As shown circuit in steady state. If charge on capacitor is `1000 muC` , find (a) battery current, (b) `R_(1),R_(2), R_(3)`. The current in `50Omega` resistor and `R_(2) is `5A`.

Explanation

Charging on capacitor = `100muC`
p.d. across `5 muF` = p.d across `10Omega = 1000/5 = 200V`
Current flowing in `10 Omega , i_(0)= 200/10 = 20A`
Battery current `i=i_(0) +5 = 20+5=25A`
`i_(0)=5+i_(1) implies 20=5+i_(1) implies i_(1) =15A`
Junction `D : 5+5 = i_(2) implies i_(2) = 10A`
`A` to `A` , path `ABDA`
`V_A - i_(0)xx10-5R_(2)+5xx50=V_(A)`
`-20xx10-5R_(2)+250=0implies R_(2) = 10 Omega`
`B` to `B` path `BDCB`
`V_(B - 5R_(2) -10xx5+15xxR_(3) = V_(B)`
`-5xx10-50+15R_(3)=0impliesR_(3) = (100)/(15) = 20/3 Omega`
`A` to `A` path `ADCXYA`
`V_A - 5xx50-10xx5+3100oxxR_(1)=V_(A)`
`-250-50+310-25R_(1)=0`
`R_(1) = 10/25 = 2/5 = 0.4 Omega`.