(a) Peak velocity $$V_0=sqrt 2 V=339.41$$ $$omega=2pi f=100pi rad/s$$ $$I_o=dfrac{V_o}{sqrt{R^2+omega^2 L^2}}=1.82 A$$ (b) The voltage is given as $$V=V_ocosomega t$$ The current is given as $$I=I_ocos(omega t-phi)$$ $$phi$$ is the phase angle between the voltage and current. The time lag is given as $$phi/omega$$. Now phase angle is given as $$tanphi=omega L/R=2pi imes50 imes 0.5/100=1.57$$. So $$phi=tan^{-1}1.57=57.5^o=57.5pi/180 radians$$ time lag is $$t=phi/omega=57.5pi/180 imes2pi imes 50=3.19 imes10^{-3} s$$