Question

(2) a) Determine the Thévenin equivalent circuit (Rth and Vtw) of the circuit shown below as seen by terminals a-b. 1 k w oa 35k 320k 25 mA 45 kn 50 V .b RTH (5 pts) VTH (15 pts) b) Determine the max power Pmax to a load R that will be connected to terminals a-b. (10 pts)

Answer 1

a) Ikr M 5kr 125mA www 45 kr 20 kr sov. (F .b For Rth :- Rth of above circuit can be is determined by short circuited and. Open circruited then is I voltage source 2, Cerrent source the circuit will be www are connected 5k2 and 20 kr Rth SK2 45ke 20kn in parelld. then the then the resultant Resistance will be. R, R₂ (5k) + (20k) Rit R₂ sk +20k 5x10²x 20x10 Req (5x10?) + (20x103) 100 X 106 Reg 4X103 = 4 kr 25 x 10 Ri 20R Req = 4k2 a ska www Rear. then the circuit will be Erth ЧКА Чs kn w b Now Reg = 4 kn and 1kn Total Connected in Series the Torad Polad equivalent resistance is sum of two resistor the. Ry - Req tlk. 4 k tolk Qol sk

= 5 k. Ry = Req + 1 k = 4k+ 1 k the circuit will be Then o a & Now, the Thevenin's equivale- 5kr - nt Rih resistance will be. 45k? Rth = skull 45 kr b sks and then is us kn Resistors are connected in parelled Thevenin's equivalent resistance Sxx 49 Riha 5k x 45k 5x10 X 45 X 10 5k + 45k (5x10)+45x10²) 3 6 Rina 225 XLO 50 x 10 4.5x10 4.5 kn ². Rth = 4.5 kr For th: N V 2 - Ik Apply nodal analysis 20kn 5kn 1 45kn 25 mA At Node 1 wwwwww Vih Osov Sov jo V + 50 5k 20k - 25m +V, V2 Ik = 0 2₂ IK2 220kn vi 1 45kr + So 5k + + -3 25 X10 5kr V, V2 IK 25 mA 54 2ok MWww With sov J uv, +4(50) + N + 20v, -2012 25 x 103 20k 25 Vi + 200 - 2012 - 3 25x 10 x 20 x 10 x 20 x 10² 500 500 - 200 250,- 2042 =

25 V - 2002 = 300 5 ( 5 V - 402) - 300 5v, - U V₂ = 300 S 54 -4 V2 = 60 0 At Node 2 :- we get VU V2 =0 Ik 45 R 45 (Vzy) = 0 45k 45 V₂-4V V₂ W-Vi Ik t. 13 0 usk 45 (12-VI) + V2 45 k 45 V₂ - 45v + V2 = 0 - 45v, +4672 = 0 We multiply equation with a sv, -4V2) 9 - 60x9 45v, - 36 V₂ 540 - Add equation ③ and equation ust, 36 V2 = 540 -45 V, +46 82 0 10 V2 540

V2 540 10 V₂ = Su v. across a terminals but voltage V₂ is not nothing a and b. ie Vih Vth of the circuit (The venin's equivalent & is Voltage of the circuit wit Van = V₂ = 54 v Vith = su v. The venin's eqreivalent a and b ciorereit across terminals Rih=4.5kr Uth=5yv to load b) maximum power RL. 57 RL = Rih only maximum power is RL. torans offered to the load Re- Rih 4.5 kr Rih=4.5kr СА IL Juan -S4V. RL=4.5kr Nth 2 Rth 54 2 (4.50 6x103 L 54 9 %10 I = 6 m A 6 m A Load current How thorough the Re is

maximum power Pmax IRL (6mjax (4.5k) 6x03 % x 4.5 x 10 36x103)" x 4.5*103 x 10 -6 36 x 10 x 4.5X10 0.162 (81) maximum power to the RL (Plmax) Van syr 4 RL 54 4X4.5K 0.162 w 4 x 4.5×10² maximum power to load RL (Pmax) = 0.162 W