A 20 V, 750 Hz source us connected to a series combination of `R = 100 Omega, C = 10 mu F` and `L = 0.1803 H`. Calculate the time in which resistance will get heated by `10^(@) C`, if thermal capacity of the material `= 2 J//.^(@)C`

Explanation

Here, `v = 750 Hz, E_(v) = 20 V`,
`R = 100 Omega, L = 0.1803 H, C = 10 mu F = 10^(-5) F`
`t = ? Dtheta = 10^(@) C`, thermal capacity `= L .^(@)C^(-1)`
`X_(L) = omega L = 2 pi v L = 2 xx (22)/(7) xx 750 xx 0.1803`
`= 849.98 ~= 850 ohm`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = (1)/(2 xx (22)/(7) xx 750 xx 10^(-5))`
`= 21.31 ohm`
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))`
`sqrt(100^(2) + (850 - 21.21)^(2)) = 834.8 ohm`
Average power, `P = E_(v) I_(v) cos phi`
`= E_(v) ((E_(v))/(Z))((R )/(Z)) = (E_(v)^(2)R)/(Z^(2))`
`P = (20^(2) xx 100)/((834.8)^(2)) = 0.0574 W`
If t is the required time, then
`P xx t =` thermal capacity
`xx` rise in temp.
`0.0574 xx t = 2 xx 10`
`t = (20)(0.574) = 348.4 s`