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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

# (a) How much power is dissipated in a short circuit of $$240 - V$$$${\rm{AC}}$$ through a resistance of $$0.250{\rm{ }}\Omega$$? (b) What current flows?

1. $$P = 230\;kW$$ is dissipated in a short circuit of $$240 - V$$$${\rm{AC}}$$ through a resistance of $$0.250{\rm{ }}\Omega$$.
2. The amount of current that flows is $$I = 960{\rm{ }}A$$.
See the step by step solution

## Step 1: Concept Introduction

If a potential difference $$\Delta V$$ is maintained across a circuit element, the power or rate at which energy is supplied to the element, as,

$$P = I\Delta V$$

Because the potential difference across a resistor is given by $$\Delta V = IR$$, we can express the power delivered to a resistor as,

\begin{align}{}P = {I^2}R\\ = \frac{{{{\left( {\Delta V} \right)}^2}}}{R}...(1)\end{align}

The energy delivered to a resistor be electrical transmission appears in the form of internal energy in the resistor.

Ohm's Law:

The current $$I$$ through a circuit element (other than a battery) can be determined by dividing the potential difference $$\Delta V$$ across the circuit element by its resistance $$R$$ as,

$$I = \frac{{\Delta V}}{R}...\left( 2 \right)$$

## Step 2: Information Provided

• The potential difference across the circuit is: $$\Delta V = 240{\rm{ }}V$$.
• The resistance of the circuit is: $$R = 0.250{\rm{ }}\Omega$$.

## Step 3: Calculation for Power

a.

The power dissipated in the circuit is found from Equation $$(1)$$as,

$$P = \frac{{{{(\Delta V)}^2}}}{R}$$

Entering the values for $$\Delta V$$ and $$R$$, it is obtained as,

\begin{align}{}P &= \frac{{{{(240\;V)}^2}}}{{0.250{\rm{ }}\Omega }}\\& = 2.30 \times {10^5}\;W\\ &= \left( {2.30 \times {{10}^5}\;W} \right)\left( {\frac{{1\;kW}}{{1000\;W}}} \right)\\ &= 230\;kW\end{align}

Therefore, the value for the power is obtained as $$P = 230\;kW$$.

## Step 4: Calculation for Current

b.

The current in the insulator is found by using Ohm's law from Equation $$(2)$$as,

$$I = \frac{{\Delta V}}{R}$$

Entering the values for $$\Delta V$$ and $$R$$, it is obtained as,

\begin{align}{}I = \frac{{240{\rm{ }}V}}{{0.250{\rm{ }}\Omega }}\\ = 960{\rm{ }}A\end{align}

Therefore, the value of the current is obtained as $$I = 960{\rm{ }}A$$.