Question

222 R 422 W 412 M 6 V 402 2V = a) Determine the value of R for maximum power transfer to R. b) Find the maximum power to R.

Answer 1

first of all I would like to eseplain about The vinen's Theorem and Maximum Power Transfer Theorem and with the help of that I am going to solve this problem:

Thevenio's Theorem at states that lineaa bilateral network any 00 in series with the load (R) as follows: aplaced by a single voltage seule (VT) connected Single gesistance CRTW) connected across 1 Rih M 'The RL To find thi We have find voltage po across the load Gemoving the head desistance (R2) from the circuit. That village is eqreal to Urh by To find Rithi We have semove all the sources and replace ithem with their respective internal resistance ie, Open Circuit (0-2) Current Source Short Voltage Source Open Circirl (x2) and find the total resistance across Bload terminal That assistance is earial to Rih.

Maximum Power Transfer Theorem: It states that for a thevenin Gruivalent circuit, the maximum power R₂ =R Th transfers to the load, when Value of that maximum 2 power P) = P²RL where I current throughe R:

Gixen circuit diagram R 7 42 OV 4.12 2V Here I am considering R? load asistance as R. -R To find Vith! follows: After Removing 'Ŕ resistor, the circuit will be as Va x Vih 42 4.2 212 Ve wo OV Note: I am denoting nodes for calculation purpose 4.2 Z 2v 714 from above circuit diagram: V=QN, Vih= Varvo Il

By applying Kch at ve node: V Ve-V6 Ve-Va 2 4- Vb 4 ****** [3] By substituting (No=2V V[1] + Vo[] X (VG = 8 4 Vc + va(-1/2) = 1/2 By applying kch at Va node: Va-6 + Va-Vc 2 so 4 Va alt +] + + Ve [-'12] Z + Ve [12] 3 3 2 Vale Ve[+] = 3 Va (?) - 3/22 V = V0[3]- 3 2 Ive

By substituting ve cin eruation o (16 (34) -3)+ V.(-) = 1/2 3.5V Va 35V By substituting Va cin ve eqration: V = (35) (372) 5) (3/2) – 3 -3 니 Vc = 21-12 9/4 म Ve= 2.250 ... Th= Vavb By substituting their values Th = 3.5 = 35 - 2 Vih -1.5V !!

To find Rini 1 RTLY 3 1 42 22 - 2 42 1 42 - TH I Here I an dividing diagram dividing diagram in to 3 parts for calculation pape

Past 0 42 Thi Equivalent Resistance of this part CR.). RA 411141 - + Ri to 2/4=14 Ri > Rican 202 7 Ri= goon w Part 22 42 By subst stituting post o with its earuivalent desistance : 92 Line RA22 TILL Esuivalent asistance of this past (Rg)- R2=82792 = 427 (series connection)

• Ra -> M Part 3 R Th 2.12 M - ny 42 2 TIT By substituting part a with its equivalent Resistance RTU 42 ARS 9 Th Ill Rth = 4.21142 L Q Rih H + 4 H2 Rith Q RTL = 22 3

Maximuern power from transfer theorem We know that maximum power transfer to load (R = ) R should be earual to Rth R = Rth =212 Then the circuit will be as follows: Rh=92 Vith 24 RL=2.2 1:50 By apply kuh cin loop L1 ViTh - I (RTW - I CR120 V Tha IC RTV+ I (R1) Vith = P (R+R) » I: Vth RihtRL By substituting their valuer 1.5 T: 1.5 2+2 4 0.375 A Q-0.375-A Maximum power to R = I²XRL -0.3752x2 to R (P = 0.28125W Maximum power