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For a \(120-\mathrm{V}, 60\) -Hz voltage, a current of \(70 \mathrm{~mA}\) passing through the body for 1.0 s could be lethal. What must be the impedance of the body for this to occur?

Short Answer

Expert verified
The impedance must be approximately 1714.29 ohms.

Step by step solution

01

Understanding Ohm's Law

Ohm's Law relates voltage (V), current (I), and impedance (Z) with the formula \( V = I \cdot Z \). In this problem, we need to find the impedance (Z) given the voltage and current.
02

Identify Given Values

We are given the voltage \( V = 120 \) volts and the current \( I = 70 \) mA. Convert the current to amperes: \( I = 70 \times 10^{-3} = 0.070 \) A.
03

Solving for Impedance

Now we apply Ohm's Law to isolate and calculate impedance. Rearrange the equation: \( Z = \frac{V}{I} \). Substituting the known values gives \( Z = \frac{120}{0.070} \).
04

Calculate the Impedance

Perform the division \( Z = \frac{120}{0.070} = 1714.29 \). Thus, the impedance is approximately \( 1714.29 \) ohms.

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Key Concepts

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

Impedance Calculation
Impedance is a crucial concept in the study of circuits, particularly in AC (alternating current) circuits. It is represented by the symbol \( Z \) and incorporates both resistance and reactance. Reactance is the opposition to change in current due to capacitance and inductance in a circuit.
For a given AC circuit, impedance is calculated using Ohm's Law, which states the relationship between voltage \( V \), current \( I \), and impedance \( Z \). The formula is:
  • \( V = I \cdot Z \)
To solve for impedance, rearrange the formula to:
  • \( Z = \frac{V}{I} \)
This formula highlights the importance of understanding impedance, as it determines how much the current is resisted when a given voltage is applied across the circuit. In our example, we found the impedance by dividing 120 volts by the current, which was converted to 0.070 A, giving an impedance of 1714.29 ohms.
Voltage and Current Relationship
The connection between voltage and current is fundamental to electronics and electrical engineering. Ohm's Law is the guiding principle that helps us understand how these quantities interrelate. According to the law:
  • Voltage \( V \) is the energy per unit charge provided by a power source, measured in volts.
  • Current \( I \) is the flow of electric charge, measured in amperes.
These elements combine to express how electrical circuits operate. If the voltage across a circuit is increased, the current will tend to increase, assuming the impedance remains constant. Similarly, if the impedance increases while keeping voltage constant, the current will decrease. This relationship is crucial for predicting how changes in one quantity affect the others, and managing these relationships is essential for designing effective and safe electrical systems. The example problem illustrates this by using known quantities to find the missing one—impedance.
Resistance in Circuits
Resistance is a part of impedance that specifically opposes the flow of direct current (DC) in a circuit. It is a measure of how much a material or component reduces the electric current through it. In our exercise, the reference to impedance includes but is not limited to resistance. Resistance is measured in ohms (\( \Omega \)).
All components in a circuit will have some resistance, and while resistance relates to direct current, it also affects alternating current regimes as a part of total impedance. Understanding resistance helps in:
  • Determining how much energy is used or lost as heat.
  • Setting safe operational parameters to prevent circuit damage.
The body impedance mentioned in the exercise serves as a kind of resistive barrier, preventing too much current from passing through, which if exceeded, could be dangerous or even lethal. This principle underlines the importance of designing circuits that account for the resistive properties of materials.

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Most popular questions from this chapter

(II) The output of an electrocardiogram amplifier has an impedance of 45\(\mathrm { k } \Omega\) . It is to be connected to an \(8.0 - \Omega\) loud speaker through a transformer. What should be the turns ratio of the transformer?

(I) The magnetic field inside an air-filled solenoid \(38.0 \mathrm{~cm}\) long and \(2.10 \mathrm{~cm}\) in diameter is \(0.600 \mathrm{~T}\). Approximately how much energy is stored in this field?

At \(t = 0 ,\) the current through a 60.0 -mH inductor is 50.0\(\mathrm { mA }\) and is increasing at the rate of 78.0\(\mathrm { mA } / \mathrm { s } .\) What is the initial energy stored in the inductor, and how long does it take for the energy to increase by a factor of 5.0 from the initial value?

To demonstrate the large size of the henry unit, a physics professor wants to wind an air-filled solenoid with self-inductance of \(1.0 \mathrm{H}\) on the outside of a \(12-\mathrm{cm}\) diameter plastic hollow tube using copper wire with a 0.81 -mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact). How long will the plastic tube need to be and how many kilometers of copper wire will be required? What will be the resistance of this solenoid?

(III) Two tightly wound solenoids have the same length and circular cross- sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid \(2 .(a)\) What is the ratio of their inductances? (b) What is the ratio of their inductive time constants (assuming no other resistance in the circuits)?

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