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

Question

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.

Answer 1

There is no difference it is just because you have chosen to consider another reference axis rotated by 90 degrees from the reference taken in the book.

By subtracting 90 degrees of any of your angles you will get the corresponding angle in the book:

• 36-90 = -54
• -54 -90 = -144
• -9.02 - 90 = -99.02