## 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
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## 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.
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## 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:
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## 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:
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## Find currents using KVL

Find resistor currents using KVL.

Solution:

$R_1$ 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:

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## Total Energy Stored - Circuit with Capacitors and Inductors

Find the total energy stored in the circuit.

Solution
The circuit contains only dc sources. Recall that an inductor is a short circuit to dc and a capacitor is an open circuit to dc. These can be easily verified from their current-voltage characteristics. For an inductor, we have $v(t)=L \frac{d i(t)}{dt}$. Since a dc current does not vary with time, $\frac{d i(t)}{dt}=0$. Hence, the voltage across the inductor is zero for any dc current. This is to say that dc current passes through the inductor without any voltage drop, exactly similar to a short circuit. For a capacitor, the current-voltage terminal characteristics is $i(t)=L \frac{d v(t)}{dt}$. Voltage drop across passive elements due to dc currents does not vary with time. Therefore, $\frac{d v(t)}{dt}=0$ and consequently the current of the capacitor is zero. This is to say that dc current does not pass through the capacitor regardless of the voltage amount. This is similar to the behavior of an open circuit. Please note that unlike dc current, ac current passes through capacitors in general.
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## Thévenin's Theorem - Circuit with Two Independent Sources

Use Thévenin's theorem to determine $I_O$.

Solution
Lets break the circuit at the $3\Omega$ load as shown in Fig. (1-27-2).
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## Thévenin's Theorem - Circuit with An Independent Source

Use Thévenin's theorem to determine $V_O$.

Solution
To find the Thévenin equivalent, we break the circuit at the $4\Omega$ load as shown below.
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## Superposition Method - Circuit With Dependent Sources

Determine $I_x$, $I_y$ and $V_z$ using the superposition method.

Solution
I. Contribution of the $-2V$ voltage source:
We need to turn off the current source by replacing it with an open circuit. Recall that we do not turn off dependent sources. The resulting circuit is shown below.
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## Superposition Problem with Four Voltage and Current Sources

Determine $V_x$ and $I_x$ using the superposition method.

Solution
I. Contribution of the $-5V$ voltage source:

To find the contribution of the $-5V$ voltage source, other three sources should be turned off. The $3V$ voltage source should be replaced by short circuit. The current source should be replaced with open circuits, as shown below.
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