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

Options

A. `3.3muF,60 V`
B. `3.3muF,98 V`
C. `1.6muF,30 V`
D. `2muF,60 V`

Answer

B

Explanation

`V_(C)^(2)+V_(R)^(2)=V^(2)`
`50_(C)^(2)+V_(R)^(2)=V^(2)`
`V_(R)^(2)=110^(2)-50^(2)`
`V_(R)=sqrt(110^(2)-50^(2))`

`V_(R)=sqrt(160xx60)=98V`
Then, `I_(v)=(sqrt(160xx60))/50 (`:. ` R=50 Omega)`
Also, `I_(v)=110/(sqrt(R^(2)+X_(c)^(2))) :. 98/50=110/(sqrt(50^(2)+X_(c)^(2)))`
Flux `X_(C)` and Use `X_(c)=1/(omega C)`, find `C`,
`C=3.3 muF`