Q:    In the following circuit find out the value of R_L and the highest power dissipation across it by Thevenin's Theorem.

Solution:

 We have to determine R_L and the highest power across it by utilizing Thevenin's theorem.

Step 1: Remove R_L

To remove R_L to calculate V_th .In this R_th is similar to R_L.

Step 2: determining V_th value

We desire to calculate V_th . By apply KVL in two loops to determine the individual loop currents.

            Here                   I₁ = 2mA

                        By applying KVL to loop 2

                                                            6kI₂ +3+ 3k (I₂ - I₁) =0

                                                                 9kI₂ -3kI₁ + 3 =0

                                                                         9kI₂ - 6 +3 =0

                                                                          I₂ = 1/3 mA =0.33mA

Now calculate the voltage across 4k & 6k to search the value of V_th

                        So,

                                    V_AB = 4k (I₁)

                                             = 4k (2m)                    //putting the value of I₂

                                    V_AB = 8 V

                                    V_BC = 6k (I₂)

                                              = 6k x 0.33m               //putting the value of I₂

                                    V_BC = 2V

            Therefore,

                                                V_th = V_AB + V_BC

                                              = 8 +2 = 10 volts

Step 3: Determining R_th

To estimate the value of R_th. Open circuiting the current source & short circuiting the voltage source.

3k is in parallel to 6k & 4k is series with these two therefore

                                    3k||6k + 4k = 3k x 6k/(3k + 6k)   +4k = 2k +4k

                                                                            = 6k = R_th =R_L                  // because R_L =R_th

                                                      R_th =R_L= 6k      

Step 4:   Calculating unknown quantity.

After estimating value of 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 two resistances.

                        For highest power dissipation

                                                R_L =R_th = 6k

                                                   P_L = I_2R ........... (i)

                                    From using Ohm's law V=IR & I = V/R

                        Therefore                    I =10/12

                                                  By putting this value in (i) we get

                                                            P_L = (10/12k) X 2 X (6k)

                                                              =   (0.83) 2 X 6 =4.1 mW