For the below resistor network Calculate the following: i1,i2, i3,i4,i5,i6, i7,
ID: 1323069 • Letter: F
Question
For the below resistor network Calculate the following: i1,i2, i3,i4,i5,i6, i7, V1, V2, V3, V4, V5, V6, V7, (P = Power)P1, P2, P3, P1, P3, P6, P7, Req, ibattery, Pcircuit. Place these values with proper units in table format at the end of your work, or points will be deducted. Within your work, box your answers as you go, or points will be deducted. If the grader can not follow/read your process and work, they have been instructed to assign you a zero for the problem - regardless of your answers.Explanation / Answer
Req
6?and 4? are parllel
resultant will be=6*4/10
=2.4?
2.4? is connected in series with 6?
resultant will be=6+2.4
=8.4?
4? and 2? are in series resultant will be=6?
6? is parllel to 8? resultant will be
=3.428?
3.428? is in series with 8?
=11.428?
11.428? is parllel to 8.4?
Req=4.84?
ibattery=12/4.84
=2.479A
Pcircuit=V*ibattery
=12*2.479
=29.75W
CURRENT VOLTAGE POWER
1) I1=12/11.428 =1.05A V1=I1R =1.05*8=8.4V P1=V1I1=1.05*8.4=8.82W
2) I2=0.45A V2=I2*8=0.45*8=3.6V P2=V2I2=3.6*0.45=1.62W
3) I3=0.6A V3=I3*4=0.6*4=2.4V P3=V3I3=2.4*0.6=1.44W
4) I4=0.6A V4=I4*2 =0.6*2=1.2V P4=V4I4=1.2*0.6=0.72W
5) I5=1.428A V5=I5*6=1.428*6=8.57V P5= V5* I5=12.24W
6) I6=0.5712A V6=I6*6=0.5712*6=3.427V P6=V6* I6=1.957W
7) I7=0.856A V7=I7*4=0.856*4=3.427V P7= V7* I7=2.933W