# Problem

Find $I_x$ and $I_y$ :

# Solution

Three resistors are in series and their equivalent, $6\Omega$, is parallel with the voltage source. So, according to the Ohm's law: $I_y=-\frac{6V}{6 \Omega}=-1 A$. The negative sign comes from the direction $I_y$.
Applying KCL at the bottom node:
$-(-2A)+I_x+I_y=0 \rightarrow I_x=-1 A$.
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# Find Equivalent Impedance - AC Steady State Analysis

Determine the driving-point impedance of the network at a frequency of $2$kHz:

# Solution

Lets first find impedance of elements one by one:

## Resistor $R$

The resistor impedance is purely real and independent of frequency.

$Z_R=R=20 \Omega$

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## Problem

Find $I_x$ and $I_y$:

# Superposition method - Circuit with two sources

Find $I_x$ using superposition rule:

# Solution

## Superposition

The superposition theorem states that the response (voltage or current) in any branch of a linear circuit which has more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are turned off (made zero).

# Find Thevenin's and Norton's Equivalent Circuits

Find Thevenin's and Norton's Equivalent Circuits:

Suppose that $R_1=5\Omega$, $R_2=3\Omega$ and $I_S=2 A$.

# Solution

The circuit has both independent and dependent sources. In these cases, we need to find open circuit voltage and short circuit current to determine Norton's (and also Thevenin's) equivalent circuits.
Suppose that $R_1=2 \Omega$, $R_2=4 \Omega$, $R_3=1 \Omega$, $I_S=5 A$ and $V_S=4 V$
$R_2$ and $R_3$ are parallel. The current of $I_S$ is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get