Balanced Wheatstone

%% Balanced Wheatstone
% Calculates the resistance in R1
% Notes and clarifications are at the bottom of the code

% Inputs and Variables
Rsub = input('Input the values of R2, R3, and R4 in brackets\n');
for i = 1:3
    R(i+1) = Rsub(i)
end

% Outputs and Equations
R(1) = R(2)*R(3)/R(4);
fprintf('The resistance of R1 is %.3f', R(1))

%{
Later versions will allow users to choose which resistor the user would
    like to solve for.
In this scenario, the circuit runs between R1 and R3 and R2 and R4. R1 and
    R3 are on the left side and R1 and R2 are on the top side.
When the user inputs values R2-4, they will take the first 3 spots of the
    array 'Rsub'. The for loop will put the resistance values in their corresponding
    place in the array 'R'
The user must input the values of the known resistances in brackets [] for
    the code to work.
%}