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Two lines in parallel: ABCD constants

Short answer

Using parallel two-port formulas with C_i = (A_i²-1)/B_i: equivalent parameters are A = D = 0.9778 angle 0.147°, B = 24.75 angle 74.06° Ω, C = 1.788×10⁻³ angle 99.57° S. The method combines A1B2 + A2B1 over B1+B2 for A, and B1B2/(B1+B2) for B.

Two lines are represented by their respective ABCD constants and are connected in parallel. The line constants are as follows: A1 = 0.980∠0.1° = D1, B1 = 45∠73.2° Ohm; A2 = 0.975∠0.2° = D2, B2 = 55∠75.1° Ohm. [The paper does not state what is to be found; presumably the equivalent ABCD constants of the combination.]

Assumption: the equivalent ABCD constants of the two lines in parallel are required.

Given: A1=D1=0.980∠0.1∘A_1 = D_1 = 0.980\angle 0.1^\circ, B1=45∠73.2∘ ΩB_1 = 45\angle 73.2^\circ\ \Omega; A2=D2=0.975∠0.2∘A_2 = D_2 = 0.975\angle 0.2^\circ, B2=55∠75.1∘ ΩB_2 = 55\angle 75.1^\circ\ \Omega.

Formulas for two networks in parallel

A=A1B2+A2B1B1+B2,B=B1B2B1+B2,D=D1B2+D2B1B1+B2C=C1+C2+(A1−A2)(D2−D1)B1+B2\begin{aligned} A &= \frac{A_1B_2 + A_2B_1}{B_1 + B_2}, \quad B = \frac{B_1B_2}{B_1 + B_2}, \quad D = \frac{D_1B_2 + D_2B_1}{B_1 + B_2} \\ C &= C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2} \end{aligned}

C1C_1 and C2C_2 are not given; since each line is symmetrical and reciprocal, C=(AD−1)/B=(A2−1)/BC = (AD - 1)/B = (A^2 - 1)/B.

Step 1: C of each line

C1=A12−1B1=8.833×10−4∠101.96∘ SC2=A22−1B2=9.062×10−4∠97.25∘ S\begin{aligned} C_1 &= \frac{A_1^2 - 1}{B_1} = 8.833\times10^{-4}\angle 101.96^\circ\ \text{S} \\ C_2 &= \frac{A_2^2 - 1}{B_2} = 9.062\times10^{-4}\angle 97.25^\circ\ \text{S} \end{aligned}

Step 2: Common terms

B1+B2=(13.01+j43.08)+(14.14+j53.15)=27.15+j96.23=99.99∠74.25∘ ΩA1B2=53.90∠75.20∘,A2B1=43.875∠73.40∘A1B2+A2B1=(13.77+j52.11)+(12.54+j42.05)=26.30+j94.16\begin{aligned} B_1 + B_2 &= (13.01 + j43.08) + (14.14 + j53.15) = 27.15 + j96.23 = 99.99\angle 74.25^\circ\ \Omega \\ A_1B_2 &= 53.90\angle 75.20^\circ, \quad A_2B_1 = 43.875\angle 73.40^\circ \\ A_1B_2 + A_2B_1 &= (13.77 + j52.11) + (12.54 + j42.05) = 26.30 + j94.16 \end{aligned}

Step 3: Equivalent constants

A=D=26.30+j94.1627.15+j96.23=0.9778∠0.147∘B=45×55∠148.3∘99.99∠74.25∘=2475∠148.3∘99.99∠74.25∘=24.75∠74.06∘ ΩC=C1+C2−(A1−A2)2B1+B2=(−1.831+j8.641)×10−4+(−1.143+j8.990)×10−4−2.8×10−7∠−111.6∘=1.788×10−3∠99.57∘ S\begin{aligned} A = D &= \frac{26.30 + j94.16}{27.15 + j96.23} = 0.9778\angle 0.147^\circ \\ B &= \frac{45\times55\angle 148.3^\circ}{99.99\angle 74.25^\circ} = \frac{2475\angle 148.3^\circ}{99.99\angle 74.25^\circ} = 24.75\angle 74.06^\circ\ \Omega \\ C &= C_1 + C_2 - \frac{(A_1 - A_2)^2}{B_1 + B_2} \\ &= (-1.831 + j8.641)\times10^{-4} + (-1.143 + j8.990)\times10^{-4} - 2.8\times10^{-7}\angle -111.6^\circ \\ &= 1.788\times10^{-3}\angle 99.57^\circ\ \text{S} \end{aligned}

Check: AD−BC=1.000AD - BC = 1.000.

Answer: A=D=0.9778∠0.147∘A = D = 0.9778\angle 0.147^\circ, B=24.75∠74.06∘ ΩB = 24.75\angle 74.06^\circ\ \Omega, C=1.788×10−3∠99.57∘C = 1.788\times10^{-3}\angle 99.57^\circ S.

(The C angles above 90° come from the given A and B values; real lines have C close to 90°.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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