Q:  Determine the voltage Vo by using Thevenin's theorem.

Solution:

We have to calculate V₀ by utilizing Thevenin's theorem .               

Step 1: Remove R_L

Here value of R_L is 8k resistor at which we desire to calculate the voltage Vo.

Step 2: Finding the value of  V_th

To calculate V_th we will use superposition method

When only voltage source is active

No current will pass in 'CA' branch therefore voltage drop across 4k & 2k resistor is zero therefore

                                                V_CD = V_th1

            Now by voltage division rule

                        V_CD =V_th1 =V_6k = (12 x6)/(3+6)

                                                    = 8 V

                        Now when only current source is active

                        3k||6k = (3k x6k)/(3k+6k)

                                       =2k

2k is in series to 2k =4k

Current flow through open circuit is zero by using ohm's law so

                                                                            V_th2 = 4k x2m

                                                                 V_th2= 8 V

                        Therefore

                                                            V_th = V_th1 +V_th2

                                                                         = 8 +8 = 16 V

Step 3: Determining R_th

                                                            3k||6k = (3k x6k)/9k= 2k

                                                            R_th = 2k+2k+4k =8k

Step 4:   Determining unknown quantity.

After determining V_th & R_th, re-inserting the load resistance R_L in the circuit in series along R_th and letting the V_th like a battery in series along these two resistances.

                                                                     V₀ = (8k/16k) x16 

                                                                    V₀= 8 V