Power Transfer Over Cable

My approach was similar to that of @Karan Chatrath , in that I swept $V_S$ and calculated the load power each time. I solved the following node equation for the load resistor voltage $V_L$ (sum of currents out of the node is zero).

$\frac{V_L - V_S}{Z} + \frac{V_L}{Z_C} + \frac{V_L}{R_L} = 0$

Then the load power is:

$P_L = \frac{|V_L|^2}{R_L}$

Cables or transmission lines with significant parasitic capacitance can see the voltage rise from the sending end toward the load. This is because a capacitive current is drawn through an inductive series impedance. In order to dissipate $40$ watts, the resistor needs $120$ volts across it. And because of this voltage rise (the Ferranti Effect), we only have to apply $111.78$ volts at the sending end in order for the load resistor to get $120$ volts.

Currents are indicated in the diagram. Using Kichhoff's current and voltage laws yields the following equations in a matrix form:

$\left[\begin{matrix} 1 & -1 & 0 & -1 & 0\\0 & 0 & -1 & 1 & -1\\0 & Z_C & 0&0&0\\0& -Z_C& Z_C&Z&0\\0&0&Z_C&0&-R_L \end{matrix} \right] \left[\begin{matrix} I\\I_{C1}\\I_{C2}\\I_Z\\I_L \end{matrix} \right] = \left[\begin{matrix} 0\\0\\V_S\\0\\0 \end{matrix} \right]$

$\implies AI = b$ $\implies I_L = \left[\begin{matrix} 0&0&0&0&1 \end{matrix} \right]A^{-1} b$

$\implies P_A = \mathrm{real} \left(I_LR_L I_L^{*}\right)$

To solve for $V_S$ I used brute force. I simply swept through different values of $V_S$ until I found the solution to some numerical tolerance.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

clear all
clc

% GIven parameters:
RL       = 360;
Z        = 8 + j*40;
ZC       = -j*400;

% A matrix as shown in the solution description:
A        = [1 -1 0 -1 0;0 0 -1 1 -1;0 ZC 0 0 0;0 -ZC ZC Z 0;0 0 ZC 0 -RL];

% Varying VS in small increments:
for VS  = 0:0.0001:120

% b vector as shown in the solution description:
b        = [0;0;VS;0;0];

% Computation of load current:
IL       = [0 0 0 0 1]*inv(A)*b;

% Active power consumed by load:
PA       = real(IL*RL*conj(IL));

% Checking if active power is close to 40 W to some tolerance:
if abs(PA - 40) <= 1e-4
  ANSWER    =  VS;
end


end

% ANSWER = 111.78