XL = ωL therefore, XL = (377)(0.5) = 188.5 Ω
We will use the voltage as a reference again; therefore E = 200 ∠0 deg V…its phasor shown here in blue…
From Ohm’s law the current in the resistor is IR and that is equal to E/R or 200 ∠0 deg/100 ∠0 deg = 2 ∠0 deg Amps…its phasor shown here in red…
From Ohm’s law: IL = E/XL = 200∠0 deg/188.5∠90 deg = -j1.06 Amps…its phasor shown here in red…1.06 Amps at -90°
From Ohm’s law: IC = E/XC = 200∠0 deg/100∠-90 deg = +j2.0 A…its phasor shown here in red…2 Amps at +90°
Again note: that in my diagram, the magnitudes of the phasors are not necessarily exactly to scale. They were meant for demonstration purposes only.