The current flowing through R 2 is:
A
A
A
A
Text Solution
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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
as shown in the second diagram.
From the second diagram, DC & CA is in series which is parallel to DA. Hence the resistance is
as shown in third diagram.
From the second diagram, ED & DA are in series which is parallel to EA.
The equivalent resistance as can be seen from the above simplified circuit is 
The current I through battery, 
From the second diagram, the part EDACD has a resistance of 
, the total current will divide equally between EDACD & EA. So consider I 1 =2A flowing through ED. Note that I 1 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 
I’’ through 
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