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We have already studied the superposition theorem for DC circuits. In this chapter we will show its application for AC circuits.

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.

To calculate the contribution of each source independently, all the other sources must be removed and replaced without affecting the final result.

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.

Now let’s explore an example.

In the circuit shown below

R_{i} = 100 ohm, R_{1}= 20 ohm, R_{2} = 12 ohm, L = 10 uH, C = 0.3 nF, v_{S}(t)=50cos(wt)
V, i_{S}(t)=1cos(wt+30°)
A, f=400 kHz.

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 i_{1}(t) using the superposition theorem.

Let’s use TINA and hand calculations in parallel to solve the problem.

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**‘ = **I**_{1}‘= **V**_{S}/(R_{i} + R_{1} + **j***
w*L) = 50/(120+**j**2*
p*4*10^{5}*10^{-5}) = 0.3992-**j**0.0836

**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

I_{1}^{“} = 0.7749 + **j** 0.2545 A

The sum of the two steps:

**I** = **I**‘ + **I**” = 0.3082 – **j** 0.3286 = 0.451 e^{– j46.9}
^{°}A

**I**_{1} = **I**_{1}^{“} + **I**_{1}‘ = 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

i_{1}(t) = 1.1865 cos (
w×
t + 8.3
°
)A

f:=400000;

Vs:=50;

IG:=1*exp(j*pi/6);

om:=2*pi*f;

sys I,I1

I+IG=I1

Vs=I*Ri+I1*(R1+j*om*L)

end;

I=[308.093m-329.2401m*j]

abs(I)=[450.9106m]

radtodeg(arc(I))=[-46.9004]

abs(I1)=[1.1865]

radtodeg(arc(I1))=[8.2749]

As we said in the DC chapter on superposition, it gets pretty complicated using the superposition theorem for circuits containing more then two sources. While the superposition theorem can be useful for solving simple practical problems, its main use is in the theory of circuit analysis, where it is employed in proving other theorems.