Solve Using Current Division Rule

Find current of resistors, use the current division rule.
Problem 1246 (1)
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:
Problem 1226 - 1
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.
Problem 1213
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
solve using voltage division rule

Solution:
Please note that the voltage division rule cannot be directly applied. This is to say that:
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Total Energy Stored - Circuit with Capacitors and Inductors


Find the total energy stored in the circuit.

find the energy stored in the circuit
Fig. (1-28-1) - 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.

Thevenin's Theorem - Circuit containing two independent sources
Fig. (1-27-1) - Circuit with two independent sources


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 .

A circuit with a voltage source
Fig. (1-26-1) - The Circuit


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.

Superposition - Circuit with dependent sources

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.
A circuit with four voltage and current sources to be solved by 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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