Cramer's Rule

       If the network is complex, the number of equation i.e. unknowns increase. In such case, the solution of simultaneous equations can be obtained by Cramer's Rule for determinants.

  Let us assume that set of simultaneous equations obtained is, as follows,

         a₁₁  x₁₁  + a₁₂  x₂  +......+a_1n  x_n  = C₁ 
         a₂₁  x₁  + a₂₂  x₂  +........+a_2n  x_n  =C₂  
         a_n1  x₁  + a_n2  x₂  +........+a_nn  x_n  =C_n 
         Where C₁,C₂, ................, C_n  are constants.
       Then Cramer's Rule says that from a system determinant ∆ or D as,

       Then obtain the subdeterminants by replacing column of by the column of constants existing on right hand side of equations i.e. C₁, C₂,  ...... C_n ;

and

        The unknowns of equations are given by the Cramer's rule as,
        X₁  = D₁/D,     X₂  = D₂/D,........., X_n  = D_n/D
       Where D₁,D₂,....., D_n and D are values of the respective determinants.
Example1  : Apply Kirchhoff's current law and voltage law to the circuit shown in the Fig. 1.
       Indicate the various branch currents.
       Write down the equations relating the various branch currents.
       Solve these equations to find the value of these currents.
       Is the sign of any of the calculated currents negative?
       If yes, explain the signification of the negative sign.

Fig 1

 Solution : Application of Kirchhoff's law :
Step 1 and 2 : Draw the circuit with all the values which are same as the given network.
                      Mark all the branch currents starting from +ve of any the source, say +ve of 50 V source.

Step 3 :  Mark all the polarities for different voltages across the resistances. This is combined with step2 shown in the network below in 1(a).

Fig. 1 (a)

Step 4 : Apply KVL for different loops.
       Loop 1 : A-B-E-F-A.,
                                    - 15 I₁ + 20 I₂ + 50  = 0                                        ...........(1)
       Loop 2 : B-C-D-E-B.,
                                   -30 (I₁ - I₂ ) - 100 + 20 I₂ = 0                               ............(2)
       Rewriting all the equations, taking constants on one side.
                    15 I₁ + 20 I₂ = 50       .....(1)       and       - 30 I₁ + 50 I₂ =  100      ........(2)
       Apply Cramer;s rule,  

       Calculating D₁,

                                      I=D₁/D = 500/1350 = 0.37 A
Calculating D₂,

                         I = D₂ /D= 3000/1350 =2.22 A
       For I₁and  I₂, as answer is positive, assumed direction is correct.
.^..    For answer is 0.37 A. For answer is 2.22 A
.                           I₁ - I₂ = 0.37-2.22 = -1.85 A
Negative sign indicates assumed direction is wrong.
i.e. I₁ - I₂ = 1.85 A flowing in opposite direction to that of the assumed direction.

Example 2 : Find i₁ and i₂ using KCL and KVL.

Fig. 2

Solution : The current distribution using KCL is as shown,

Fig. 2 (a)

 Key point : KVL should not be applied to the loop consisting current source.
       From branch DE,
                                i₁  = 5+ 3 i₂                               .............(1)
       Applying KVL to the loop BCDEFGB without current source,
                               -1 x (5 + 3 i₂ ) + 5 i₂ = 0            ............(2)
.^..                               2 i₂ = 5
.^..                                i₁ = 2.5 A
from equation (1)        i₁ =  12.5 A