### 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  x +......+a1n  x = C
a21  x + a22  x +........+a2n  x =C2
an1  x + an2  x +........+ann  x =C
Where C1,C2, ................, C 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,  ...... Cn ;
and

The unknowns of equations are given by the Cramer's rule as,
X = D1/D,     X = D2/D,........., X = Dn/D
Where D1,D2,....., Dn 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.
.                           I1 - I2 = 0.37-2.22 = -1.85 A
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,
i= 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