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### College Physics (Urone)

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?

1. Value of the current needed to transmit the given power is $$I = 4.00{\rm{ }}kA$$.
2. Power loss in the transmission line is$${P'} = 16.0{\rm{ }}MW$$..
3. Percent loss of the of the power in transmission line is $$Percent{\rm{ }}loss = 16.0\%$$.
See the step by step solution

## Step 1: Relation between power and potential difference.

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.

## Step 2: Given data.

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: $$R = 1.00{\rm{ }}\Omega$$.

## Step 3: Calculation of current using potential difference and power.

(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 $$4.00{\rm{ }}kA$$.

## Step 4: Calculation of power loss in the transmission line.

(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 $$16.0{\rm{ }}MW$$.

## Step 5: Calculation of power loss in the transmission line in percentage

(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 $$16.0\%$$.