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### College Physics (Urone)

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 $$5.00\;{\rm{V}}$$.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as,

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

## Step 2: Significance of electrical resistance and current

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.

## Step 3: Determination of the applied electrical voltage

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 $$5.00\;{\rm{V}}$$.