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

Two resistor are connected in series across a `5 V` rms source of alte

1.	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 A. `1 V`B. `2V`C. `3V`D. `4V`
Answer» Correct Answer - D `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`

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