The Basics

The Basics

Solving complex numbers:

Complex number is a number that can be expressed in the form z= r+bi. (r is the real part and b is the imaginary part), the real part and the imaginary part of a complex number cannot be combined. Solving complex numbers with the use of common operators are easy. 

Complex number operations:

Sample problem

z1=x1+j(y1)
z2=x2+j(y2)

Addition:

Add the real part and the imaginary part.
          z1+z2= (x1+x2) + j(y1+y2)

Subtraction:
Subtract the real part and the imaginary part.
          z1-z2= (x1-x2) + j(y1-y2)   

Multiplication:
Multiply the real part then add the angles. (This method can be only used when the complex number is converted to Polar form.)
          x1*x2∟ᴓ1+ ᴓ2

Division:
Divide the real part then subtract the angles. (This method can be only used when the complex number is converted to Polar form.)
          x1/x2∟ᴓ1- ᴓ2

FIRST ORDER CIRCUITS

What is first order circuits?

     -can only contain one energy storage element (a capacitor or an inductor). the circuit will also contain resistance.

TWO TYPE OF FIRS ORDER CIRCUIT

   *RC circuit
   *RL circuit

                  SOURCE-FREE CIRCUITS 

-is one where all independent source have been disconnected from the circuit after some switch action. the voltages and currents in the circuit typically will have some transient response due to initial conditions

FOR RL Circuits

LEARNING'S:

    Wow! i learned a lot for this awesome circuits thank's for our GOOD looking teacher because of him i simply learned the process of electricity flowing to our small mansion. I learn how determine the currents passing through our appliances and how many voltage needed on it. 

MAXIMUM POWER TRANSFER

What is maximum power transfer?

Maximum Power transfer

-States that to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals

The condition of maximum power transfer does not result in maximum efficiency.if we define the efficiency n as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from  the above circuit diagram that

 Another example:

INDUCTOR

What is inductor ?

Inductor

-also called a coil or reactor,is a passive two-terminal electrical components which resists change in electric current passing through it

Formula:

        SAMPLE INDUCTOR

INUCTOR IN SERIES AND PARALLEL

 FOR PARALLEL

   sample figure:

CAPACITOR

CAPACITOR

-are one of the standard components in electronic circuits and a passive two-terminal electrical component used to store energy in an electric field.

Formula:

 -  Where Q is the charge on the conductor

 - V is the potential applied across the conductor

 - C is proportionally constant called as capacitance

if capacitor are connected in series, the formula is given by:

  

If capacitor are connected in parallel,the capacitance formula is given by:

 sample circuits:

    

Thevenin's Theorem

Thevenin's Theorem

-States that"Any linear circuit containing several voltages and resistance can be replaced by just a single voltage in series with a single resistor".In other words, it is possible to simplify and Linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load

Example:

as far as the load resistor RL is connected, any network consisting of resistive circuit elements and energy source can be replaced by one single Equivalent resistence Rs and equivalent voltage Vs, where Rs is the source resistance value looking back into the circuit and Vs is the open circuit at the terminals.

For example:

 Consider the circuit from the previous section

First, we have to remove the center 40Ω resistor and short out all the emf's connected to the circuit, or open circuit any current sources. the value of Rs is found by calculating the total resistance at the terminal A and B with the emf's removed, and the value of the voltage required Vs is the total voltage across terminals A and B with an open circuit and no load resistor Rs connected.then we get the following circuit.

We now need to reconnect the two voltages back into the circuit, and as Vs=Vab the current flowing around the loop is calculated as:

so the voltage drop across the 20Ω resistor can be calculated as:

Vab=20 - (20Ω x 0.33A)= 13.33V 

then the thevenins Equivalent circuit is shown below with the 40Ω resistor connected.

and from this current flowing in the circuit is given as:

I=(13.33V/6.67 + 40)=0.286A

SUPERPOSITION

WHAT IS SUPERPOSITION?

-is a another algorithm circuit analysis method, with a prescribed procedure, like the node voltage or loop current method

-it applies on;y in circuits with 2 or more independent sources.

The superposition procedure

1. make a partial circuit by choosing one source to keep and turning of all the other independent sources.

2. Solve this "partial circuit" for the quantities of interest. Use whatever techniques are applicable to the partial circuit.

3. Go back to the original and create another circuit by choosing another source to keep (different than the first one) and turning off all the other sources in the circuit.

4. Solve the second partial circuit for the quantities of interest.

5. Repeat the process until each independent source has been used in partial circuit

6. Sum up all the partial results to get total results for the quantities of interest 

If there are n independent sources in a circuit, then when using superposition, you would expect to solve n partial circuit and find n partial results for the whatever you are looking for 

Example of the superposition method

Consider the circuit below. It has one voltage source and one current source, and we are asked to find the voltage VR2.

1. Choose one source. It doesn't matter which, so start with the voltage source. Turn off the current source (make it open circuit) to create the first partial circuit.

2. Solve the partial circuit

3. Now turn off the voltage source (by using it out) and keeping the current source.

4. Solve the partial circuit

5. Sum the partial solution to get total solution: