Solution
Given:
V = 120 ( cos π 4 + i sin π 4 ) , Z = 1 + i 3 2 V=120\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right),\quad Z=\frac{1+i\sqrt{3}}{2} V = 120 ( cos 4 π + i sin 4 π ) , Z = 2 1 + i 3 Using Ohm's law V = I Z V=IZ V = I Z
I = V Z I=\frac{V}{Z} I = Z V First write Z Z Z
Z = 1 2 + 3 2 i Z=\frac{1}{2}+\frac{\sqrt{3}}{2}i Z = 2 1 + 2 3 i ∣ Z ∣ = ( 1 2 ) 2 + ( 3 2 ) 2 = 1 , arg ( Z ) = tan − 1 ( 3 ) = π 3 |Z|=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=1,\quad \arg(Z)=\tan^{-1}(\sqrt{3})=\frac{\pi}{3} ∣ Z ∣ = ( 2 1 ) 2 + ( 2 3 ) 2 = 1 , arg ( Z ) = tan − 1 ( 3 ) = 3 π So:
Z = cos π 3 + i sin π 3 Z=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3} Z = cos 3 π + i sin 3 π Now divide in polar form:
I = 120 1 [ cos ( π 4 − π 3 ) + i sin ( π 4 − π 3 ) ] = 120 [ cos ( − π 12 ) + i sin ( − π 12 ) ] \begin{aligned}
I
&=\frac{120}{1}\left[\cos\left(\frac{\pi}{4}-\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}-\frac{\pi}{3}\right)\right]\\
&=120\left[\cos\left(-\frac{\pi}{12}\right)+i\sin\left(-\frac{\pi}{12}\right)\right]
\end{aligned} I = 1 120 [ cos ( 4 π − 3 π ) + i sin ( 4 π − 3 π ) ] = 120 [ cos ( − 12 π ) + i sin ( − 12 π ) ] I = 120 ( cos ( − π 12 ) + i sin ( − π 12 ) ) \boxed{I=120\left(\cos\left(-\frac{\pi}{12}\right)+i\sin\left(-\frac{\pi}{12}\right)\right)} I = 120 ( cos ( − 12 π ) + i sin ( − 12 π ) ) Equivalent form:
I = 120 ( cos π 12 − i sin π 12 ) \boxed{I=120\left(\cos\frac{\pi}{12}-i\sin\frac{\pi}{12}\right)} I = 120 ( cos 12 π − i sin 12 π ) 