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Expert-verifiedDuring open-heart surgery, a defibrillator can be used to bring a patient out of cardiac arrest. The resistance of the path is\({\bf{500}}\;{\bf{\Omega }}\), and a \({\bf{10}}\,{\bf{mA}}\)current is needed. What voltage should be applied?
The applied electric voltage is \(5.00\;{\rm{V}}\).
The given data can be listed below as,
Whenever an electric current flows through a particular electrical resistance, the electrical resistance offers a restriction to moving the current smoothly. The value of the electrical resistance can be obtained with the help of the concept of Ohm’s principle.
According to Ohm’s law, the expression to calculate the applied electrical voltage is expressed as,
\(V = IR\)
Here, \(V\) is the applied electrical voltage.
Substitute all the known values in the above equation.
\(\begin{aligned}V = \left[ {\left( {10.0\;{\rm{mA}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{A}}}}{{1\;{\rm{mA}}}}} \right)\left( {500\;{\rm{\Omega }}} \right)} \right]\\ = 5.00\;{\rm{A}} \cdot {\rm{\Omega }}\\{\rm{ = }}\left( {5.00\;{\rm{A}} \cdot {\rm{\Omega }}} \right)\left( {\frac{{1\;{\rm{V}}}}{{1\;{\rm{A}} \cdot {\rm{\Omega }}}}} \right)\\ = 5.00\;{\rm{V}}\end{aligned}\)
Thus, the applied electrical voltage is \(5.00\;{\rm{V}}\).
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