In the circuit given in fig. `(V_C)=50 V and R=50 Omega`. The values of C and `(V_R)` are

Options

A. (A) `3.3 mF`, `60 V`
B. (B) `104 muF`, `98 V`
C. (C) `52 muF`, `98 V`
D. (D) `2 muF`, `60 V`

Answer

B

Explanation

`V_(C )^(2)+V_(R )^(2) =V^(2)`

`50^(2) +V_(R)^(2) = 110^(2) implies V_(R ) = sqrt(160 xx 60 ) = 98 V`
then, `(I_V) = (sqrt(160 xx 60))/(50) R = 50 Omega`
alos `(I_V) =(110)/(sqrt(R^(2)+X_(C )^(2)) :. 98/50 =(110)/(50^(2)+X_(C )^(2))`
Flux, `(X_C)` and now using `(X_C)=(1)/(omegaC)`, we get `C=104 (mu)F`.