Correct Answer - A::B::C::D
(a) `V_R=IR=80V, V_C=100V`
and `V_L=IX_X=160V`
`:. V=sqrt(V_R^2+(V_L-V_C)^2)=100V`
(b) since the current is lagging behind, there should be an inductor in the box.
`X_C=V_C/I=100Omega`
Now `0.8=R/Z80/sqrt((80)^2+(X_L-100)^2)`
solving we get
`X_L=160Omega`
or `omega L=160`
`:. (2pifL)=160`
`:. L=160/(2pif)=160/((2pi)xx50)`
`=1.6/piH`.