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.
AC Circuit – Calculate Total Current in RMS for Parallel RL Circuit
acinductorparallelresistors
Related Solutions
Firstly, you can ignore the 1 ohm resistor because if the other components are exactly resistive (at some value of \$\omega\$ then the addition of the 1 ohm won't change this.
So, find the impedance of the inductor, capacitor and 10 ohm resistor - that's step 1. Next, once you have the impedance you should get rid of the complex terms in the denominator by multiplying top and bottom by the denominator's complex conjugate.
This leaves you with a complex number divided by a real number.
The next trick is to equate only the imaginary part of the numerator to zero and the rest is easy - you'll find a solution for \$\omega\$ and this will be defined by L, C and R.
Here is a question that seeks to find the impedance of an L, C and R but with R in series with L. It's very close to what you are looking for.
The circuit you analyse with your formula is essentially this one, with the switch open: two independent braches, each with two resistors in series.
simulate this circuit – Schematic created using CircuitLab
The current in the two braches is independennt, each current is V / ( R + R ).
Note that the potential (voltage) at both sides of the switch is the same, so we can close the switch, without effect on the circuit. Now we have your circuit, except that R1 is represented by TWO parallel resistors, each R, so the equivalent is R/2.
To summarize, you analyzed your circuit as if it were the one I show, which is different from your circuit in the value of R1.
The comment from The Photon gives another view, which amounts to the same thing.
Best Answer
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: