A person with body resistance between his hands of accidentally grasps the terminals of a 14-kV power supply. (a) If the internal resistance of the power supply is 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 mA or less?

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03-09-2022
Physics

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A person with body resistance between his hands of accidentally grasps the terminals of a 14-kV power supply. (a) If the internal resistance of the power supply is 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 mA or less?

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rafikiedu08 rafikiedu08 · Answer 1 · 2022-09-07T10:08:15-04:00

A person who unintentionally grabs the terminals of a 14-kV power supply while holding body resistance between his palms. (A) What is the current flowing through the person's body if the power supply's internal resistance is? (b) Where does his body's power go when it is dispersed? (c) If the power supply's internal resistance is to be increased in order to make it safer, what internal resistance should it have for the maximum current in the aforementioned scenario to be 1.00 mA or less?

employing ohm's law

a) The equation V = IR, where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) E = IR + Ir = (731.71 0.0036) Power = IV = v = = = 0.1107 W

To learn more about ohm's law visit : https://brainly.com/question/15233790

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