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Q85PE

Expert-verifiedFound in: Page 736

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?**

- \(P = 230\;kW\) is dissipated in a short circuit of \(240 - V\)\({\rm{AC}}\) through a resistance of \(0.250{\rm{ }}\Omega \).
- The amount of current that flows is \(I = 960{\rm{ }}A\).

**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,**

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

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

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 \(P = 230\;kW\).

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 \(I = 960{\rm{ }}A\).

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