Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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 \(Percent{\rm{ }}loss = 16.0\% \).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \(16.0\% \).

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Relation between power and potential difference.

Step 2: Given data.

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

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

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

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Relation between power and potential difference.

Step 2: Given data.

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

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

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

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics