In Fig. switch `S` is closed for a long time. At `t = 0`, if it is opened, then

Options

A. total heat produced in resistor `R` after opening the switch is `(1)/(2) (Lepsilon)/(R^(2))`
B. total heat produced in resistor `R_(1)` after opening the switch is `(1)/(2) (Lepsilon^(2))/(R^(2)) ((R_(1))/(R_(1) + R_(2)))`
C. heat produced in resistor `R_(1)` after opening the switch is `(1)/(2) (R_(2)L epsilon^(2))/((R_(1) + R_(2)) R^(2))`
D. no heat will be produced in `R_(1)`

Answer

C

Explanation

Just before opening the switch, the current in the inductor is
`(E)/(R )*` Energy stored in it `= (1)/(2) L((epsilon)/(R ))^(2)`
This energy will dissipate in resistance `R_(1)` and `R_(2)` in the ratio
`(1)/(R_(1))` and `(1)/(R_(2))*`