# AC Equivalent Circuit

In the discussion to follow, keep in mind that
the term equivalent refers only to the fact that for the same applied potential, the same impedance and input current will result.
In a series ac circuit, the total impedance of two or more elements in series is often equivalent to an impedance that can be achieved with fewer elements of different values, the elements and their values being determined by the frequency applied. This is also true for parallel circuits. For the circuit of Fig. 1(a), $$\begin{split} Z_T &= {Z_C Z_L \over Z_C+ Z_L} \\ &= {(5 Ω \angle -90^\circ) (10 Ω \angle 90^\circ) \over 5 Ω \angle -90^\circ+ 10 Ω \angle 90^\circ} \\ &={ 50 \angle 0^\circ \over 5 \angle 90^\circ}\\ &=10 Ω \angle -90^\circ\\ \end{split}$$
Fig. 1: Defining the equivalence between two networks at a specific frequency
The total impedance at the frequency applied is equivalent to a capacitor with a reactance of 10 Ω, as shown in Fig. 1(b). Always keep in mind that this equivalence is true only at the applied frequency. If the frequency changes, the reactance of each element changes, and the equivalent circuit will change-perhaps from capacitive to inductive in the above example.
Fig. 2: Finding the series equivalent circuit for a parallel R-L network.
Another interesting development appears if the impedance of a parallel circuit, such as the one of Fig. 2(a), is found in rectangular form. In this case, $$\begin{split} Z_T &= {Z_L Z_R \over Z_R+ Z_L} \\ &= {(4 Ω \angle 90^\circ) (3 Ω \angle 0^\circ) \over 3 Ω \angle 0^\circ+ 4 Ω \angle 90^\circ} \\ &={ 12 \angle 90^\circ \over 5 \angle 53.13^\circ}\\ &=2.40 Ω \angle 36.87^\circ\\ &=1.920 Ω+j 1.440Ω\\ \end{split}$$ which is the impedance of a series circuit with a resistor of 1.92 Ω and an inductive reactance of 1.44 Ω, as shown in Fig. 2(b). The current I will be the same in each circuit of Fig. 1 or Fig. 2 if the same input voltage E is applied. For a parallel circuit of one resistive element and one reactive element, the series circuit with the same input impedance will always be composed of one resistive and one reactive element. The impedance of each element of the series circuit will be different from that of the parallel circuit, but the reactive elements will always be of the same type; that is, an R-L circuit and an R-C parallel circuit will have an equivalent R-L and R-C series circuit, respectively. The same is true when converting from a series to a parallel circuit. In the discussion to follow, keep in mind that
the term equivalent refers only to the fact that for the same applied potential, the same impedance and input current will result.
To formulate the equivalence between the series and parallel circuits, the equivalent series circuit for a resistor and reactance in parallel can be found by determining the total impedance of the circuit in rectangular form; that is, for the circuit of Fig. 3(a).
Fig. 3: Defining the parameters of equivalent series and parallel networks.
$$Y_p = {1 \over R_p} + {1 \over \pm X_p} = {1 \over R_p} \mp {1 \over X_p}$$ and $$\begin{split} Z_p &= {1 \over Y_p} = {1 \over ({1 \over R_p} \mp {1 \over X_p})} \\ &= {1/R_p \over (1/ R_p)^2 + (1/X_p)^2} \pm j {1/X_p \over (1/ R_p)^2 + (1/X_p)^2} \\ \end{split}$$ Multiplying the numerator and denominator of each term by $R_p^2 X_p^2$ results in $$\begin{split} Z_p &= {R_p X_p^2 \over X_p^2 + R_p^2}\pm j{R_p^2 X_p \over X_p^2 + R_p^2}\\ &= R_s + j X_s\\ \end{split}$$ and $$\bbox[10px,border:1px solid grey]{R_s = {R_p X_p^2 \over X_p^2 + R_p^2}} \tag{1}$$ with $$\bbox[10px,border:1px solid grey]{X_s = {X_p R_p^2 \over X_p^2 + R_p^2}} \tag{2}$$ For the network of Fig. 2, $$\begin{split} R_s &= {R_p X_p^2 \over X_p^2 + R_p^2}\\ &= {(3 Ω) (4 Ω)^2 \over(4 Ω)^2 + (3 Ω)^2}\\ &={48 Ω \over 25} = 1.920 Ω\\ \end{split}$$ and $$\begin{split} X_s &= {X_p R_p^2 \over X_p^2 + R_p^2}\\ &= {(4 Ω) (3 Ω)^2 \over(4 Ω)^2 + (3 Ω)^2}\\ &={36 Ω \over 25} = 1.440 Ω\\ \end{split}$$ which agrees with the previous result.
The equivalent parallel circuit for a circuit with a resistor and reactance in series can be found by simply finding the total admittance of the system in rectangular form; that is, for the circuit of Fig. 3b), $$\begin{split} Z_s &= R_s \pm j X_s\\ Y_s &= {1 \over Z_s} = {1 \over R_s \pm j X_s}\\ &= { R_s \over R_s^2 + X_s^2} \mp j { X_s \over R_s^2 + X_s^2}\\ &= G_p \mp j B_p = { 1 \over R_p} \mp j {1 \over X_p}\\ \end{split}$$ or $$\bbox[10px,border:1px solid grey]{R_p = { R_s^2 + X_s^2 \over R_s} } \tag{3}$$ with $$\bbox[10px,border:1px solid grey]{X_p = { R_s^2 + X_s^2 \over X_s} } \tag{4}$$ For the above example, $$R_p = { R_s^2 + X_s^2 \over R_s} \\ R_p = { (1.92 Ω)^2 + (1.44 Ω)^2 \over 1.92 Ω} \\ = { 5.76 Ω \over 1.92} = 3.0 Ω$$ and $$X_p = { R_s^2 + X_s^2 \over X_s} = { 5.76 Ω \over 1.44} = 4.0 Ω$$
Example 1: Determine the series equivalent circuit for the network of Fig. 4.
Fig. 4: Example 1.
Solution: $R_p = 8 kΩ$
$X_p = | X_L - X_C | = | 9 kΩ - 4 kΩ| = 5kΩ$
and $$\begin{split} R_s &= { R_p X_p^2 \over X_p^2 + R_p^2}\\ &={(8 kΩ)(5 kΩ) ^2 \over (8 kΩ)^2+(5 kΩ) ^2 }\\ &= { 200 k Ω \over 89 } = 2.247 kΩ \end{split}$$ with $$\begin{split} X_s &= { R_p^2 X_p \over X_p^2 + R_p^2}\\ &={(8 kΩ)^2(5 kΩ) \over (8 kΩ)^2+(5 kΩ) ^2} \\ &= { 320 k Ω \over 89 } = 3.596 kΩ\\ \end{split}$$ The equivalent series circuit appears in Fig. 5.
Fig. 5: The equivalent series circuit for the parallel network of Fig. 4.