From $$e=220sin 100pi t$$ $$e_{0}=220V$$, $$omega = 100pi$$, $$L=(1/pi)H$$ $$X_{I}=omega L=100pi imes dfrac{1}{pi}=100Omega$$ $$ herefore I_{0}=dfrac{e_{0}}{X_{L}}=dfrac{220}{100}=2.2A$$ As current lags behind the e.m.f by a phase angle $$pi /2$$ $$ herefore I=I_{0}sin (omega t-pi/2)$$ $$I=2.2sin (100pi t-pi/2)$$ Reading of a.c. galvanometer, $$I_{v}=dfrac{I_{0}}{sqrt{2}}=dfrac{2.2}{sqrt{2}}$$ $$I_{v}=dfrac{2.2}{sqrt{2}}dfrac{sqrt{2}}{sqrt{2}}= 1.1 imes 1.414=1.555A$$