Answer to Question #167087 in Electric Circuits for Vinit Kumawat

Question

If the currents in the three parallel branches of an ac circuit are given as I1 = 25∠30°A, I2 = 25e-jΠ/6A and I3 = (50 + j 50)A, express the total current in the form

(i) I ∠Φ and

(ii) Im sin (ωt + Φ).

Answer 1

$I_1=25\angle30\degree\space A=25(cos\dfrac{\pi}{6}+jsin\dfrac{\pi}{6})=(21.65+j12.5)\space A$

$I_2=25e^{-j\pi/6}\space A=25(cos\dfrac{-\pi}{6}+jsin\dfrac{-\pi}{6})=(21.65-j12.5)\space A$

$I_3=50+j50\space A$

(a) $I=I_1+I_2+I_3$

$\Rightarrow I=(21.65+21.65+50)+j(12.5-12.5+50)$

$\Rightarrow I=93.3+j50$

Magnitude of $I=\sqrt{(93.3)^2+(50)^2}$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=105.85$

Position of $I$ with respect to $x$ axis is $\theta=tan^{-1}\dfrac{50}{93.3}$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=28.18\degree$

$I=105.85\angle28.18\degree\space A$

(b) $I_m=105.85\space A$

$\phi=28.18\degree$

Assuming frequency of AC supply to be 50 Hz

$\omega=2\pi f=2\pi\times50=314s^{-1}$

$I=I_m(\omega t+\phi)$

$I=105.85sin(314t+28.18\degree)\space A$