Q:  In the given circuit, determine the value of voltage V_o by using Thevenin's theorem.

Solution:

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

Step 1: Removing R_L from the circuit

Here R_L is equal to 6k resistor at which we desire to calculate the voltage Vo.

Step 2: calculate V_th value

                        By apply KVL to the circuit

Here                                I₂ = 2mA

For loop 1

                                                2k( I₁ -I₂) +4kI₁ = 6

                                                   2kI₁ - 2kI_{2 +}4kI₁   =6

                                                           6kI₁ -2kI₂ =6

Now put value of I₂

                                                            6kI₁ - 4 =6

                                                                 I₁ = 10/6k

                                                                I₁ =1.6mA

Voltage around 4k resistor will be

                                                             V_4k = 1.6m x4k=20/3

                                                                         V_4k = 6.66V

Voltage across 2k resistor will be

                                                            V_2k = 2m x2k

                                                              V_2k = 4V

Therefore

                                                             V_th = V_4k +V_2k

                                                                    =20/3 + 4 = 6.66 +4 =10.66 volts

Step 3: Calculating R_th

                                                            4k||2k = 4k x2k/6k

                                                                      =8/6 k =4/3 k=1.33 k

They are in series therefore

                                                 R_th=1.33k +2k=10/3k =3.33k

Step 4:  Determining unknown quantity.

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

                                    V₀ = 32/3 x6 x 1/(10/3+6 )

                                                = 64 x3/10 +18

                                                =192/28   =48/7V