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

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