# A parallel RL circuit being energized by a sinusoidal a.c. voltage of $v=100\sin(1000t+36°)$. Find total current in RMS value

The question is as follows: $$\\bullet\$$In my book it is done as:
\\begin{align}\\ i_R &=\frac{V_m}{R}\sin(1000t+36°)\\ &=10\sin(1000t+36°)A\\ &=10\cos(1000t-54°)....(a)\\ \end{align}\\\
\\begin{align}\\ i_L&=\frac{V_m}{\omega L}\sin(1000t+36°-90°)\\ &=10\sin(1000t-54°)\\ &=10\cos(1000t-144°)....(b)\\ \end{align}\\\
From (a) and (b) we get, $$\I_R=10\angle-54°\$$ and $$\I_L=10\angle-144°\$$
$$\\therefore\$$ net r.m.s. current
\\begin{align}\\ I&=I_R+I_L =10\angle-54°+10\angle-144°\\ &{\begin{aligned}\\ =10(\cos54°-j\sin54°)&+10(\cos144°-j\sin144°)\\ \end{aligned}\\}\\ &=5.88-j8.1-8.1-j5.88\\ &=(-2.22-j13.97)A\\ \end{align}\\\ i.e.$$\I=14.14\angle-99°\$$A
$$\\bullet\$$I did it like:
\\begin{align}\\ i_R &=\frac{V_m}{R}\sin(1000t+36°)\\ &=10\sin(1000t+36°)A....(a)\\ \end{align}\\\
\\begin{align}\\ i_L&=\frac{V_m}{\omega L}\sin(1000t+36°-90°)\\ &=10\sin(1000t-54°)....(b)\\ \end{align}\\\
From (a) and (b) we get, $$\I_R=10\angle36°\$$ and $$\I_L=10\angle-54°\$$
$$\\therefore\$$ net r.m.s. current
\\begin{align}\\ I&=I_R+I_L =10\angle36°+10\angle-54°\\ &{\begin{aligned}\\ =10(\cos36°+j\sin36°)&+10(\cos54°-j\sin54°)\\ \end{aligned}\\}\\ &=8.1+j5.88+5.88-j8.1\\ &=(13.98-j2.22)A\\ \end{align}\\\ i.e.$$\I=14.15\angle-9.02°\$$A
I don't know why the angle is different (though the magnitude is same). Please check this which method is faulty.