


As can be seen from the first circuit above, CB&AB are in series which is in parallel with CA. So, the net resistance is
R′=26Ω=3Ω as shown in the second diagram.
From the second diagram, DC&CA is in series which is parallel to DA. Hence the resistance is
R′′=((6+3)6×3)Ω=2Ω as shown in third diagram.
From third diagram, both ED&DA are in series which is parallel to EA.
The equivalent resistance as can be seen from the above simplified circuit is Req=2Ω.
The current I through battery, I= 28=4A.
From the second diagram, the part EDACD has a resistance of 4Ω, the total current will divide equally between EDACD&EA. So consider I1=2A flowing through ED. Note that I1 will be the same current flowing in the first diagram through ED.
From first diagram, when the resistances are in parallel the current is divided in the inverse ratio. Thus,
I′ through CD=93×2=32A
I′′ through R2=21(32)=31A
