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Q77PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**In this problem, you will verify statements made at the end of the power losses for Example \(20.10\). (a) What current is needed to transmit \(100{\rm{ }}MW\)of power at a voltage of \(25.0{\rm{ }}kV\)? (b) Find the power loss in a \(1.00{\rm{ }} - {\rm{ }}\Omega \) transmission line. (c) What percent loss does this represent?**

- Value of the current needed to transmit the given power is \(I = 4.00{\rm{ }}kA\).
- Power loss in the transmission line is\({P'} = 16.0{\rm{ }}MW\)..
- Percent loss of the of the power in transmission line is \(Percent{\rm{ }}loss = 16.0\% \).

**The power, or rate at which energy is delivered to a circuit element, is **\(P = I\Delta V\)** if a potential difference **\(\Delta V\)** is maintained across the element.**

**We can represent the power applied to a resistor as **\(\begin{align}{}P = {I^2}R\\P = \frac{{{{(\Delta V)}^2}}}{R}\end{align}\)** since the potential difference across a resistor is given by **\(\Delta V = IR\)**.**

**A resistor's inherent energy manifests as the energy transmitted to it through electrical transmission.**

The power transmitted through the transmission line is:

\(\begin{align}{}P &= (100{\rm{ }}MW)\left( {\frac{{{{10}^6}\;W}}{{1{\rm{ }}MW}}} \right)\\P &= 1.00 \times {10^8}\;W\end{align}\)

The voltage at which the power is transmitted is:

\(\begin{align}{}\Delta V & = (25.0{\rm{ }}kV)\left( {\frac{{1000\;V}}{{1{\rm{ }}kV}}} \right)\\\Delta V &= 25.0 \times {10^3}\;V\end{align}\)

The resistance of the transmission line is: \(R = 1.00{\rm{ }}\Omega \).

(a)

The power transmitted through the transmission line is found from equation \((1)\):

\(P = I\Delta V\)

Solving for \(I\), we get:

\(I = \frac{P}{{\Delta V}}\)

Entering the values for \(P\) and \(\Delta V\)we obtain:

\(\begin{align}{}I &= \frac{{1.00 \times {{10}^8}\;W}}{{25.0 \times {{10}^3}\;V}}\\I &= 4.00 \times {10^3}\;A\\I &= \left( {4.00 \times {{10}^3}\;A} \right)\left( {\frac{{1{\rm{ }}kA}}{{1000\;A}}} \right)\\I& = 4.00{\rm{ }}kA\end{align}\)

Therefore, the current value is obtained as \(4.00{\rm{ }}kA\).

(b)

The power loss in the transmission line is found from equation \((2)\):

\({P'} = {I^2}R\)

Entering the values for \(I\) and \(R\), we obtain:

\(\begin{align}{}{P'} &= {I^2}R\\{P'} &= {\left( {4.00 \times {{10}^3}\;A} \right)^2}(1.00{\rm{ }}\Omega )\\{P'} &= 16.0 \times {10^6}\;W\\{P'}& = \left( {16.0 \times {{10}^6}\;W} \right)\left( {\frac{{1{\rm{ }}MW}}{{{{10}^6}\;W}}} \right)\\{P'}& = 16.0{\rm{ }}MW\end{align}\)

Hence, the power loss value is obtained as \(16.0{\rm{ }}MW\).

(c)

The power loss in the transmission line in percentage is:

\(\begin{align}{}Percent{\rm{ }}loss &= \left( {\frac{{{P'}}}{P}} \right)(100\% )\\Percent{\rm{ }}loss &= \left( {\frac{{16.0{\rm{ }}MW}}{{100{\rm{ }}MW}}} \right)(100\% )\\Percent{\rm{ }}loss& = 16.0\% \end{align}\)

Therefore, the power loss percentage value is obtained as \(16.0\% \).

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