Question

7) In the following circuit, find the current lo using superposition. (5 points) 6/0°Α -j2 Ω 6Ω j4 Ω +) 30/30° V 8Ω

Answer 1

7) -jar th 343 To w 62 30*30*1 Erst 6co A super position theorem @ States that - for a linear system the respose (voltage or current) in any branch of a biltateral linear corcent having more than one independent source equals the algebraic sum of the responses caused by Each independent source let Io Ivt Ii where 'Iv' is the current due to voltage source and Ii is the current due to Current for voltage source response open the Current source source a w 62 30<30' bet point a has voltage a -25 so, By nodal at a sum of total current leave 8e}(Ir 4j3 point node a x + a + 2-30<30 8-2 41 6 x [5-2; + 7 + 1 / ] = 5630 2 5<30° -362-37.8² 21 = 13.89(67.gov 13.89(67089 -1.68581.24 8-2) 2 i Iv 8-23

Now Let take current due to current source short the voltage source 6co How by nodal method at y'voltage y' + y? y' 8-21 = 630 уі 6 III j43 2. ty' - nu 62 :Y'[++ - 2 + 1 2 + 1] = 626 زه Y' = 6co ·36 <-3708 =16.6753708 v Now, Ii = 16.67 <37.8" Y' 8-2j 8-23 ! Ii 2 2.021 (51-83' A Now total current Io = Ivt Ii = 1:685581684 +2.021351.83 =3.58 265.44 . ܢ Io = 3-58565.44.) A
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