Refer to Fig. `7.49`.

Long time after closing the switch , find the current through the ` 5 Omega` resistor.

Options

A. `(4)/(5) A`
B. `(2)/(5) A`
C. `(6)/(5) A`
D. `(8)/(5) A`

Answer

C

Explanation

Initially, When all the capacitors are uncharged, they will act as conducting wires. Hence, all the resistances `( "except" 5 Omega)` will be short - circuited. So no resistance occurs between `A and F`. Thus current through the `5 Omega` resistor will be `(10 // 5)A = 2 A`.
In steady state, distibution of current is shown.
Loop `ABCEFMA , 10 = 2I_(1) + 3I_(1) + 5(I_(1) + I_(2))`
or `10 I_(1) + 5 I_(2) = 10` .....(I)
Loop `ADGHFMA , 10 = 3I_(2) + 7I_(2) + 5(I_(1) + I_(2))`
or `5I_(1) + 15I_(2) = 10` .......(ii)
From Eqs. (i) and (ii) , `I_(1) = (4)/( 5) A, I_(2) = (2)/(5) A`
Hence , current through the `5 Omega` is `I_(2) + I_(2) = (6//5) A`. Potential
difference across the ` 2 mu F` capacitor is
`2 I_(1) = 2 xx (4)/( 5) = (8)/( 5) V`
Energy in it is given by
`U_(1) = (1)/(2) xx 2((8)/(5))^(2) = ( 64 // 25) mu J`
Potential difference across is `3 muF` capacitor is
`3 I_(1) = 3 xx (4)/(5) = (12)/(5) V`
Energy in it is given by
`U_(2) = (1)/(2) xx 3 ((12)/(5))^(2) = (216 // 25) muJ`
Potential difference across the `3 mu F` capacitor is
` 3I_(2) = 3 xx (2)/(5) = (6)/(5) V`
Energy in it is given by
`U_(3) = (1)/(2) xx 3((6)/(5))^(2) = ( 54//25) mu J`
potential difference across the `7 muF` capacitor is
`7 I_(2) = 7 xx (2)/(5) = (14)/(5) V`
Energy in it is given by
`U_(4) = (1)/(2) xx 7 ((14)/(5))^(2) = (686// 25) muJ`
Thus , whole of the energy stored in all capacitors will be dissipated after the switch is opened. So the energy dissipated is `U_(1) + U_(2) + U_(3) + U_(4) = 40.8 muJ`.