the voltage and \(R\) is the resistance. Therefore, \(I = 14kV / 12k\Omega = 1.1667 A\).
02

Calculate the power dissipated in his body

The power dissipated in his body can be calculated using the formula \(P = I^2R\), where \(P\) is the power, \(I\) is the current and \(R\) is the resistance. Substituting \(I\) with the current calculated in step 1 and \(R\) with the body resistance: \(P = (1.1667A)^2 * 10k\Omega = 13.61 kW\).
03

Calculate the necessary internal resistance

We are asked to find the internal resistance that would reduce the current to 1.00mA or less. Applying Ohm's law: \(\ I = V/R\), we solve for \(R\) to get \(R = V/I\). Substituting \(V\) with 14kV and \(I\) with 1.00mA, we find \(R = 14M\Omega\). Since the internal resistance is part of this total resistance, we subtract the body resistance (10k\Omega) to find the necessary internal resistance: \(14M\Omega - 10k\Omega = 13.990M\Omega\).

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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 concept in electricity and circuits that relates the voltage across a conductor to the current flowing through it and the resistance of the conductor. The law is mathematically expressed as \( I = \frac{V}{R} \), where \( I \) is the current in amperes (A), \( V \) is the voltage in volts (V), and \( R \) is the resistance in ohms (惟).

Ohm's Law is incredibly useful because it allows us to calculate the unknown quantity if we know the other two. In the exercise, we used this formula to determine the current flowing through a person's body when they accidentally touch the terminals of a high-voltage power supply. Since the total resistance includes both the body's resistance and the internal resistance of the power supply, we first needed to sum these resistances to apply Ohm's Law effectively.

Understanding Ohm鈥檚 Law helps in designing circuits that operate efficiently and safely, by ensuring the current stays within desired levels even when subjected to high voltages.
Power dissipation
Power dissipation in electrical circuits is the process by which an electric circuit or component converts electrical energy into heat energy, often leading to a rise in temperature of the component or surroundings. Calculating the power dissipation is important because excessive heat can damage components. Power is calculated using the formula \( P = I^2R \), where \( P \) is the power in watts (W), \( I \) is the current in amperes, and \( R \) is the resistance in ohms.

In the exercise, we calculated the power that was dissipated in the person's body due to the current flowing through it. We know the current from our Ohm's Law calculation and the body's resistance, allowing us to find the power dissipation.

This understanding of power dissipation can prevent electrical hazards by ensuring components like resistors don't overheat, thus maintaining safety and longevity of electrical devices.
Circuit Resistance
Circuit resistance is the total resistance encountered by electric current as it travels through a circuit. It takes into account the inherent resistance of the components along with the wires and connections. In practical terms, circuit resistance affects how much current flows for a given voltage, as described by Ohm's Law.

Effective management of circuit resistance is crucial for ensuring that electrical components operate within safe limits. In the context of our exercise, by increasing the internal resistance of the power supply, one can reduce the current to safer levels even at high voltages. This demonstrates how adjusting resistance can be used to achieve desired circuit performance and safety requirements.

Understanding and calculating circuit resistance helps in designing circuits and in troubleshooting to ensure they function correctly and safely in various electrical applications.

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Most popular questions from this chapter

A copper wire has a square cross \(2.3 \mathrm{~mm}\) on a side. The wire is \(4.0 \mathrm{~m}\) long and carries a current of \(3.6 \mathrm{~A}\). The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

A \(25.0-\mathrm{cm}\) -long metal rod lies in the \(x y\) -plane and makes an angle of \(36.9^{\circ}\) with the positive \(x\) -axis and an angle of \(53.1^{\circ}\) with the positive \(y\) -axis. The rod is moving in the \(+x\) -direction with a speed of \(6.80 \mathrm{~m} / \mathrm{s}\). The rod is in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.120 \mathrm{~T}) \hat{\imath}-(0.220 \mathrm{~T}) \hat{\jmath}-(0.0900 \mathrm{~T}) \hat{\boldsymbol{k}} .\) (a) What is the magni- tude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

\(\mathrm{A} \cdot 540 \mathrm{~W} "\) electric heater is designed to operate from \(120 \mathrm{~V}\) lines. (a) What is its operating resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{~V}\), what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads \(12.6 \mathrm{~V}\). You cut off a \(20.0 \mathrm{~m}\) length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a \(40.0 \mathrm{~m}\) length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assure you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?
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