# Answered Essay: Applying Kirchoff's current law at each of the nodes (black dots), and using the current directions indicated b

matlab language  Applying Kirchoff’s current law at each of the nodes (black dots), and using the current directions indicated by the arrows, we get the following system of equations: i_1 – i_2 =0 i_2 + i_3 – i_4 i_4 – i_3 – i_5 = 0 i_5 – i_1 = 0 Applying Kirchoff’s voltage law around the three loops in the circuit, we get the equations: V – i_2R_2 – i_4R_4 – i_5R_5 = 0 V_1 – i_2R_2 – V_3 – i_5R_5 = 0 V_3 – i_4R_4 = 0 This gives 7 equations for 5 unknowns (the currents). From this point on, we drop the final equation from each group. Write a script that puts these equations into matrix form. Then solve the equations to find the current through each branch for the following parameters: V_1 =10V R_2 = 3 Ohm V_3 = 4V R_4 = 4 Ohm R_5 = 3 Ohm We now want to find by how much we can increase the voltage at V_1 before melting a wire (which occurs when more than 5A of current passes through any of them). Using the appropriate type of loop and a step size of 0.1V, increase the voltage at V_1 until the maximum current flowing through any of the wires exceeds 5A. Then print the largest acceptable value of V_1 to the screen, in the form: The maximum voltage that can be applied is ? volts. Where? is replaced by the number of volts.

Part 1:

​The matrix is formed to solve the system of linear equation of the form AX=B where X is a 5-by-1 vector containing current values i1,i2,i3,i4 and i5. Below is the matlab code with comments describing matrices and after that output is attached.

Matlab code:

% ###################### 1 #########################
% variable to calculate are currents i1, i2, i3, i4, i5
syms i1 i2 i3 i4 i5
% given values
v1 = 10;
r2 = 3;
v3 = 4;
r4 = 4;
r5 = 3;

% given 5 equations
eqn1 = i1-i2 == 0;
eqn2 = i2+i3-i4 == 0;
eqn3 = i4-i3-i5 == 0;
eqn4 = v1-i2*r2-i4*r4-i5*r5 == 0;
eqn5 = v1-i2*r2-v3-i5*r5 == 0;

% equation matrix AX = B
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5], [i1, i2, i3, i4, i5]);

% solving equation AX = B, X is a vector containing 5 current values
X = linsolve(A,B)
A
B

Output part 1:

matrix A matrix B: ​matrix X( solution containing currents i1, i2, i3, i4, i5): ​Part 2:

​In part 2, voltage V1 is set as an variable and matrix is solved similarily as in part 1 but V1 as an variable. Which produces solution in terms of variable V1 which is found to be: Putting V1 = 10 in the above solution gives answer same as in part 1. Now initialize V1= 10 V and calculate currents by putting V1 = 10 in above solution. Now in a while loop increase V1 by 0.1 in each step , calculate new currents by putting this V1 value in above solution matrix until either of i1, i2, i3, i4 or i5 gets 5A. Finally print V1.

Matlab Code:

%##################### 2 #######################
% now put v1 also an variable
syms i1 i2 i3 i4 i5 v1
% rest constants as it is
r2 = 3;
v3 = 4;
r4 = 4;
r5 = 3;

% define equations
eqn1 = i1-i2 == 0;
eqn2 = i2+i3-i4 == 0;
eqn3 = i4-i3-i5 == 0;
eqn4 = v1-i2*r2-i4*r4-i5*r5 == 0;
eqn5 = v1-i2*r2-v3-i5*r5 == 0;

% equation matrix
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5], [i1, i2, i3, i4, i5]);

% solve system in terms of v1
sol = linsolve(A,B)

% initial value of v1
v1_init = 10;
% put v1= 10 in system of equation solution
currents = subs(sol,v1,v1_init);
% while currents less than 5
while double(currents(1))<5 && double(currents(2))<5 && double(currents(3))<5 && double(currents(4))<5 && double(currents(5))<5
v1_init = v1_init+0.1;
currents = subs(sol,v1,v1_init);
end

% print final value of v1 when any current gets 5
fprintf(‘The maximum volatge that can be applied is %2.2f volts.’,v1_init);
% print currents at that time
currents

Output part 2: Currents at this voltage:  Calculate your paper price
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