SSC JE Electrical 2018 with solution SET-2
Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.
Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).
RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5
RN = 7.66
The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.
According to voltage Division Rule
VAB = V1R3 ⁄ (R1 + R3)
= 24 × 8 ⁄ (4 + 8)
VAB = 16 V
Now Norton Current IN is
IN = VAB ⁄ RN
IN = 16 ⁄ 7.66
IN = 2.08