The maximum value of first current is 10 A and it leads the reference quantity by 45°. The second current can be written as
i2 = 5 cos (ωt − π/2) = 5sin [90 + (ωt − π/2)] = 5 sin ωt
Hence, its maximum value is 5 A and is in phase with the reference quantity.
∴ Im1 = 10 (cos 45° + j sin 45°) = (7.07 + j 7.07)
Im2 = 5 (cos0° + j sin0°) = (5 + j0)
The maximum value of resultant current is
Im = (7.07 + j7.07) + (5 + j0) = 12.07 + j7.07 = 14 ∠ 30.4°
∴ R.M.S. value = 14/√2 = 10 A