BLOG 18 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE THEVENIN THEOREM IN AC CIRCUIT.

WHAT IS THEVENIN THEOREM IN AC CIRCUIT?

Thévenin's Theorem for AC circuits with sinusoidal sources is very similar to the theorem we have learned for DC circuits. The only difference is that we must consider impedanceinstead of resistance. Concisely stated, Thévenin's Theorem for AC circuits says:

Any two terminal linear circuit can be replaced by an equivalent circuit consisting of a voltage source (V_Th) and a series impedance (Z_Th). 

In other words, Thévenin's Theorem allows one to replace a complicated circuit with a simple equivalent circuit containing only a voltage source and a series connected impedance. The theorem is very important from both theoretical and practical viewpoints.

It is important to note that the Thévenin equivalent circuit provides equivalence at the terminals only. Obviously, the internal structure of the original circuit and the Thévenin equivalent may be quite different. And for AC circuits, where impedance is frequency dependent, the equivalence is valid at one frequency only.

Using Thévenin's Theorem is especially advantageous when:

· we want to concentrate on a specific portion of a circuit. The rest of the circuit can be replaced by a simple Thévenin equivalent.

· we have to study the circuit with different load values at the terminals. Using the Thévenin equivalent we can avoid having to analyze the complex original circuit each time. 

We can calculate the Thévenin equivalent circuit in two steps:

1. Calculate Z_Th. Set all sources to zero (replace voltage sources by short circuits and current sources by open circuits) and then find the total impedance between the two terminals.

2. Calculate V_Th. Find the open circuit voltage between the terminals. 

EXAMPLE

In this circuit, the load is the series-connected RL and CL. These load components are not part of the circuit whose equivalent we are seeking. Find the current in the load using the Norton equivalent of the circuit.

v₁(t) = 10 cos wt V; v₂(t) = 20 cos (wt+30°) V; v₃(t) = 30 cos (wt+70°) V;

v₄(t) = 15 cos (wt+45°) V; v₅(t) = 25 cos (wt+50°) V; f = 1 kHz.

First find the open circuit equivalent impedance Z_eq by hand (without the load).

Numerically

 Z_N = Z_eq = (13.93 - j5.85) ohm. 

Below we see TINA's solution. Note that we replaced all the voltage sources with short circuits before we used the meter.

 

Now the short-circuit current:

The calculation of the short-circuit current is quite complicated. Hint: this would be a good time to use Superposition. An approach would be to find the load current (in rectangular form) for each voltage source taken one at a time. Then sum the five partial results to get the total.

We will just use the value provided by TINA: 

i_N(t) = 2.77 cos (w×t-118.27°) A

 

Putting it all together (replacing the network with its Norton equivalent, reconnecting the load components to the output, and inserting an ammeter in the load), we have the solution for the load current that we sought:

 

By hand calculation, we could find the load current using current division:

Finally

I = (- 0.544 - j 1.41) A

and the time function

i(t) = 1.51 cos (w×t - 111.1°) A

LEARNING

•     In solving thevenin theorem first we find the ZTH and VTH.
•     In finding ZTH we need to turn off all the sourcing.
•     We can use the series and parallel in solving the ZTH.
•     In solvong the VTH we can use the mesh and nodal analysis.
•     We can use the basic law in solving the VTH.

BLOG 17 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE SOURCE TRANSFORMATION IN AC CIRCUIT.

WHAT IS SOURCE TRANSFORMATION IN AC CIRCUIT?

   -Source transformation is simplifying a circuit solution, especially with mixed sources, by transforming a voltage into a current source, and vice versa. Finding a solution to a circuit can be difficult without using methods such as this to make the circuit appear simpler. Source transformation is an application of Thévenin's theorem and Norton's theorem.Performing a source transformation consists of using Ohm's law to take an existing voltage source in series with a resistance, and replace it with a current source in parallel with the same resistance. Remember that Ohm's law states that a voltage on a material is equal to the material's resistance times the amount of current through it (V=IR). Since source transformations are bilateral, one can be derived from the other. [2] Source transformations are not limited to resistive circuits however. They can be performed on a circuit involving capacitors and inductors, as long as the circuit is first put into the frequency domain. In general, the concept of source transformation is an application of Thévenin's theorem to a current source, or Norton's theorem to a voltage source.

Specifically, source transformations are used to exploit the equivalence of a real current source and a real voltage source, such as a battery. Application of Thévenin's theorem and Norton's theorem gives the quantities associated with the equivalence. Specifically, suppose we have a real current source I, which is an ideal current source in parallel with an impedance. If the ideal current source is rated at I amperes, and the parallel resistor has an impedance Z, then applying a source transformation gives an equivalent real voltage source, which is ideal, and in series with the impedance. This new voltage source V, has a value equal to the ideal current source's value times the resistance contained in the real current source V=I*Z. The impedance component of the real voltage source retains its real current source value.

In general, source transformations can be summarized by keeping two things in mind:

Ohm's Law

Impedances remain the same

LEARNING

• In solving the source transformation, if the voltage source change to current source  use the formula of I = V / Z.
• In solving the source transformation, if the current source change to the voltage source use the formula of V =  I Z.
• Then we can use the kirchhoff's voltage law and we can use the basic law in solving the source transformation.

BLOG 16 - FUNDAMENTAL CIRCUIT

THIS BLOG CONTAIN ABOUT THE SUPERPOSITION THEOREM IN AC CIRCUIT.

WHAT IS SUPERPOSITION THEOREM IN AC CIRCUIT.

    -The superposition theorem states that in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently. The theorem is valid for any linear circuit. The best way to use superposition with AC circuits is to calculate the complex effective or peak value of the contribution of each source applied one at a time, and then to add the complex values. This is much easier than using superposition with time functions, where one has to add the individual time functions.

• When removing a voltage source, its voltage must be set to zero, which is equivalent to replacing the voltage source with a short circuit.

• When removing a current source, its current must be set to zero, which is equivalent to replacing the current source with an open circuit. 

• Superposition Theorem 

•  Since AC circuits are linear, the superposition theorem applies to AC circuits the same way it applies to dc circuits. 

•  The theorem becomes important if the circuit has sources operating different frequencies. 

• In this case, we must have a different frequency‐domain circuits for each frequency. 

•  The total response must be obtained by adding the individual responses in the time domain. 

•  İts incorrect to try to add the responses in the phasor or frequency domain.

EXAMPLE

Notice that both sources have the same frequency: we will only work in this chapter with sources all having the same frequency. Otherwise, superposition must be handled differently.

Find the currents i(t) and i₁(t) using the superposition theorem.

First substitute an open circuit for the current source and calculate the complex phasors I', I1' due to the contribution only from VS.

The currents in this case are equal: 

I' = I₁' = V_S/(R_i + R₁ + j*w*L) = 50/(120+j2*p*4*10⁵*10^-5) = 0.3992-j0.0836

I' = 0.408 e^j 11.83° A

Next substitute a short-circuit for the voltage source and calculate the complex phasors I'', I1'' due to the contribution only from IS.

In this case we can use the current division formula:

I'' = -0.091 - j 0.246 A

and

I₁^" = 0.7749 + j 0.2545 A

The sum of the two steps:

I = I' + I" = 0.3082 - j 0.3286 = 0.451 e^- j 46.9° A

I₁ = I₁^" + I₁' = 1.174 + j 0.1709 = 1.1865 e^j 8.28° A 

The time functions of the currents: 

i(t) = 0.451 cos (w×t - 46.9°) A

i₁(t) = 1.1865 cos (w×t + 8.3°) A

LEARNING

•   I learn about the ac superposition theorem is we need to turn of the one of the independent source.
•   In Alternating Current it has they same procedure in Direct Current.
•   When removing source, we short the voltage source and the current source we open it.
•   We can use the mesh and nodal analysis to solve the superposition