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Asked by patel140903 | 10 Jul, 2020, 09:36: AM
Capacitors are in μF and resistances are in ohms. Hence time constant for charging is order of micro seconds.
Hence after 1 min , capacitors are almost fully charged and behave like open circuit .
Hence batteries of EMF 2V connected in the line F-H-A are disconnected, and we get the actial circuit as shown in above figure.

Current distribution i1 , i2 and i3 are considered as per the marking shown in the figure.

Let us apply Kirchoff's voltage law to the loop G-D-C-B-G

2 i2 + 2 i1 = 6      or  i2 + i1 = 3  ...........................(1)

Let us apply Kirchoff's voltage law to the loop I-E-D-G-I

2 i2 = - 3  ........................ (2)

Hence from eqn.(1) and (2) , we get ,  i2 = -1.5 A   and i1 = 4.5 A

Ratio of current in the branch BC to current in the branch GD = 4.5 : 1.5 = 3 : 1
Answered by Thiyagarajan K | 16 Oct, 2020, 09:20: PM

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