The discussion for parallel ac circuits will be very similar to that for dc circuits. In dc circuits, conductance (G) was defined as being equal to 1/R. The total conductance of a parallel circuit was then found by adding the conductance of each branch. The total resistance $R_T$ is simply $1/G_T$. In ac circuits, we define admittance (Y) as being equal to $1/Z$.
The unit of measure for admittance as defined by the SI system is siemens, which has the symbol S.
Admittance is a measure of how well an ac circuit will admit, or allow, current to flow in the circuit. The larger its value, therefore, the heavier the current flow for the same applied potential.
The total admittance of a circuit can also be found by finding the sum of the parallel admittances. The total impedance $Z_T$ of the circuit is then $1/Y_T$; that is, for the network of Fig. 1: $$\bbox[10px,border:1px solid grey]{Y_T = Y_1 + Y_2 + Y_3 + ...+ Y_N} \tag{1}$$
Fig. 1: Parallel ac network.
or, since $Z = 1/Y$, $$\bbox[10px,border:1px solid grey]{{1 \over Z_T} = {1 \over Z_1} + {1 \over Z_2} + {1 \over Z_3} + ...+ {1 \over Z_N}} \tag{2}$$ For two impedances in parallel, $${1 \over Z_T} = {1 \over Z_1} + {1 \over Z_2}$$ the total resistance of two parallel resistors are now applied, the following similar equation will result: $$Z_T = {Z_1Z_2 \over Z_1 + Z_2}$$ For three parallel impedances, $$\bbox[10px,border:1px solid grey]{Z_T = {Z_1 Z_2 Z_3 \over Z_1Z_2 + Z_2Z_3 + Z_1Z_3}} \tag{3}$$ As pointed out in the introduction to this section, conductance is the reciprocal of resistance, and $$\bbox[10px,border:1px solid grey]{Y_R = {1 \over Z_R} = { 1 \over R \angle 0^\circ} = G \angle 0^\circ} \tag{4}$$

### What is Susceptance

The reciprocal of reactance (1/X) is called susceptance and is a measure of how susceptible an element is to the passage of current through it. Susceptance is also measured in siemens and is represented by the capital letter B.
For the inductor $$\bbox[10px,border:1px solid grey]{Y_L = {1 \over Z_L} = { 1 \over X_L \angle 90^\circ} = { 1 \over X_L} \angle -90^\circ} \tag{5}$$ Defining $$\bbox[10px,border:1px solid grey]{B_L = { 1 \over X_L}} \, (siemens, S) \tag{6}$$ we have $$\bbox[10px,border:1px solid grey]{Y_L = B_L \angle -90^\circ} \tag{7}$$ Note that for inductance, an increase in frequency or inductance will result in a decrease in susceptance or, correspondingly, in admittance.
For the capacitor, $$\bbox[10px,border:1px solid grey]{Y_L = {1 \over Z_C} = { 1 \over X_C \angle -90^\circ} = { 1 \over X_C} \angle 90^\circ} \tag{8}$$ Defining $$\bbox[10px,border:1px solid grey]{B_C = { 1 \over X_C}} \, (siemens, S) \tag{9}$$ we have $$\bbox[10px,border:1px solid grey]{Y_C = B_C \angle 90^\circ} \tag{10}$$ For the capacitor, therefore, an increase in frequency or capacitance will result in an increase in its susceptibility.
For parallel ac circuits, the admittance diagram is used with the three admittances, represented as shown in Fig. 2.
Note in Fig. 2 that the conductance (like resistance) is on the positive real axis, whereas inductive and capacitive susceptances are in direct opposition on the imaginary axis.
For any configuration (series, parallel, series-parallel, etc.), the angle associated with the total admittance is the angle by which the source current leads the applied voltage. For inductive networks, $\theta_T$ is negative, whereas for capacitive networks, $\theta_T$ is positive.
Example 1: For the network of Fig. 3:
a. Find the admittance of each parallel branch.
c. Calculate the input impedance.
a. $$\begin{split} Y_R &= G \angle 0^\circ = {1 \over R} \angle 0^\circ = {1 \over 20} \angle 0^\circ \\ &=0.05 \angle 0^\circ = 0.05 + j 0 S Y_L &= B_L \angle -90^\circ = {1 \over X_L}\angle -90^\circ = {1 \over 10}\angle -90^\circ \\ &=0.1 S \angle -90^\circ = 0 - j 0.1 S \end{split}$$ b.
$$\begin{split} Y_T &= Y_R + Y_L = (0.05 S + j 0) + (0 j + 0.1 S)\\ &= 0.05 S + j 0.1 S = G + j B_L\\ \end{split}$$ c.
$$\begin{split} Z_T &= {1 \over Y_T} = {1 \over 0.05 S + j 0.1 S} \\ &={1 \over 0.112 S \angle -63.43^\circ } \\ &= 8.93 Ω \angle 63.43^\circ\\ \end{split}$$ d. The admittance diagram appears in Fig. 4.
On many occasions, the inverse relationship $Z_T = 1 / Y_T$ or $Y_T = 1 / Z_T$ will require that we divide the number 1 by a complex number having a real and an imaginary part. This division, if not performed in the polar form, requires that we multiply the numerator and denominator by the conjugate of the denominator, as follows: $$\begin{split} Y_T &= {1 \over Z_T}={1 \over 4Ω + j6Ω}\\ &=({1 \over 4Ω + j6Ω})({4Ω - j6Ω \over 4Ω - j6Ω})\\ &={4Ω - j6Ω \over 4^2 + 6^2} = &={4Ω - j6Ω \over 52} \\\\ Y_T&={4Ω \over 52} S- j{6 Ω \over 52}S\\ \end{split}$$ To avoid this laborious task each time we want to find the reciprocal of a complex number in rectangular form, a format can be developed using the following complex number, which is symbolic of any impedance or admittance in the first or fourth quadrant:
$$\begin{split} {1 \over a_1- jb_1}&=({1 \over a_1- jb_1}) ({a_1 \pm jb_1 \over a_1 \pm jb_1})\\ &= {a_1 \pm jb_1 \over a_1^2 + b_1^2} = {a_1 \over a_1^2 + b_1^2} \pm j{b_1 \over a_1^2 + b_1^2}\\ \end{split}$$
a. $Z = R - j X_C = 6Ω - j 8 Ω$ $$\begin{split} Y &= {1 \over 6Ω- j8Ω}\\ &={6 \over 6^2 + 8^2} + j{8 \over 6^2 + 8^2}\\ &={6 \over 100} S + j{8 \over 100} S\\ \end{split}$$ b. Z = 10 Ω + j 4 Ω + (-j 0.1 Ω) = 10 Ω + j 3.9 Ω $$\begin{split} Y &= { 1 \over Z} = {1 \over 10Ω+ j3.9Ω}\\ &={10 \over 10^2 + 3.9^2} - j{3.9 \over 10^2 + 3.9^2}\\ &={10 \over 115.21} S - j{3.9 \over 115.21} S\\ &=0.087 S-j0.034 S\\ \end{split}$$