Two resistor are connected in series across a `5 V` rms source of alternating potential. The potential difference across `6 Omega` resistor is `3V`. If `R` is replaced by a pure inductor `L` of such magnitude that current remains same. Then the potential difference across `L` is

Options

A. `1 V`
B. `2V`
C. `3V`
D. `4V`

Answer

D

Explanation

`V_(rms)(6 Omega)=3V=6I_(V) :. I_(V)=0.5A`
`I_(V)=1/2=5/(sqrt(6^(2)+X_(L)^(2))), X_(L)=8 Omega`
Now, `V_(L)=I_(V), X_(L)=1/2xx8=4V`