Given, initial quality factor (Qi)=100 Let initial inductance (xLi)=x Final inductance (xLf)=2x and initial resistance (Ri)=R Final resistance (Rf)=‌
R
2
Final quality factor =Qf Since, Qi=‌
XL
R
and Qf=‌
2XL
R∕2
⇒‌‌Qf=‌
4XL
R
=4Qi=4×100 Hence, final quality factor will be 400 .