Elementary Electrical Engineering Question Solve ,AMIE Section A October 2013

Elementary Electrical Engineering Question Solve April 2011 1b

April 2011 1(b).
Find the potential difference between C and D

I₁=4v/(6Ω+4Ω)                       I₂=8v/(6Ω+10Ω) 

   =0.4A                                         = 0.5A          

 V₁=0.4x6 = 2.4v                     V₂ =0.5x6=3v

I₁ = 4v/(6Ω+4Ω)                       I₂ = 8v/(6Ω+10Ω) 

   = 0.4A                                      = 0.5A 

         

 V₁= 0.4x6 = 2.4v                     V₂ = 0.5x6=3v

From Fig2

V_CA = -V₁   , V_AB = -10v ,     V_BD = V_{2 ,}

V_CD = V_CA+V_AB+V_BD 

         = -2.4v -10v+3v

         = -9.4v      (Ans.)        

Here is the Simulation..............

                               

jito79@gmail.com     01912346855

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

jito79@gmail.com , 01912346855

Elementary Electrical Engineering Question Solve October 2012 4a

October_2012_Question (4a)

Find  the I using Norton’s Theorem

For  In replacing 5Ω resistor by short circuit (fig. 5)

Converting 2A current source with 4Ω resistor into equivalent voltage source.(4Ωx2A=8V) (fig4)

From fig4 we can write

I₁ +I_n-4A=0

Or, I_n=4-I₁=4-{(12-8)/10}=3.6A

For R_n voltage source=short circuit & current source= open circuit (fig3).

From (fig3)

R_n=(4+6)Ω=10Ω

Now Norton equivalent circuit (fig2)

I={(R_nxI_n)/(R_n+5Ω)}=(3.6x10)/(10+5)=2.4A  (Ans)


Here is Proteus Simulation

jito79@gmail.com  01912346855