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A bather touches a radio whose case is live, and the path through the body to the drainpipe has a resistance of 4000 ohms. Ventricular fibrillation becomes possible at about 100 mA. What is the smallest case voltage that poses this danger?

A40 V
B4.0 V
C400 V
D120 V
Answer & Solution
Correct answer: C. 400 V
1. The threshold for ventricular fibrillation begins around 100 mA of current through the trunk. 2. Convert that current to amperes: 100 mA = 0.100 A. 3. The voltage needed to drive a current through a resistance is V = IR. 4. Substitute: V = 0.100 A x 4000 ohms. 5. Multiply: V = 400 V. 6. 40 V and 4.0 V are decimal-shift traps that would push well under 100 mA through this path. 7. 120 V is the trap of assuming mains voltage must be the danger point; through 4000 ohms it drives only 30 mA, painful but below the fibrillation range. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section 20.6 Electric Hazards and the Human Body_
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