1
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.