ooo T-Mobile LTE 10:30 AM Part A Determine the equivalent resistance of the \"la
ID: 1416437 • Letter: O
Question
ooo T-Mobile LTE 10:30 AM Part A Determine the equivalent resistance of the "ladder" of equal 235-n resistors shown in In other words, what resistance would an ohmmeter read if connected between points a and B? Express your answer to three significant figures and include the appropriate units. RValve Units Submit My Answers Give Up Part B What is the current through resistora if a 70.0-v battery is connected between points a and B? Express your answer to three significant figures and include the appropriate units. 1 Value, Units My Answers Give Up Part C What is the current through resistor Express your answer to three significant figures and include the appropriate units. Value Units Submit My Answers Give Up Part D What is the current through resistor e? Express your answer to three significant figures and include the appropriate units. | Value UnitsExplanation / Answer
1 ans
part A:
the resistance R,R and R connected series
R'=3R
the resistance R' and R are parallel
R"=3R^2/4R=0.75 R
the R",R and R are connected series
R"'=2.75R
the R''' and R are parallel
R""=0.73R
the resistance R'''' ,R and R are connected as series connection
the effective resistance Req=2.73 R
235=2.73 R
R=86.1 ohms
part B:
the current b/w A and B
I =V/Req=70/235=0.3 A
part C
the current at point b
Ib={(1/R''')/[1/R+1/R'''+1/R]}I
={(1/2.75*86.1)/[1/86.1+236.8+1/86.1]}0.3
=(0.004)/[0.02]0.3=0.1 A
part D
the current at point c
Ic=0.3 A
2 ans
applying kirchhoffs first law
i1+i3=i2.......................................(1)
applying kirchhoffs second law for first loop
-2i1-22i1+2i2+28i2-12i1+24=0
applyting kirchhoffs second law for second loop
-6+2i3+16 i3-28i2-2i2+11i3+12=0
=>36 i1+30 i2=24 =>16i1+15i2=12
=>16i1+15[i1+i3]=12
=>31i1+15i3=12..........................................(2)
=>29i3-30i2=-6 =>29i3-30[i1+i3]=-6
=>30i1+i3=6.............................................(3)
now solve the eq2 &eq3 we get i1
eq2 =>.31i1+15i3=12
eq3*15 => 450i1+15i3=90
........................................................
419 i1=78
i1=0.19 A
3 ans
c1=0.5*10^-6F c2=1.4*10^-6 F
potential difference V=18 v
part A:
the potential diffrerence on capacitor V1=(1/c1)V/(1/c1+1/c2)
=(1/0.5)*18/(1/0.5+1/1.4)
=2*18/2+0.7143=13.3 V
the potential difference on capacitor V2=(1/c2)*v/(1/c1+1/c2)
=0.7413*18/2+0.7413=4.7 v
part B:
the charge on the capacitor 1=>Q1=c1V1=13.3*0.5*10^-6=6.6*10^-6 C
the charge on the capacitor 2=>Q2=c2v2=1.4*10^-6*4.7=6.6*10^-6 C
part c:
the potential difference in capacitor =>V1=Q1/C1=9/0.5=18 v
the potential difference in capacitor =>v2=Q2/C2=25.2/1.4=18 v
part D:
the charge Q=v*(c1+c2)=18*1.9*10^-6=34.2*10^-6 F
the charge on capacitor Q1=c1/c1+c2*Q1=[0.5/0.5+1.4]34.2*10^-6=9*10^-6 F
the charge on capacitor Q2=1.4/1.9*34.2*10^-6=25.2*10^-6 F
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