Dear anshu ,

Total resisitance of the circuit  = (R₂ || R₃) + R₁

      R_eq  = R₂ R₃/R₂ + R₃ + R₁ ;  solving we get , R_eq  = 3 Ω

    Now time constant of the circuit  = R_eq  c  =  18 ms

now apply the transient current equation ,  i  = (V/R_eq) (e^{-t / Time constant})

  we get , i  = (10/3) ( e^-t/18ms)  A

now we can see that current in R2 and R3 divides equally as they are eual . so , current in R2  =  i/2

Applying joule 's heat formula  -

   Total heat dissipated  =  ∫i_R² R dt  =  ∫ (100/9 ( e^-t/18ms)²  * 2* dt     =   (200/9)  ∫ e^-2t/18ms   dt   ( limits from  0 to infi )

    =    ( 200/9) * (9 ms) =  200 mJ = 2 *10^₆µJ

Approved