Answer to Question #167094 in Electric Circuits for Vinit Kumawat

Question

Draw the phasor diagram and find the sum of the currents: i₁=141.42Sin(ωt +  π/2) , i₂ = -70.71 Cos ωt , i₃= 50 e^j π/3 , i₄=100∠-2 π/3 and i₅= 50 (Cosπ/3 + j Sinπ/3) 

Answer 1

Given Currents are-

$I_1=141.42sin(\omega t+\dfrac{\pi}{2})=141.42coswt$

$I_2=-70.71coswt$

$I_3=50e^{j\frac{\pi}{3}}=50(cos\dfrac{\pi}{3}+jsin\dfrac{\pi}{3})=50(\dfrac{1+\sqrt{3}j}{2})=25+25\sqrt{3}j$

$I_4=100<\dfrac{-2\pi}{3}\\=-50-86.6j$ ( convert the polar to rectangular form)

$I_5=50(cos\dfrac{\pi}{3}+jsin\dfrac{\pi}{3})=50(\dfrac{1+\sqrt{3}j}{2})=25+25\sqrt{3}j$

Adding All the currents-

$I=I_1+I_2+I_3+I_4+I_5$

$=141.42cos\omega t-70.71cos\omega t+25+25\sqrt{3}j-50-86.6j+25+25\sqrt{3}j$

$=141.42cos\omega t-70.71cos\omega t+50+50\times 1.732j-50-86.6j$

$=70.71cos\omega t+86.6j-86.6j$

$=70.71cos\omega t+0=70.71coswt=70.71sin(\omega t+\dfrac{\pi}{2})$

Hence The sum of currents is 70.71$sin(\omega t+\dfrac{\pi}{2})$ .