Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


College Physics (Urone)
Found in: Page 736

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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 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 \(\Delta V = 12.0\;V\).

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.