Sol. Initially the switch was in position 1. Therefore, initially potential difference across capacitor was equal to emf of the battery i.e. 60 volt.

∴ Initially energy stored in the capacitor was

U = CV 2 = × 0.1 × 60 2 J

= 180 J

When switch is shifted to position 2, capacitor begins to discharge and energy stored in it is dissipated in the form of heat across resistances. Let at some instant discharging current through the capacitor be I as shown in Fig.

According to Kirchhoff's laws,

i 1 + i 2 = i ….. (1)

6i 1 – 3i 2 = 0 ….. (2)

From above two equations,

i 1 = and i 2 = i

But thermal power generated in a resistance R is P = i 2 R where i is current flowing through it.

Therefore, heat generated P 1 , P 2 and P 3 across 4 Ω , 6 Ω and 3 Ω resistances is in ratio

4i 2 : 6i 1 2 : 3i 2 2 or P 1 : P 2 : P 3 = 4 : : = 6 : 1 : 2

But total heat generated is P 1 + P 2 + P 3 = U

∴ Heat generated across 4 Ω isP 1 = 120 J

Heat generated across 6 Ω is P 2 = 20 J

Heat generated across 3 Ω is P 3 = 40 J

Since, during discharging, no current flows through 10 Ω ,

therefore heat generated across it is equal to zero.