A capacitor of capacitance `3 mu F` is first charged by connecting it across a `10 V` battery by closing key `K_(1)`, then it is allowed to get discharged through `2 Omega` and `4 Omega` resistors by opening `K_(1)` and closing the key `K_(2)`. The total energy dissipated in the `2 Omega` resistor is equal to

Options

A. `0.5 mJ`
B. `0.05 mJ`
C. `0.15 mJ`
D. none of these

Answer

B

Explanation

(b) Initial energy stored on capacitor :
`U = (1)/(2) xx 3 xx 10^(-6) xx (10)^(2) = 150 xx 10^(-6) J`
When capacitor is allowed to charge, the whole energy stored on capacitor will be dissiapted as heat in `2 Omega` and `4 Omega` resistors. Heat dissipated in either resistor will be directly proportional to the value of the resistor. So heat dissiapted in `2 Omega` resistor is
`H = (2 U)/(2 + 4) = (U)/(3) = 50 xx 10^(-6) J`
`H = 0.050 mJ`