Question

3. 2 A 18 V 612 412 w w a oH+ - ob 3A 512 +) 10 V Find the Norton equivalent for the circuit at terminal a and b. Calculate the In Ans. _A

Answer 1

Answei:- чл. - Twtwob 61 we can redraw the By using source to so that our analysis becomes simple. ce transformahen Sounce transformation with . 3A wonent sour and un parallel to it. A { us we can write as v 12v R & ur V: ZXR V. 3x412V R:42 after source transformation on / TA(+ m ob 6.1 / 10 v redrawing the circuit (4+6):10.1 13V+12V30V 512 s

first we have to find the Rth by setting the independent sourcs equal to zero Toa o'c 52 } 1 02 Rus - equivalent usistanu wowss a&b terminals RN z 10/15 Rixre Req z 10 X5 10+5 RIHR I RN *5 »> 3.33.1 we must find In by (Isc) short Now circuiting the a&b terminals. ya 300 Bzz PN 2 P Rz z PN = Pse On IA I cor saņ ņ menos Apply super mesh Analysis I. 12-23 224 super mesh, an loop &, kve 51, - lov + 300 + 10 27-22720 5P + 101, -10 12 = -20

151; -10.12 2-20 > 0 Apply KVL an loop 283 Supermesh : +10 (17-2, ] -30 - 0 -102, + 1082 z 30 > sowing eq. Or we get I, 22A 12 25 A 82-8, 22A 13 - 12-2A 25A-2A [PM ²Pg ² I3 z3A Now Norton equivalent of circuit is IN2 3A (1) { 3.337 RN