Question

(a) Find the Norton equivalent circuit of the circuit below after increasing the magnitude of the voltage source to 38 V.

(a) Find the Norton equivalent circuit of the circuit below after increasing the magnitude of the voltage source to 38 V. 52 1102202_1_2 19V 4.1 211 382 o

Answer 1

a) finding In (or) short circuit a and b current between more 104 va un 20-4 Ivo bi Ix a un 38V ♡ 40 qfx » $80 write kch at Node a :- Va-38 Va-Vb 20 8 Vab va (to +4+2) va (2+5+1) Ub o = 38 x 2 8 Narb = 7 8VQ-V6=76 → Wb38 + w badan + up to y = 25 -76 30 write kch at Node b

= 38-va Vb-38 bave 3 - 2 ( 38-Na 20 Six = 38_Na Ubygg easy us ++ - - - - us (3*2*3*2") - V4 (+ + 3)-75730 V5 236 - 5 ho 35 v$ 400V4 = 4o Nbx35 - 5 Na 152 35 Vb - lova 1524 35 Vb-lova = 608 -> ☺ multiply ear o by to and ear by 8 10 (8va-V6=76) 8dva-lovb=760 - (+)-> nad & (35V5-10Va = 608) -> 280 V3 – 80NQ = 4864 270 V6 = 5624 Vb=5624 = 20.829 V 270 = 20.82gN

.: IN= Vb - 201829 = 10.4148 .: INE 20.829 = 1044A IN = 10.414 A finding of RN short circuit the 381 source and connect a carrent source of it between a and b and voltage across a and b is Vab .: TRN = Naby = Vabae | 52 20-2 loe 42 aIx 2 (

write kaf at vx:- - o | vw (+ ) - 2. 뺐 뾿 26 Nu =vy (Wx= vy write kat at vy :- Xx+1 = " + 크리트양 A Cs) +1 = 09 (+) Puri@ 2 ( -) +1 = V9 ) +) | (2 ) 5200

1 =116979) vy 520 vy=520 = 3V 73 write kvL in outer loop (ho) Vab- 2x1 - Vy zo 1. Nab=a+vy =213 Vab = 5 V :: RN = Vab -1-52 (RN=50 Nooton's beorem IA IA IRN-522 By RN52 10:44 10.44) I