Kirchoff Electronic Circuits - Yousef's Notes
Kirchoff Electronic Circuits

Kirchoff Electronic Circuits

#Rules

  • Current flow described by a system of linear equations
  • A voltage source such as a battery forces a current to flow the network.
  • When current passes through resistor some voltage is used by by Ohm’s law.
  • The current flowing in loops with direction arbitrary. If a current turns out to be negative, then the actual direction is opposite to the figure.
  • If current direction shown is away from the positive (longer) around the negative (shorter) side, voltage is positive; otherwise, it’s negative.

#Example 1

#Loop 1:

$4I_1 + 4I_1 + 3I_1 = 11I_1$ and direction in loop 1 is opposite to loop 2, so$3I_2$ is negative. Voltage 30 is in loop 1 direction then is positive

$$ 11I_1 - 3I_2 = 30 $$

#Loop 2:

$I_2 + I_2 + 3I_2 + I_2 = 6I_2$ direction in loop 2 is opposite to loop 1, so $3I_1$ is negative and direction in loop 2 is also opposite to loop 3, so $I_3$ is negative. Voltage 5 is in loop 2 direction then is positive

$$ -3I_1 + 6I_2 - I_3 = 5 $$

#Loop 3

$I_3 + I_3 + I_3 = 3I_3$ direction in loop 3 is opposite to loop 2, so $I_2$ is negative Both voltages: 20 + 5 are in opposite direction then they are negative

$$ -I_2 + 3I_3 = -25 $$

#Equations

$$ \begin{align*} 11I_1 - 3I_2 = 30 \\ -3I_1 + 6I_2 - I_3 = 5 \\ -I_2 + 3I_3 = -25 \end{align*} $$

#Matrix Form

$$ \begin{bmatrix} 11 & -3 & 0 \\ - 3 & 6 & -1 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} 30 \\ 5 \\ - 25 \end{bmatrix} \rightarrow \begin{bmatrix} 11 & -3 & 0 & | & 30 \\ - 3 & 6 & -1 & | & 5 \\ 0 & -1 & 3 & | & -25 \end{bmatrix} $$

#RREF

$$ RREF \rightarrow \begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & -8 \end{bmatrix} $$

#Solution

$$ \boxed{I_1 = 3 \quad I_2 = 1 \quad I_3 = -8} $$

#Example 2 (video)

put video here