Elementary Electrical Engineering Question Solve April 2013 4a

April_2013_Question (4a)

Find  the Vo using Norton’s Theorem

For  In replacing 4Ω resistor by short circuit (fig. 2)

KCL at node 1    (120-V1)/6 - V1/3 -In=0 

                       or,120-3V1-6In = 0   .............(1)

KCL at node 2    In+6-V2/2 = 0

                        or,  6In+36-3V2 = 0    

                       or, 36-3v1+6In = 0     ............(2)           (V1=V2 )

(1) - (2) we get     12In = 84  or,     In = 7V

From Fig3 we get

Rn= {(6x3)/(6+3)} + 2 = 4Ω

Now Norton equivalent circuit (fig4)

From Fig3 we get

I = {(R_nxI_n)/(R_n+4Ω)} = (7x4)/(4+4)

 = 28/4 = 3.5A

Vo = 3.5 x 4 = 14V  (Ans)

Here is the Simulation by Proteus

http://www.youtube.com/watch?v=XNe5vdO0j9I&feature=youtu.be

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