A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is:
(a) `4.4 A`
(b) `11sqrt(2) A`
(c) `2.2 A`
(d) `11 A`

Options

A. `4.4 A`
B. `11sqrt(2)A`
C. `2.2 A`
D. `11A`

Answer

C

Explanation

Here `E_(0)=E_(rms)xx sqrt(2)`
`= 220xx sqrt(2)V`
`|Z|=sqrt(R^(2)+X_(C )^(2))=sqrt((100)^(2)+(100)^(2))`
`|Z|=sqrt(2+(100)^(2))`
`|Z|=100 sqrt(2)Omega`
`therefore` Maximum value of displacement current `I_(D)=(E_(0))/(|Z|)`
`therefore I_(d)=(220sqrt(2))/(100sqrt(2)) " " therefore I_(d)=2.2 A`