A circuit consists of 3 parallel branches the branch current are given as i1= 10 sin (wt), i2=20 sin(wt+600) and i3=75 sin(wt -300). Find the resultant current and express it in the form i=Im sin(wt+0). If the supply frequency is 50Hz. Calculate the resultant current when i) t=0 ii) t=0.001s.

Solution:

$i_1=10 \sin (wt) \\ i_2=20 \sin (wt+600) \\ i_3=75 \sin (wt-300) \\ i_1=\dfrac{10}{\sqrt{2}} \angle 0^0=7.071 \angle 0^0=7.071+j0 \\ i_2=\dfrac{20}{\sqrt{2}}\angle60^0=14.142\angle60^0=7.071+j12.247 \\ i_3=\dfrac{75}{\sqrt{2}}\angle-30^0=53.033\angle-30^0=45.928-j26.5165$

Now,

$i=i_1+i_2+i_3 \\ =7.071\angle0^0+14.142\angle60^0+53.033\angle-30^0 \\ 60.07+j14.2695 \\ =61.741\angle-13.362^0 \\ i=({\sqrt{2}\times61.741})(\sin(wt-13.362)) \\ i=87.315 \sin(wt-13.362^0) \\ I_m=87.315 A \\ Given \ \ f=50Hz \\ t=0 i=87.315 \sin\bigg[2 \pi \times 50 \times 0 - 13.36 \times \dfrac{\pi}{180}\bigg] \\ =-0.3553 A \\ t=0.001 s \\ i=87.315 \sin \bigg[2\pi \times 50 \times 0-0.001-13.36 \times \dfrac{\pi}{180}\bigg] \\ i=-0.1234 A$