Purchase Solution

Ohm"s Law: Shocking voltage.

Not what you're looking for?

Ask Custom Question

Part 1

A person notices a mild shock if the current along a path through the thumb and index finger exceeds 87.6μA. Calculate the maximum allowable voltage without shock across the thumb and index finger with a dry-skin resistance of 3.25E+5Ω.

Part 2

Calculate the maximum allowable voltage with a wet-skin resistance of 1810 ohm.

Purchase this Solution
Solution Summary

Two parts of a problem related to Ohm's law are solved and explained.

Solution Preview

The Ohm's law gives the relation between the current I, voltage V and resistance R as

...

Purchase this Solution
Free BrainMass Quizzes
Intro to the Physics Waves

Some short-answer questions involving the basic vocabulary of string, sound, and water waves.

Introduction to Nanotechnology/Nanomaterials

This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.

Basic Physics

This quiz will test your knowledge about basic Physics.

The Moon

Test your knowledge of moon phases and movement.

Variables in Science Experiments

How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.

View More Free Quizzes
Related BrainMass Solutions

  • Three problems on circuit analysis

    Voltage rise in battery - voltage drop in 1 ohm resistance - voltage drop in 2 ohm resistance = 0 14 - I1 - 2(I1+I2) = 0 Or 3I1 + 2I2 = 14 ...........(1) We apply Kirchhoff's loop law to the loop BCED starting from point C.

  • Circuit Calculation

    From the circuit, we have I_1 + I_2 = 1A by Kirchoff's Current Law and I_1 * 20 Ohm = I_2 * 10 Ohm + I_2 * 10 Ohm + 10V by Kirchoff's Voltage Law.

  • current in series circuit

    What is the voltage across the 80 ohm resistor? c. What is the voltage across the 40 ohm resistor? a.

  • Basic Electric Circuit - Resistance

    Using Ohm's law (V=IR), the current from the first source is 7V/2.75Ohm=2.55A (because the I2 arrow is pointing to the left in the figure, it is -2.55A using the passive sign convention). The voltage at point B is 2.55A*0.75Ohm=1.91 V. 2.

  • Ohm's Law

    Solving we get n = 5 2.6 As per Ohm,s law : I = V/R where I = Current in the circuit, V = Net voltage in the circuit and R = Total resistance in the circuit.

  • capacitor acts as an open circuit

    Answer is (A) 9.0 Ohms Apply Ohm's Law: Voltage = Current * Resistance, V = I * R Call the emf as V Volts and the original Resistance as R Ohms.

  • Electric Circuits

    Hi, For this problem, we will use Ohm's law. a. The voltage will be equal to: V = IR V = (0.01A)(100 ohm) V = 1V b.

  • Voltage Drop Across the Resistor

    37625 Voltage Drop Across the Resistor Determine the voltage drop (from top to bottom) across the 16 ohm resistor in the circuit shown. Determine the voltage drop across the 4 ohm resistor. What is the voltage V?

  • Circuit analysis problems

    From above we conclude that the voltage VC (0-) = VC (0+) .......(2) Before the switch S is closed at t=0 (t=0-) Applying Kirchhoff's loop law to the loop ABDGA : +32 - Vc - V = 0 ........(3) Given V(0-) = 15 V.

View More Related Solutions