(a) During surgery, a current as small as \(20.0\mu A\) applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is \(300{\rm{ }}\Omega ,\)what is the smallest voltage that poses this danger? (b) Does your answer imply that special electrical safety precautions are needed?

Ohm's law states that the voltage across a conductor is proportional to the current flowing through it between two points.

When determining the current \(I\) through a circuit element (other than a battery), divide the potential difference \(\Delta V\) across the circuit element by its resistance \(R\):

\(I = \frac{{\Delta V}}{R}\)

Ventricular fibrillation is caused by the following current:

\(\begin{aligned}{}I& = (20.0{\rm{ }}\mu A)\left( {\frac{{1\;A}}{{{{10}^6}{\rm{ }}\mu A}}} \right)\\ &= 2.00 \times {10^{ - 5}}\;A\end{aligned}\)

The exposed heart's resistance is: \(R = 300{\rm{ }}\Omega \).

a.

Equation is used to find the minimum voltage that can trigger ventricular fibrillation. \(\Delta V = IR\)

Substituting the values from other equations for\(I\)and\(R\)

\(\begin{aligned}{}\Delta V &= \left( {2.00 \times {{10}^{ - 5}}\;A} \right)(300{\rm{ }}\Omega )\\ &= 6.00 \times {10^{ - 3}}\;V\\ &= \left( {6.00 \times {{10}^{ - 3}}\;V} \right)\left( {\frac{{1000{\rm{ }}mV}}{{1\;V}}} \right)\\& = 6.00{\rm{ }}mV\end{aligned}\)

Therefore, smallest voltage is \(\Delta V = 6.00{\rm{ }}mV\).

(b)

Yes, as shown by our response, more electrical safety measures are necessary. This is because static charges that accumulate on hand gloves can produce very small voltages.

Consequently, extra electrical safety measures are required.