i 1 = 14.42 s i n ( w t + π 2 ) ; i 2 = − 70.71 c o s ( w t ) ; i 3 = 50 e i π / 3 i_1=14.42sin(wt+\frac{\pi}{2}); i_2=-70.71cos(wt); i_3=50e^{i\pi/3} i 1 = 14.42 s in ( wt + 2 π ) ; i 2 = − 70.71 cos ( wt ) ; i 3 = 50 e iπ /3
i 4 = 100 ∠ − 2 π / 3 ; i 5 = 50 ( c o s ( π 3 ) + i s i n s ( π 3 ) ) i_4=100 \angle-2\pi/3 ; i_5=50(cos(\frac{\pi}{3})+isins(\frac{\pi}{3})) i 4 = 100∠ − 2 π /3 ; i 5 = 50 ( cos ( 3 π ) + i s in s ( 3 π ))
As i ( t ) = 1 m s i n ( w t + ϕ ) i(t)=1_msin(wt+\phi) i ( t ) = 1 m s in ( wt + ϕ ) can be written as i ( t ) = 1 m 2 ∠ ϕ ⟹ r m s v a l u e i(t)=\frac{1_m}{\sqrt{2}}\angle\phi \implies rms \space value i ( t ) = 2 1 m ∠ ϕ ⟹ r m s v a l u e
i 1 = 141.42 2 ∠ − π / 2 i_1=\frac{141.42}{\sqrt{2}} \angle-\pi/2 i 1 = 2 141.42 ∠ − π /2
i 2 = 70.71 2 ∠ − π / 2 i_2=\frac{70.71}{\sqrt{2}}\angle-\pi/2 i 2 = 2 70.71 ∠ − π /2
The polar form of representation i(t) is r . e i θ ⟹ r ∠ θ r.e^{i\theta} \implies r\angle\theta r . e i θ ⟹ r ∠ θ
i 3 = 50 ∠ π / 3 i_3=50\angle\pi/3 i 3 = 50∠ π /3
i 4 = 100 ∠ − 2 π / 3 i_4=100\angle-2\pi/3 i 4 = 100∠ − 2 π /3
i 5 = 50 ∠ π / 3 i_5=50\angle\pi/3 i 5 = 50∠ π /3
With these values, the phasor diagram is as shown below
Sum of current, i T = i 1 + i 2 + i 3 + i 4 + i 5 i_T=i_1+i_2+i_3+i_4+i_5 i T = i 1 + i 2 + i 3 + i 4 + i 5
i T = 100 ∠ π / 2 + 50 ∠ − π / 2 + 50 ∠ π / 3 + 100 ∠ − 2 π / 3 + 50 ∠ π / 3 i_T=100\angle\pi/2+50\angle-\pi/2+50\angle\pi/3+100\angle-2\pi/3+50\angle\pi/3 i T = 100∠ π /2 + 50∠ − π /2 + 50∠ π /3 + 100∠ − 2 π /3 + 50∠ π /3
i T = 50 j + 100 ( 1 2 + j 2 3 ) + 100 ( − 1 2 − j 2 3 ) i_T=50j+100(\frac{1}{2}+j\frac{\sqrt{2}}{3})+100(-\frac{1}{2}-j\frac{\sqrt{2}}{3}) i T = 50 j + 100 ( 2 1 + j 3 2 ) + 100 ( − 2 1 − j 3 2 )
i T = 50 2 s i n ( w + π 2 ) i_T=50\sqrt{2}sin(w+\frac{\pi}{2}) i T = 50 2 s in ( w + 2 π )
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