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Q86PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**What voltage is involved in a \(1.44 - kW\) short circuit through a \(0.100 - \Omega \) resistance?**

\(\Delta V = 12.0\;V\)is involved in a \(1.44{\rm{ }}kW\) short circuit through a \(0.100{\rm{ }}\Omega \) resistance.

**The power or rate at which energy is delivered to a circuit element, as, **\(P = I\Delta V\)**, if a potential difference **\(\Delta V\)** is maintained across the element.**

**We may describe the power delivered to a resistor as **\(\begin{align}{c}P = {I^2}R\\ = \frac{{{{\left( {\Delta V} \right)}^2}}}{R}...(1)\end{align}\)** since the potential difference across a resistor is given by **\(\Delta V = IR\)**.**

**A resistor's internal energy manifests as the energy transmitted to it by electrical transmission.**

The power dissipated in the circuit is: \(\begin{align}{}P &= \left( {1.44{\rm{ }}kW} \right)\left( {\frac{{1000{\rm{ }}W}}{{1{\rm{ }}kW}}} \right)\\ &= 1.44 \times {10^3}{\rm{ }}W\end{align}\).

The resistance of the circuit is: \(R = 0.100{\rm{ }}\Omega \).

The potential difference across the circuit is found by solving Equation \((1)\) for \(\Delta V\) –

\(\Delta V = \sqrt {PR} \)

Entering the values for \(P\) and \(R\) –

\(\begin{align}{}\Delta V &= \sqrt {\left( {1.44 \times {{10}^3}\;W} \right)(0.100{\rm{ }}\Omega )} \\ &= 12.0\;V\end{align}\)

Therefore, the value for the voltage is obtained as \(\Delta V = 12.0\;V\).

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