Question

1. Use the following circuit diagram (10 points)

The resistance between two points A and B is denoted by R_AB

R_AB = 0.2 Ω; R_BC = 0.2 Ω; R_CD = 0.2 Ω; R_EF = 0.2 Ω; R_FG = 0.1 Ω;

R_GH = 0.1 Ω; R_HI = 0.2 Ω; R_JK = 0.3 Ω; R_KL = 0.3 Ω;

The current between two points F and B is denoted by I_FB

I_FB = 10 A; I_CH = 4 A; I_ID = 8 A; I_GK = 6; I_LI = 12 A

The current directions are marked by arrows. For example, I_FB flows from node F to node B.

Find the following currents:

Current Name	Current Value	Direction
IFB	10 A	node F to node B
IAB
IBC
ICD
IEF
IFG
IGH
IHI
IJK
IKL

Find the following resistance values

Resistor Name	Resistor Value
RBF
RCH
RDI
RGK
RIL

B 10A 4A KM Rch SA Roi ERBE WW= 1201 G H I 12 A GA M 120V - REK RIL + J K

Answer 1

- IBC IBA = - TAB ICB B IDC =-I CD .D c A K LOA 4А &A -120v + IHI G I E ^ I IGH I FG AGA IEF 12A rov + L J к IKL IJK At node-Ki I & IJK = 6+12 = 6 +12 = 10 A IKL = 12A At node - Ii IHI+12=8 IHI -= =-4 A At node - Hi- IGH = 4+ IH I IGH - 4 +(-4) IGH - DA

At node -Gi- IFG + 6 = IGH IfG = 0-6 = -6A At node-Fi 10+(-6) IEF = 10+ IF G IEF=4A At node -ci ICB = 4+ IDC & IDC= 8A ICB = 4 +8 = 12A • IBCs - IcR = -12A & -Igc = ICD = -8A At node-B! IBA=10+ Ic B = 10+12 IBA=22A .: FAB=- IBA == 22A

Current Value Direction (from node) Current Name IFB IAB IOA F to B - 22A A to B IBC -12A B to c -8A c to D 4А — GA ICD IEF I FG IGH IHI IJK l E to f F to G G to H OA -UA H to 10A J to k IKL 12A K to L - Resistor Name Resistor value RBF 122 3002 RCH RDI RGK 15.2 202 RIL 102