I_1 = (V_1 - V_2)/Z + V_1 Y_c/2 1/0.03 + j0.04 = 12 - j16 [I_1 I_2] = [12 - j15.

Question

I_1 = (V_1 - V_2)/Z + V_1 Y_c/2 1/0.03 + j0.04 = 12 - j16 [I_1 I_2] = [12 - j15.9 - 12 + j16 -12 + j16 12 - j15.9] [V_1 v_2] V = [1.0 0.8 - j0.2] Then [12 - j15.9 - 12 + j16 -12 + j16 12 - j15.9] [1.0 0.8 - j0.2] = [5.60 - j0.70 -5.58 + j0.88] Therefore the power injected at bus 1 is S_1 = V_1 I*_1 = 1.0 times (5.60 + j0.70) = 5.60 + j0.70 S_2 = V_2 I*_2 = (0.8 - j0.2) times (-5.58 - j0.88) = -4.64 + j0.41 I = [5.0 -4.8] Then [12 - j15.9 -12 + j16 -12 + j16 12 - j15.9]^-1 [5.0 -4.8] = [0.0738 - j0.902 -0.0738 -j1.098] Therefore the power injected is S_1 = V_1 I*_1 = (0.0738 - j0.902) times 5 = 0.37 - j4.51 S_2 = V_2 I*_2 = (-0.0738 - j1.098) times (-4.8) = 0.35 + j5.27

Explanation / Answer

clear
clc
linedata=[ 1 2 0.03 0.04 0.1];
fb = linedata(:,1); % From bus number...
tb = linedata(:,2); % To bus number...
r = linedata(:,3); % Resistance, R...
x = linedata(:,4); % Reactance, X...
b = linedata(:,5); % Ground Admittance, B/2...
z = r + i*x; % Z matrix...
y = 1./z; % To get inverse of each element...
b = i*b; % Make B imaginary...
nbus = max(max(fb),max(tb)); % no. of buses...
nbranch = length(fb); % no. of branches...
ybus = zeros(nbus,nbus); % Initialise YBus...

% Formation of the Off Diagonal Elements...
for k=1:nbranch
ybus(fb(k),tb(k)) = -y(k);
ybus(tb(k),fb(k)) = ybus(fb(k),tb(k));
end

% Formation of Diagonal Elements....
for m=1:nbus
for n=1:nbranch
if fb(n) == m | tb(n) == m
ybus(m,m) = ybus(m,m) + y(n) + b(n);
end
end
end
ybus; % Bus Admittance Matrix
zbus = inv(ybus); % Bus Impedance Matrix
ybus
busdata= [ 1 1.0 5.0;
2 0.8-j*0.2 -4.8];
busno=busdata(:,1);
V= busdata(:,2);
I=busdata(:,3);
V;
I;
nbus=max(busno);
for j=1:nbus
i=ybus*V;
end
disp('calculation of current')
i %Flow of current in the network due to application of voltage at both ends
  
for k=1:nbus
for j=1:nbus
s1=V(1)*conj(i(1));
s2=V(k,:)*conj(i(j,:));
end
end
disp('power injected at bus 1')
s1 %calculation of reactive power at bus 1(S1=V1I1*)
s2 %calculation of reactive power at bus 1(S1=V1I1*)
for i=1:nbus
V2=inv(ybus)*I;
end
disp('voltage injected due to flow of current')
V2 % Voltage induced due to the flow of current I given in data')
for k=1:nbus
for j=1:nbus
S1=V2(1)*conj(I(1));
S2=V2(k,:)*conj(I(j,:));   
end
end
disp('power injected in the network due to voltage induced and current applied')
S1 %Power injected into the circuit from bus 1
S2 %power injected into the circuit from bus 2
  

%I didn't clearly understand the question so i programmed for the data which is given in the question. If you are having any doubts regarding the program just comment in the comment box i will try to clear it out.%

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