A person notices a mild shock if the current along a path through the thumb and index finger exceeds 82 µA. Find the maximum allowable potential difference without shock across the thumb and index finger for a dry-skin resistance of 2.4 × 105 Ω. Answer in units of V.

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Khoso123 Khoso123 · Answer 1 · 2020-02-27T05:27:21+01:00

Answer:

[tex]V_{voltage}=19.68V[/tex]

Explanation:

Given data

Current I=82µA=82×10⁻⁶A

Resistance R=2.4×10⁵Ω

to find

Voltage

Solution

From Ohms law we know that:

[tex]V_{voltage}=I_{current}*R_{resistance}\\V_{voltage}=82*10^{-6}*2.4*10^{5} \\V_{voltage}=19.68V[/tex]

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ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-04-05T13:12:42+02:00

The maximum allowable potential difference without shock across the thumb and index finger for a dry-skin is 19.68V

Given that the resistance of the dry skin is R = 2.4 × 10⁵ Ω

And the maximum allowed current is I = 82 µA = 82 × 10⁻⁶ A

According to Ohm's Law, the voltage drop across a resistance is given by:

V = IR

So the maximum allowable potential will be:

V = 82 × 10⁻⁶A × 2.4 × 10⁵Ω

V = 19.68V

So, the voltage or potential difference must not exceed 19.68V to prevent the shock.

Learn more about Ohm's Law:

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