# Superposition Theorem (ac)

You will recall from Chapter 8 that the superposition theorem eliminated the need for solving simultaneous linear equations by considering the effects of each source independently. To consider the effects of each source, we had to remove the remaining sources. This was accomplished by setting voltage sources to zero (short-circuit representation) and current sources to zero (open-circuit representation). The current through, or voltage across, a portion of the network produced by each source was then added algebraically to find the total solution for the current or voltage.
The only variation in applying this method to ac networks with independent sources is that we will now be working with impedances and phasors instead of just resistors and real numbers.
The superposition theorem is not applicable to power effects in ac networks since we are still dealing with a nonlinear relationship. It can be applied to networks with sources of different frequencies only if the total response for each frequency is found independently and the results are expanded in a non-sinusoidal expression, as appearing in next chapters.
One of the most frequent applications of the superposition theorem is to electronic systems in which the dc and ac analyses are treated separately and the total solution is the sum of the two. It is an important application of the theorem because the impact of the reactive elements changes dramatically in response to the two types of independent sources. In addition, the dc analysis of an electronic system can often define important parameters for the ac analysis.
We will first consider networks with only independent sources to provide a close association with the analysis of Chapter 8.
Example 1: Using the superposition theorem, find the current I through the 4-Ω reactance ($X_{L_{2}}$ ) of Fig .1.
Fig. 1: Example 1.
Solution: For the redrawn circuit (Fig .2),
Fig. 2: Assigning the subscripted impedances to the network of Fig. 1.
$$Z_1 = +j X_{L_{1}} = j 4 Ω$$ $$Z_2 = +j X_{L_{2}} = j 4 Ω$$ $$Z_3 = -j X_{C} = - j 3 Ω$$ Considering the effects of the voltage source $E_1$ (Fig. 3),
Fig. 3: Determining the effect of the voltage source E1 on the current I of the network of Fig. 1.
we have $$\begin{split} Z_{23} &= { Z_2 Z_3 \over Z_2 + Z_3} \\ &= { (j 4 Ω)(-j3Ω) \over j4Ω - j3Ω} \\ &={ 12 Ω \over j} = -j12 Ω= 12 Ω \angle -90^\circ\\ I_{S1} &= { E_1 \over Z_{23} + Z_1} = {10 V \angle 0^\circ \over -j12Ω + j4Ω}\\ &= { 10V \angle 0^\circ \over 8 Ω \angle -90^\circ} = 1.25 A \angle 90^\circ \\ \text{and} I' &= { Z_3 I_{S1} \over Z_2 + Z_3} \\ &= { (-j3Ω)(j1.25A) \over j4Ω -j3Ω} = {3.75 \over j1} \\ &=3.75A \angle -90^\circ\\ \end{split}$$ Considering the effects of the voltage source E2 (Fig. 4),
Fig. 4: Determining the effect of the voltage source E1 on the current I of the network of Fig. 1.
we have $$Z_{12} = {Z_1 \over N } = {j 4Ω \over 2} = j2Ω$$ $$I_{S_{2}} = { E_2 \over Z_{12} + Z_3} = { 5 V \angle 0^\circ \over j2Ω - j3Ω}\\ ={5 V \angle 0^\circ \over 1Ω \angle -90^\circ } = 5 A \angle 90^\circ$$ and $$I'' ={I_{S_{2}} \over 2} = 2.5 A \angle 90^\circ$$ The resultant current through the $4-Ω$ reactance $X_{L_{2}}$ is $$I = I' - I''\\ =3.75 A \angle -90^\circ - 2.5 A \angle 90^\circ\\ =-j 3.75 A - j 2.50 A= -j 6.25 A\\ I = 6.25 A \angle 90^\circ$$
For dependent sources in which the controlling variable is not determined by the network to which the superposition theorem is to be applied, the application of the theorem is basically the same as for independent sources. The solution obtained will simply be in terms of the controlling variables.
Example 2 Using the superposition theorem, determine the current $I_2$ for the network of Fig. 5. The quantities $\mu$ and h are constants.
Fig. 5: Example 2.
Solution: With a portion of the system redrawn (Fig. 6),
Fig. 6: Assigning the subscripted impedances to the network of Fig. 5.
Fig. 7: Determining the effect of the voltage-controlled voltage source on the current $I_2$ for the network of Fig. 5.
$$Z_1 = R_1 = 4Ω$$ $$Z_2 = R_2 + j X_L = 6+ j 8 Ω$$ For the voltage source (Fig. 7), $$I' = {\mu V \over Z_1+Z_2} = {\mu V \over 4Ω+ 6+ j 8 Ω} \\ ={\mu V \over 10 Ω+ j 8 Ω} = {\mu V \over 12.8 Ω \angle 38.66^\circ} \\ =0.078 \mu V/Ω \angle -38.66^\circ$$ For the current source (Fig. 8),
Fig. 8: Determining the effect of the current-controlled current source on the current I2 for the network of Fig. 5
$$I'' = {Z_1 (hI) \over Z_1+Z_2} = {(4 Ω) (hI) \over 12.8 Ω \angle 38.66^\circ}\\ = 4 (0.078) hI \angle -38.66^\circ = 0.312 hI \angle -38.66^\circ$$ The current $I_2$ is $$I_2 = I' + I''\\ I_2=0.078 \mu V/Ω \angle -38.66^\circ \\ +0.312 hI \angle -38.66^\circ\\$$ For $V = 10 V \angle 0^\circ$, $I = 20 mA \angle 0^\circ$, $\mu= 20$, and $h = 100$, $$\begin{split} I_2 &= 0.078(20)(10 V \angle 0^\circ)/Ω \angle -38.66^\circ \\ & + 0.312(100)(20 mA \angle 0^\circ) \angle -38.66^\circ\\ &= 15.60 A \angle -38.66^\circ + 0.62 A \angle -38.66^\circ\\ I_2 &=16.22 A \angle -38.66^\circ \end{split}$$