Question

3. Find V, from following circuit using Thevenin's theorem. 3 kΩ 6 kΩ 3 kΩ ' 4 mA 12V 12V 4. Find Vo from following circuit using Norton's theorem only. 12V 2 ΚΩ 3 mA ( 4 ΚΩ 4 ΚΩ 12 ΚΩ

Answer 1

Given cirunit is 3km + $6 kr ЗКл vo UMA + 12V 124 the circuit thevenins theorem open To find voltage through load where we need to find it + Vo=vx-12 66 YMA Va 12 y i=0 Зc li VX - VX =<6KYmA)+12) -{6K UMA + - Vaa 24+12 112v UMA Vx=-2 6 volts Vo= Va-12 - 26-12 -38 VOLH Jo

Consider the Circuit 12V 2 кл + 3mA uku vos UK •12k To find vo with Norton's theorem Short Circuit the load & find Isc 12V 2K 3mA UK 312k ISC Now we can make use of superporition theorem Consider source 3mA only 2K M 3 mA : 12K => 3MA Чк Isc. Isca Isc, = -( 3 x 3х4k GK Isci -2 MA Now consider voltage dourie 12 y 127 (ta 2k (ISC2 312k ? 12k ISC2 UK Isc 12 Iscz = 12k E IMA Isc = Isc, t Isc2= -2mA+loma = - IMA Vo = Isc X 4K = (-1m)(4k) = -4 volts G