Equivalent ABCD of lines in parallel

Two lines with constants $(A_1, B_1, C_1, D_1)$ and $(A_2, B_2, C_2, D_2)$ are connected in parallel: same $V_S$ and $V_R$ at both ends; the currents add.

          IS1  +------------+  IR1
   IS  +--->---| A1 B1 C1 D1|--->---+  IR
 o-->--+       +------------+       +-->--o
 VS    |  IS2  +------------+  IR2  |    VR
       +--->---| A2 B2 C2 D2|--->---+
               +------------+
 o------------------------------------o

Step 1: Write each line in "admittance" form

From $V_S = A_kV_R + B_kI_{Rk}$:

$$I_{Rk} = \frac{V_S - A_kV_R}{B_k}, \quad k = 1, 2$$

Total receiving current:

$$I_R = I_{R1} + I_{R2} = V_S\left(\frac{1}{B_1} + \frac{1}{B_2}\right) - V_R\left(\frac{A_1}{B_1} + \frac{A_2}{B_2}\right)$$

Step 2: Solve for $V_S$ (gives A and B)

$$V_S\,\frac{B_1 + B_2}{B_1B_2} = I_R + V_R\,\frac{A_1B_2 + A_2B_1}{B_1B_2}$$ $$V_S = \frac{A_1B_2 + A_2B_1}{B_1 + B_2}V_R + \frac{B_1B_2}{B_1 + B_2}I_R$$ $$A = \frac{A_1B_2 + A_2B_1}{B_1 + B_2}, \qquad B = \frac{B_1B_2}{B_1 + B_2}$$

Step 3: Sending current (gives C and D)

$I_{Sk} = C_kV_R + D_kI_{Rk}$, and $I_S = I_{S1} + I_{S2}$:

$$I_S = (C_1 + C_2)V_R + D_1\frac{V_S - A_1V_R}{B_1} + D_2\frac{V_S - A_2V_R}{B_2}$$

Substituting $V_S$ from Step 2 and simplifying with $A_kD_k - B_kC_k = 1$:

$$D = \frac{D_1B_2 + D_2B_1}{B_1 + B_2}, \qquad C = C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2}$$

Result

$$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} \frac{A_1B_2 + A_2B_1}{B_1 + B_2} & \frac{B_1B_2}{B_1 + B_2} \\ C_1 + C_2 + \frac{(A_1 - A_2)(D_2 - D_1)}{B_1 + B_2} & \frac{D_1B_2 + D_2B_1}{B_1 + B_2} \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

Special case: for two identical lines, $A = A_1$, $B = B_1/2$, $C = 2C_1$, $D = D_1$. The result still satisfies $AD - BC = 1$.