What is admittance, and how is it used to solve parallel AC circuits?

In one line

Admittance \( Y = 1/Z \) measures how easily a branch allows AC to flow; in a parallel circuit, branch admittances simply add, making parallel AC circuits easy to solve.

Explained simply

Resistance tells you how hard it is for current to flow; conductance (\( 1/R \)) tells you how easy it is. For parallel paths, "ease" adds up: two open doors side by side let people through twice as easily. Admittance is the AC version of this "ease", including phase.

Technically, \( Y = G + jB \), where \( G \) is conductance and \( B \) is susceptance, all in siemens (S). With \( Z = R + jX \): \( Y = \dfrac{R}{R^2+X^2} - j\dfrac{X}{R^2+X^2} \). So an inductive branch has a negative imaginary part, a capacitive branch a positive one.

How it works

1. Find each branch impedance \( Z_1, Z_2, \dots \).
2. Convert each to admittance: \( Y = 1/Z \) (multiply by the conjugate, or divide in polar form).
3. Total admittance \( Y = Y_1 + Y_2 + \dots \) (add real parts and imaginary parts).
4. Total current \( \bar I = \bar V\bar Y \); branch currents \( \bar I_k = \bar V\bar Y_k \).
5. Power factor \( = \cos\phi = G/|Y| \); lagging if the net susceptance is negative (inductive).

Branch currents can also be found directly as \( V/Z_k \) and added as phasors; the admittance method gives the same result more neatly.

Key formula

\( Y = \dfrac{1}{Z} = G + jB, \quad |Y| = \sqrt{G^2 + B^2}, \quad I = V|Y| \)

\( Y \) = admittance (S), \( G = R/(R^2+X^2) \) = conductance (S), \( B \) = susceptance (S), \( Z \) = impedance (\( \Omega \)), \( V \) = voltage (V), \( I \) = current (A).

Quick example

Branch 1: \( Z_1 = 3 + j4\,\Omega \), so \( Y_1 = (3 - j4)/25 = 0.12 - j0.16 \) S. Branch 2: \( Z_2 = 6 - j8\,\Omega \), so \( Y_2 = (6 + j8)/100 = 0.06 + j0.08 \) S. Total \( Y = 0.18 - j0.08 \) S, \( |Y| = 0.197 \) S. On 100 V: \( I = 19.7 \) A, lagging by \( \tan^{-1}(0.08/0.18) = 24.0^\circ \), pf \( = 0.18/0.197 = 0.914 \) lagging.

Common mistakes

• Taking \( G = 1/R \) for a branch that also has reactance; \( G = R/(R^2+X^2) \).
• Adding branch impedances in parallel instead of admittances.
• Getting the sign of susceptance wrong for inductive and capacitive branches.

How MU asks it

Asked as "Define admittance, conductance and susceptance" (3–5 marks) or "Two impedances are connected in parallel across 230 V; find branch currents, total current and power factor" (7–10 marks). For full marks: convert each branch to admittance with steps, add, find current, state power factor with lag/lead, and draw the phasor diagram with voltage as reference (the common quantity in parallel circuits).