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Found in: Page 736

College Physics (Urone)

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.

See the step by step solution

Step 1: Concept Introduction

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.

Step 2: Information Provided

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: $$R = 0.100{\rm{ }}\Omega$$.

Step 3: Calculation of Voltage

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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