Question

P2.82. Find the Thévenin and Norton equivalent circuits for the circuit shown in Figure P2.82. Atarotas ant boitiaoqian gates i vino writo sulsy 5622 homoa ноа 24 123 C di ba b 3 A ure 3612 29 og ob. Cinta D282

Answer 1

Solution to the given problem is:

2 Given ciocuit is → 242 3A Theverin Equivalent → for the The veuin Equivalent' cirauit we meed to find theveuin's resistance and therenin's Voltage. for thevenin's resistomce Rth acroes a-b, source and open cireuit we short-chrault voltage current soure. amy remember , VOltage Source to be short- circuited and current Sourre to be open- circuited Should one. be independnt So, for Rtn acroes a-b, open circuit current Source, So we have, 2928

(24 +6)x 6 30 X6 Ryn = 24+6+6 36 mow, Calculate for The vemin'g Vo ltage Vtu 1 we a-b. Voltage acrous FVtn = Vab 24-e ЗА Current twoougu 62 resietor å > 2 Am P8. X 24 A m pa : 11 24+12 i.e. for circuit , Say, ER2 , R1 I xR2 Ry +R2 current tho ough Rs 1) 4 Curreut tuoouga Rz z RAR* RI Ry +R2 So by using tuie formula, we find that Curreut througu 6e vesistor i8 2 Ampe.

6x 2) V 6 d=Vaba Vtu a vo Hage .· Vo lHage across 12 VOI4S. :. Thevenin'g volHage Vin = 12 VOHS. Thevewin Equivalent circuit ůe > Rtn 552 Vth= 12 V 9.

Norton Equivalent → 3n Norton Equivalent we Caloulate Norton's resistance (R) ļ Norton's Currrent (IN), Norton's resig tance is Calculated in the lame way as Therenin's resistance Ry = Rm = 5e. So tor Nortons current we short ciocuit a-b and Caloulate tue Short circuit current troougu it.

.e. 242 Ca-b Short circuited). 62 3A 6-2 (as mo cunnent )3A IN 2425 ell floro througu 6 2 mear Short -cineuited wire). -X 24 24+6 2.4 Amps. 3 x24= 30 . Norton's current = IN = 2.4 Ampe,

coait . Nootornis Equivalent +) RN 52 Rpt IN = 2.4 Ampa 9.