Thevenin’s theorem

                            Thevenin’s theorem

"The current in a load impedance connected between two terminal of a network of generators and linear impedance is same as if this load impedance is connected  to a single voltage generator whose e.m.f. is equal to open circuit (when there is no load) voltage between the terminals". 

The impedance will be equal to the impedance of the network into the terminal when all the generators are replaced by their internal impedance.

 Note : It is a reduction technique by which we can reduce a complicated network containing several generators and impedance into a single equivalent voltage generator.

   In fig. (a) N is the network containing a number of generators and impedance with output terminals A and B. ZL  is the load impedance. Let E’ be the open circuit voltage across A and B when all the generators are replaced by their internal impedance. The network N will produce the same current in external load impedance ZL as a single voltage generator of e.m.f. E’ and internal impedance Z’ would do (Fig (b)).

  

          To established the theorem reduce the network to an equivalent T-section arrangement of impedance Z1 ,Z2 and Z3 as shown in fig.(c) . Here E is the source of e.m.f., Current I1 is supplied by the source and IL be the current flowing in the load impedance ZL . According to Thevenin’s theorem, the circuit with e.m.f., E’ and impedance Z’  in fig.(b) will be equivalent to the circuit with impedance ZL and e.m.f.  E in fig.(c).

To find the expression of the load current IL , apply Kirchoff’s second law in the mesh of fig.(c), we have –

                    E = I1 (Z1 + Z3) – IL Z3  ………………….(1)

 and             0 = IL (Z2 + Z3 + ZL) – I1 Z3  ……………(2)

 from equation (2), we get ----

                   I1 =  IL (Z2 + Z3 + ZL) / Z3

Substituting this value of I1 in equation (1), we get -----

                  E = IL (Z1 + Z3) (Z2 + Z3 + ZL) / Z3   - IL Z3

                  E =  IL {(Z1 + Z3) (Z2 + Z3 + ZL) / Z3  - Z3 }

                  IL =  E / {(Z1 + Z3) (Z2 + Z3 + ZL) / Z3  - Z3 }

  I_L  = E / {(Z₂(Z₁ + Z₃) + Z₃² +Z₁ Z₃ + Z_L (Z₁ + Z₃) – Z₃² ) / Z₃ }

  I_L = E Z₃ / {Z₂(Z₁ + Z₃) + Z₁ Z₃ + Z_L (Z₁ + Z₃)}

         Multiply and divide by (Z₁ + Z₃), we get ---

  I_L  = E(Z₃ /(Z₁ + Z₃)) / { Z₂ + (Z₁Z₃/(Z₁ + Z₃)) + Z_L }  ……(3)

From fig.(c) , the open circuit voltage at terminal A and B when load  is disconnected isgiven by---

  E^’ = E Z₃/ (Z₁ + Z₃) = E { Z₃/ (Z₁ + Z₃)}  …………(4)

The impedance between the terminal A and B after disconnected the load Z_L in fig.(c) is given by ---

   Z^’ = Z₂ + Z₁Z₃/(Z₁ + Z₃)                ………………..(5)

Substituting the values of E^’ and Z^’ from equation (4), (5) in equation (3) …..

                      I_L = E^’ / (Z^’ + Z_L)

This is the expression of current for network in fig.(b).

                                                         https://youtu.be/UCHIvvAvekA