The capacitors are initially uncharged. In a certain time the capacitor of capacitance `2muF` gets a charge of `20 muC`. In that time interval find the heat produced by each resistor individually.

Answer

B

Explanation

`q_2=20muC`
`:.q_1=10muC` (as they are in parallel)
Energy stored at this instant,
`U=1/2 q_1^2/C_1+1/2q_2^2/C_2`
`=1/2xx((10^-5)^2)/10^-6+1/2xx((2xx10^-5)^2)/(2xx10^-6)`
`=1.5xx10^-4J`
`=0.15mJ`
In chargingg of a capacitor `50%` of the energy is stored and rest `50%` is dissipated in the form of heat.
Therefore, `0.15mJ` will be dissipated in the form of heat across all the resistors. In series in direct ratio of resistance `(H=i^2Rt)` and in parallel in inverse ratio of resistance
`:. H_2=0.075mJ, H_3=0.05mJ`
and `H_6=0.025mJ`