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LSC Week 1 Numerical Values for Currents and Voltages Questions

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User Generated
Subject
Engineering
School
Lone Star College
Type
Worksheet
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Name: Lan Nguyen
ID: 7442814
If your last digit of LoneStar Id is even solve problem A, but if it is odd solve problem B. Note: For all the
problems SHOW your WORK. If necessary, to add something do it. Complete the exam and upload the
file to D2L-> Course Activities-> Assignment Folders-> Homew ork_Week_01_day_1_2
Problem A: The numerical values for the currents and voltages in the circuit in Figure are given in Table.
Explain your answers.

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Use text, equations, laws, number etc. to write or describe how do you find:
Ia: -4 A (because in series circuit Ia=Ib=-4A)
Vc: =-16V (because P is absorbed so it is should be P=-64/4=-16V)
Vd: = -80V (In KVL, Va+Vb+Vc-Vd=0 so Vd= -40-24-16=-80V)
Id: =-1.5A (In KCL, Id=Ib-If= -4-(-2.5)=-1.5A)
Ie: = 2.5A (because in series circuit, I will be the same, however Ie=-If so Ie=2.5A )
Vf: = 120V (in KVL, Ve-Vd-Vf=0 so Vf=40+80=120V)
Pa (absorbed/supply): absorbed because P is positive
Pa= I*V=-40*-4=160 (+vi sign)
Pb(absorbed/supply): absorbed because P is positive
Pb= V*I= -24*-4=96W (+vi sign)
Pc (absorbed/supply): absorbed because P is positive (P=64W) (-vi sign)
Pd (absorbed/supply): Supply because P is negative
Pb= V*I= -80*-1.5=120W with (-vi sign=-120W)
Pe (absorbed/supply): absorbed because P is negative
Pb= V*I= 40*2.5=100W (+vi sign)

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Name: Lan Nguyen ID: 7442814 If your last digit of LoneStar Id is even solve problem A, but if it is odd solve problem B. Note: For all the problems SHOW your WORK. If necessary, to add something do it. Complete the exam and upload the file to D2L-> Course Activities-> Assignment Folders-> Homework_Week_01_day_1_2 Problem A: The numerical values for the currents and voltages in the circuit in Figure are given in Table. Explain your answers. Use text, equations, laws, number etc. to write or describe how do you find: Ia: -4 A (because in series circuit Ia=Ib=-4A) Vc: =-16V (because P is absorbed so it is should be P=-64/4=-16V) Vd: = -80V (In KVL, Va+Vb+Vc-Vd=0 so Vd= -40-24-16=-80V) Id: =-1.5A (In KCL, Id=Ib-If= -4-(-2.5)=-1.5A) Ie: = 2.5A (because in series circuit, I will be the same, however Ie=-If so Ie=2.5A ) Vf: = 120V (in KVL, Ve-Vd-Vf=0 so Vf=40+80=120V) Pa (absorbed/su ...
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