A.C. Circuits (8hrs) · Basics of Electrical & Electronics Engineering

admittance

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admittance

1. What Admittance Is

Admittance (YY, siemens SS, mhomho) = reciprocal of impedance, complex ease with which AC current flows:

Y=1/Z=G+jBY = 1/Z = G + jB

where

GG

= conductance (real part, SS),

BB

= susceptance (imaginary, SS). For Z=R+jXZ=R+jX,

Y=1/(R+jX)=(R−jX)/(R2+X2)=G+jBY =1/(R+jX)= (R-jX)/(R²+X²) = G + jB

So

G=R/(R2+X2),B=−X/(R2+X2)G = R/(R²+X²), B = -X/(R²+X²)

.

For pure elements:

  • RR:
YR=1/R=GY_R =1/R = G

(real, B=0B=0)

  • LL:
YL=1/(jXL)=−j/XL=−j/(ωL)=jBLY_L =1/(jX_L)= -j/X_L = -j/(ωL) = jB_L

with

BL=−1/(ωL)B_L= -1/(ωL)
  • CC:
YC=jωC=jBCY_C = jωC = jB_C

with

BC=ωCB_C= ωC

2. Why Use Admittance

  • Parallel circuits are easier with YY: Yeq=ΣYiY_{eq}= Σ Y_i for parallel (like conductance), while ZZ series adds. So for parallel R,L,CR,L,C,
Y=G+j(BC+BL)=1/R+j(ωC−1/ωL)Y = G + j(B_C + B_L) =1/R + j(ωC -1/ωL)

then

Z=1/YZ=1/Y

.

  • Magnitude
∣Y∣=√(G2+B2)|Y|=√(G²+B²)

, phase

θY=−θZ=tan−1(B/G)θ_Y = -θ_Z = tan^{-1}(B/G)

.

3. Worked Examples

(a) Series RL: Z=3+j4 ⇒ Y=1/(3+j4)= (3-j4)/25=0.12 -j0.16 S ⇒ G=0.12 S, B=-0.16 S ✓

(b) Parallel RC: R=10Ω, C=100µF at 50Hz: G=0.1 S, B_C= ωC=0.0314 S ⇒ Y=0.1+j0.0314=0.105∠17.4° S, |Z|=1/|Y|≈9.52Ω ✓

(c) Parallel RLC at resonance: B= ωC -1/(ωL)=0 ⇒ Y=G=1/R pure real ⇒ |Z| maximal for parallel.

(d) Need to find current via Y: I = V·Y (since Y=1/Z): V=10∠0° V, Y=0.12-j0.16 ⇒ I=1.2 -j1.6 =2∠-53° A ✓

4. Exam Points

  • Define Y=1/Z=G+jBY=1/Z=G+jB, GG, BB — 5 marks, very frequent.
  • Show Yeq=ΣYiY_{eq}=ΣY_i for parallel — 5 marks, frequent.
  • Compute YY for R,L,CR,L,C and series/parallel — 5-7 marks, very frequent.
  • Relation BC=ωCB_C=ωC, BL=−1/ωLB_L=-1/ωL — 3-5 marks, frequent.
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