SSC JE Electrical Engineering 24 Jan 2018 Evening Paper

Question : 120 of 200

Marks: +1, -0

Deter mine the Nor t on’s cur r ent (in A) and Norton’s resistance (in Ω) respectively, for the given electrical circuit across the load resistance, RL .

2.09, 7.66
2.34, 3.45
4.43, 3.26
2.34, 2.55

Validate

Solution:

R_\th=4∥8+5

={4×8}/12+5

= 7.66 Ω

{I_A-24}/4+V_A/8+V_A/5=0

10V_A-240+5V_A++8V_A=0

V_A=240/23V

I_S=240/{23×5}=2.09A

So, Norton’s Current is 2.09A & resistance is 7.66Ω.

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