What voltage is involved in a \(1.44 - kW\) short circuit through a \(0.100 - \Omega \) 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 \(\Delta V = 12.0\;V\).