BITSAT 2020 Paper

At steady state the capacitor will be frilly charged and thus there will be no current in the 10Ω resistance. So the effective circuit becomes Net current from the 6 V battery,
I=

6

( 2×3 2+3 )+ 2.8 1

=

6

1.2+2.8

=

3

2

=1.5A
Between A and B, voltage is same in both resistances, 2I1=3I2 where
I1+I2=I=1.5
⇒2I1=3(1.5−I1)
⇒I1=0.9A