# 7.10 Solving a circuit using simultaneous equations.

Consider the circuit that was solved using the superposition theorem in the previous section. Using this alternative method the directions of the currents and the voltage drops are guessed to begin with. If these guesses prove to be incorrect, the values obtained from the calculations will turn out to be negative and so they can then be corrected.

Kirchhoff's voltage law can be applied to any loop in the circuit.
(Kirchhoff's voltage law states that around any loop in the circuit, the sum of the e.m.f. s is equal to the sum of the voltage drops.)

• Applying Kirchoff's voltage law to loop 1, gives E1 = V1 + V2 .
• Applying Kirchoff's voltage law to loop 2, gives E2 = V3 + V2 .

using Ohm's law, V = IR, to replace V1, V2 and V3 in the above expressions gives.

• E1 = I1R1 + I2R2 .
• E2 = I3R3 + I2R2 .

Putting in the values for the resistors and the e.m.f.s gives;

• 10 = 4I1 + 6I2      (1) .
• 15 = 10I3 + 6I2     (2) .

Applying Kirchhoff's current law to the junction:

• I2 = I1 + I3 .
• Therefore I3 = I2 - I1     (3) .

substituting for I3 in equation (2) gives;

• 15 = 10(I2 - I1) + 6I2 .
• 15 = 10I2 - 10I1 + 6I2 .
• 15 = - 10I1 + 16I2   (4) .

multiplying equation (1) by 10
& multiplying equation (4) by 4 gives:

• 100 = 40I1 + 60I2   (5) .
• 60 = -40I1 + 64I2   (6) .

adding equation (5) to equation (6) .

• 160 = 124I2 .
• I2= 160/124 = 1.29 A .

substituting this value of I2 into equation (1) .

• 10 = 4I1 + 6 x 1.29 .
• 10 = 4I1 + 7.74 .
• 10 - 7.74 = 4I1 .
• 2.26 = 4I1 .
• I1 =2.26/4 = 0.565 A .

putting the values for I1 and I2 into eq (3).

• I3 = I2 - I1 .
• Therefore I3 = 1.29 - 0.565 = 0.725 A .

This gives the same answers (to within 0.001A due to rounding errors) as the superposition theorem.
In this case all the current values were positive, indicating that the initial assumed directions of current flow were correct.
The voltage drops could now be found using ohms law as shown in the superposition theorem example.