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College Physics (Urone)
Found in: Page 733

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Short Answer

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 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 \(5.00\;{\rm{V}}\).

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