Example:(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak +V,power consumption rate of a 60.0-W AC light bulb?For (a):-VoSolving the equation Vrms= for the peak voltage Vo and substituting the knownvalue for Vms gives:Vo=2-vZVrms=1.414(120 V)= 170 V.This means that the AC voltage swings from 170V to -170 V and back 60 times every second. Anequivalent DC voltage is a constant 120 V.For (b):Peak power is peak current times peak voltage. Thus,Figure 2. The potential difference Vbetween the terminals of an AC voltageSo the power swings from zero to 120 W onhundred twenty times per second (twice eachcycle), and the power averages 60 W.fluctuatesshown.ThePo= loVo = 2(; loVo) = 2Pave.sourceasmathematical expression for V is given by V= Vosin2nftPo = 2(60.0 W) = 120 W.AveragepowerActivity:Given a transmission line having a line resistance of 5.000 sove for the following.Use the rubrics as your guide in presenting your solution (refer to attachment A.1 atthe next page).I,Vo7.(a) What current is needed to transmit 250 MW of power at 400 kV?(b) What is the power dissipated by the transmission lines if they have a resistanceFigure 3. AC power as a function of time.Since the voltage and current are in phasehere, their product is non-negative andfluctuates between zero and I,Vo. Averagepower is (1/2)1,Vo-of 5.000?(c) What percentage of the power is lost in the transmission lines?

Answered: Image /qna-images/answer/bbc14670-87c5-4ba7-9eec-be3971ef46c2.jpg

(a) What current is needed to transmit 250 MW of power at 400 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of 5.000? (c) What percentage of the power is lost in the transmission lines?

(a) What current is needed to transmit 250 MW of power at 400 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of 5.000? (c) What percentage of the power is lost in the transmission lines?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Example:
(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak +V,
power consumption rate of a 60.0-W AC light bulb?
For (a):
-Vo
Solving the equation Vrms= for the peak voltage Vo and substituting the known
value for Vms gives:
Vo=2-vZVrms=1.414(120 V)= 170 V.
This means that the AC voltage swings from 170
V to -170 V and back 60 times every second. An
equivalent DC voltage is a constant 120 V.
For (b):
Peak power is peak current times peak voltage. Thus,
Figure 2. The potential difference V
between the terminals of an AC voltage
So the power swings from zero to 120 W on
hundred twenty times per second (twice each
cycle), and the power averages 60 W.
fluctuates
shown.
The
Po= loVo = 2(; loVo) = 2Pave.
source
as
mathematical expression for V is given by V
= Vosin2nft
Po = 2(60.0 W) = 120 W.
Average
power
Activity:
Given a transmission line having a line resistance of 5.000 sove for the following.
Use the rubrics as your guide in presenting your solution (refer to attachment A.1 at
the next page).
I,Vo
7.
(a) What current is needed to transmit 250 MW of power at 400 kV?
(b) What is the power dissipated by the transmission lines if they have a resistance
Figure 3. AC power as a function of time.
Since the voltage and current are in phase
here, their product is non-negative and
fluctuates between zero and I,Vo. Average
power is (1/2)1,Vo-
of 5.000?
(c) What percentage of the power is lost in the transmission lines?

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