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