Switch S of circuit shown in Fig 5.200 is in position 1 for a long time. At instant t = 0, it is thrown from position 1 ot 2 Calcuate thermal power `P_1 (t) P_2 (t)` generated across resistanace `R_1 and R_2,` respectively.

Explanation

Initally, the switch was in position 1 for a long time,
therefore, initially the capacitor was fully charged and potential
difference across capacitor at t = 0 was equal to emf E of the battery. Initial charge on capacitor is `q_0 = CE`
When the switch is thrown to position 2, the capacitor starts
to discharge through resistance `R_1 and R_2` To calculate thermal power `P_1(t) and P_2(t)` generated across `R_1 and R_2` respectively,
current I at time t through the circuit must be known. we know
that I is given by
`I = (E )/((R_1+R_2))e^(-t//(R_1+R_2)C`
Hence, thermal power across `R_1 is `
`P_1 = I^2R_1 = (E^2R_1)/((R_1+R_2)^2) e^(-2t(R_1+R_2)C)`
Similarly, thermal power across `R_2 is `
`P_2 = I^2R_2 = (E^2R_2)/((R_1_R_2)^2) e^(-2t//R_1+R_2)C`