Please help step by step solution for this circuits problem. Solution You can so

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You can solve this circuit using KVL or ( mesh current method ) so what u need to do first is put ur currents directions, which will be clockwise for both. after this u see the node above the 2k ohm, name it Vth and u cud name the 1 above the 4k ohm v1 for example if ur solving using KCL (NVM). So for the 2 loops u got there name the left loop current i1 and the right loop current i2. (Both of them clockwise) ok, so now start KVL for the first loop (i1) i1(2k) + 6 + (i1-i2)(4k)=0 ** notice the i1 - i2 is because the resistance is between the 2 loops,and we didnt say i2- i1 cus we r doing kvl for the loop i1 , so u start with it ( not necessary tho) make a few simplifications and its > i1(6k) = i2(4k) - 6 >> name this equation (1) k so now start KVL for 2nd loop (i2) -12 + i2(2k) + (i2-i1)(4k) after few simplifications , get i2 in 1 side i2= (12 + i1(4k))/6k >> name this equation (2) now subs (2) in (1) u will get i1 and i2-------- i1 = 0.6 mA , i2 = 2.4 mA now what u need is vth , so get an equation for it * V=IR * Vth= 2k(i2) subs with ur i2 and gratz u got ur vth :D which is 4.8V Its really easy, all u need is practise alot and u will solve it in no time ^^ oh btw u can name vth >> vo, it doesnt really matter.

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