Question

Problem 6 The following dc measurements were made on the resistive network shown in (Figure 1). Measurement 1 Measurement 2 V =4V Vi = 20 mV 11 =5 mA I = 20 MA V2=0 V = 40 V 12 = -200 mA 12 = 0 A variable resistor R, is connected across port 2 and adjusted for maximum power transfer to R Suppose that V = 6.25 mV. Figure 2500 + v Resistive network

Answer 1

Problem 6e- measurement 1 Iz 250 Y IN (Norton current) V2=0 Concept - Measurement 1 satisfy Norton theorem Norton theorems - It state that any corcuit can be replaced by a constant current source (Norton current) Basallel with Resistance (Norton Resistance) • Norton current is determined by making short ciscuit (V2=0) across which we have to determine Noston equivaleut ( IN & R) · Hence, measurement 1 satisfy Norton theorem in which U2=o » Iz=-IN = -200MA = - IN IN = 200 mA ( Nortos Current) Measurent 2 25ou luz Izzo [open arcuit

cond can Eith the venin th asurement 2 satisfy Thevenin theorem, evenin theorem: - It state that any circuit replace og a constant voltage source (Uth) in series with equivalent Resistance (Ⓡ). inevenin voltage is determined by opening the curcuit across which we have to determine the venin equivalent ( Uth and R) → V2 = 'some value = Vth 240V I2-0 Hence, Thevenin voltage =you - (uth) Equivalent Resistance R) = Thevenin voltage Norton current 404 200mA 4ox 103 - = 40 1o 200x10-31 40 x loan = 2002 200 Log > R=2003 Maximum Power Transfer theorem: It state that For maximum Power transfered to load, Thevenon Resistance (R) and Load Resistance (Ro) must be equal. -> RO - R = 2002 . Pmax = VE 140)2 40X40 - 2 watt 4 x 200 4x200 Hence, maximum Power Transfered to Ro (Pmax) = 2 0