While taking a bath, a person touches the metal case of a radio. The path through the person to the drainpipe and ground has a resistance of \({\bf{4000}}\;{\bf{\Omega }}{\bf{.}}\).What is the smallest voltage on the case of the radio that could cause ventricular fibrillation?

Ohm’s law states that the amount of current flowing in the circuit\(\left( {\bf{I}} \right)\) is directly proportional to the potential difference applied in the circuit\(\left( {\bf{V}} \right)\).

\({\bf{V = IR}}\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Here,\(\left( {\bf{R}} \right)\)is the resistance of the circuit.

The passage through the person to the drainpipe and the ground has the resistance:\(R = 4000\;{\rm{W}}.\)

The minimum current that might cause ventricular fibrillation is:

\(\begin{array}{c}I = (100\;\;{\rm{mA}})\left( {\frac{{1\;\;{\rm{A}}}}{{1000\;\;{\rm{mA}}}}} \right)\\ = 0.100\;\;{\rm{A}}\end{array}\)

For the given values of\(I\)and\(R\), equation (1) becomes-

\(\begin{array}{c}V = (0.100\;{\rm{A}})(4000\;{\rm{W)}}\\ = 400{\rm{\;V}}\end{array}\)

Therefore, the smallest voltage that could cause ventricular fibrillation is \(V = 400\;{\rm{V}}\).