91Ó°ÊÓ

According to Ohm's law, the current flowing through a conductor\(\left( {\bf{I}} \right)\)between two points is proportional to the voltage between those points\(\left( {\bf{V}} \right)\).

It is given as-

\(\begin{array}{l}{\bf{I}} \propto {\bf{V}}\\{\bf{I = }}\frac{{\bf{V}}}{{\bf{R}}}\\{\bf{V = IR}}\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\end{array}\)

The AC power source's voltage is:\(V = 120\;{\rm{V}}\)

The resistance of a person standing on a rubber mat is:

\(\begin{array}{c}{R_1} = (300\;{\rm{k}}\Omega )\left( {\frac{{1000\;\Omega }}{{1\;{\rm{k}}\Omega }}} \right)\\ = 3.00 \times {10^5}\;\Omega \end{array}\)

The resistance of person standing on wet grass is:

\(\begin{array}{c}{R_2} = (4000\;{\rm{k}}\Omega )\left( {\frac{{1000\;\Omega }}{{1\;{\rm{k}}\Omega }}} \right)\\ = 4.00 \times {10^6}\;\Omega \end{array}\)

Using Ohm's law from Equation, the current flowing through the individual is determined.

Substituting the values from other equations for\(V\)and\({R_1}\)

\({I_1} = \frac{V}{{{R_1}}}\)

\(\begin{array}{c}{I_1} = \frac{{120\;{\rm{V}}}}{{3.00 \times {{10}^5}\;\Omega }}\\ = 4.00 \times {10^{ - 4}}\;{\rm{A}}\\ = 0.400\;{\rm{mA}}\end{array}\)

Therefore, the current passes through a person is \({{\rm{I}}_{\rm{1}}}{\rm{ = 0}}{\rm{.400\;mA}}{\rm{.}}\) This current is below the person's threshold of sensation, therefore he or she will not be aware of it.

Using Ohm's law from Equation, the current flowing through the individual is determined.

Substituting the values from other equations for\({\rm{\Delta V}}\)and\({{\rm{R}}_{\rm{2}}}\)

\({I_2} = \frac{V}{{{R_2}}}\)

\(\begin{array}{c}{I_2} = \frac{{120\;{\rm{V}}}}{{4.00 \times {{10}^6}\;\Omega }}\\ = 3.00 \times {10^{ - 5}}\;{\rm{A}}\\ = 0.03\;{\rm{mA}}\end{array}\)

Therefore, the current passes through a person is \(0.03\;{\rm{mA}}\). This current is below the person's threshold of sensation, therefore he or she will not be aware of it.