# 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.

# Solve Using Current Division Rule

Find current of resistors, use the current division rule.

Suppose that $R_1=2 \Omega$ , $R_2=4 \Omega$ , $R_3=1 \Omega$ , $I_S=5 A$ and $V_S=4 V$

Solution:
$R_2$ 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

# Mesh Analysis - Supermesh

Solve the circuit and find the power of sources:

$V_S=10V$ , $I_S=4 A$ , $R_1=2 \Omega$ , $R_2=6 \Omega$ , $R_3=1 \Omega$ , $R_4=2 \Omega$ .

Solution:
There are three meshes in the circuit. So, we need to assign three mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.

# Solve By Source Definitions, KCL and KVL

Find the voltage across the current source and the current passing through the voltage source.

Assume that $I_1=3A$ , $R_1=2 \Omega$ , $R_2=3 \Omega$ , $R_3=2 \Omega$ , $I_1=3A$ , $V_1=15 V$ ,

Solution
$R_1$ is in series with the current source; therefore, the same current passing through it as the current source:

# Find Voltage Using Voltage Division Rule

Determine voltage across $R_2$ and $R_4$ using voltage division rule.
Assume that
$V_1=20 V$ , $R_1=10 \Omega$ , $R_2=5 \Omega$ , $R_3=30 \Omega$ and $R_4=10 \Omega$

Solution:
Please note that the voltage division rule cannot be directly applied. This is to say that:
$R_1$ and $V_1$ are parallel. So the voltage across $R_1$ is equal to $V_1$ . This can be also calculated using KVL in the left hand side loop: