Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: BIO The average bulk resistivity of the human body (apart from surface resistance of the skin) is about . The conducting path between the hands can be represented approximately as a cylinder long and in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of ? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

Short Answer

Expert verified

Answer

  1. The resistance between the hands if the skin resistance is negligible is .1019Ω
  2. The potential difference between the hands is needed for a lethal shock current of 100mAIs 102 V
  3. The power is dissipated in the body10.19W.

Step by step solution

01

Define the ohm’s law, resistance and power . 

Consider the expression for the resistance is:

R=VI=ρLA

Here, ρis resistivity Ω·m, L is length in A and is area in m2.

Consider the formula for the power dissipated in the circuit.

P=I2R ….. (1)

02

Determine the resistance.

Consider the given values of resistivity, length and the diameter.

ρ=5.0Ω·mL=1.6md=0.10m

Substitute the values in resistance formula:

R=ρLAR=ρLπd22

Substitute the values and solve as:

R=5.0Ω·m1.6mπ0.1022=1019Ω

Hence, the resistance between the hands if the skin resistance is negligible is 1019Ω.

03

Determine the potential difference

(b)

Consider the value of the current.

I=100mA

Consider the expression value of the voltage from the Ohm’s law.

V=100mA1019Ω=102V

Hence, the potential difference between the hands is needed for a lethal shock current of 100mAis 102 V .

04

Determine the power.

(c)

Consider the value of the current.

I=100mA

Substitute the values in the equation (1) to determine the power.

P=100mA21019Ω=10.19W

Hence, power is dissipated in the body.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the circuit shown in Fig. E26.49, C = 5.90 mF, Ԑ = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

A cylindrical rod has resistance R. If we triple its length and diameter, what is its resistance in terms of R

Suppose a resistor R lies alongeach edge of a cube (12 resistors in all)with connections at the corners. Find theequivalent resistance between two diagonally opposite corners of the cube (pointsa and b in Fig. P26.84).

A 5.00-A current runs through a 12-gauge copper wire (diameter

2.05 mm) and through a light bulb. Copper has8.5×108free 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?

An electron at point in figure has a speed v0=1.41×106m/s. Find (a) the magnetic field that will cause the electron to follow the semicircular path from to and (b) The time required for the electron to move fromAtoB.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free