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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Find currents using KVL


Find resistor currents using KVL.
Resistive Circuit

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.

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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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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Turning Sources Off


Turning off a source, which is usually used in solving circuits with superposition method, means setting its value equal to zero. For a voltage source, setting the voltage equal to zero means that it produces zero voltage between its terminals. Therefore, the voltage source must insure that the voltage across two terminals is zero. Replacing the source with a short circuit can do that. Thus, voltage sources become a short circuit when turned off.

For a current source, setting the current equal to zero means that it produces zero current. Therefore, the current source must insure that no current flows through its branch. An open circuit can do that. Hence, to turn off a current source it should be replaced by an open circuit.

How about dependent sources? The voltage/current of a dependent source is dependent on other variables of the circuit. Therefore, dependent sources cannot be turned off.

Example I: Turn off sources one by one.

turning sources off example 1-1

Example 1



Solution:
I) The voltage source:

turning sources off example 1-2

Turning off the voltage source


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Solving Quadratic Equations II: Taking Square Roots


A quadratic equation can be solved by taking the square root of both sides of the equation. This method uses the square root property,
y^2=z \to y=\pm \sqrt{z}
Before taking the square root, the equation must be arranged with the x2 term isolated on the left- hand side of the equation and its coefficient reduced to 1. There are four steps in solving quadratic equations by this method:
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