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Q8PE

Expert-verifiedFound in: Page 733

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**During 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,**

- The resistance of the path is, \(R = 500\;{\rm{\Omega }}\).
- The needed electric current is, \(I = 10.0\;{\rm{mA}}\).

**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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