Question

PROBLEM: Simulate the resistance and the current across a circuit Consider the following circuit Where V- voltage from a battery source (unit of measure: Volts). I-Current (unit of measure: Amperes) And Ri, R. Rs are resistors connected in parallel (unit of measure: Ohms) When resistors are connected in parallel, the equivalent resistance, Ra, can be computed: By Ohm's Law, Components are designed to perform at a nominal value, however, due to manufacturing inconsistencies all components are subject to variation from its desired value. Referring to the diagram above, consider the following: Voltage is specified to be 18 V, tolerance is t 0.5 V Resistors Ri and R2 is specified to be 2 2 and Rj is specified to be 4 2. The tolerance on Ri and R2 is 0.1 ? ( 5% of nominal) The tolerance on Rs is ± 0.2 ? (±5% of it's nominal) 2. Simulation of the current, I, across this circuit. Steps: For N-200,000, Randomly generate the Voltage, V, where V -Unif [17.5,18.5]. Using your simulated Reg vector from part 1 and the voltage, V, create the current vector, I The equation relating I, V, and Req is on the from page For you to hand in: a. A well labelled histogram of I. Based on the nominal value of V and theoretical value for Ros, compute (by hand) the theoretical value of I b. Provide the mean current and the standard deviation based on your simulation mean of your simulation compare to the theoretical value of I? c. . How does the Find the 95% prediction interval for the current across this system by lopping off the smallest 2.5% of the values, find this cut point, then lop off the largest 2.5% of the values, determine this cutpoint. d. Thus 95% of your simulated values fall between these two cut points. Matlab pretile(12.5); will provide the cut point between the lowest 2.5% of values and the remaining 97.5% of valu Prctile(1.975); will provide the cut point between the lowest 97.5% of values and the upper 2.5% of values cs e. Provide supporting matlab code.

Answer 1

A) Histogram of I: A histogram is a plot that lets you discover, and show, the underlying frequency distribution (shape) of a set of continuous data. The histogram of the graph tells us it follows a normal distribution and the peak of the histogram is around 22.5. It shows that values of the current are symmetrically dispersed on either side of the peak value.

B) V = 18V

A = 1/R1+1/R2+1/R3 = 1/2+1/2+1/4 = 5/4

Req = 1/A = 4/5 = 0.80 ohm

I = V/req = 18/0.8 = 22.5 A

C) Mean and standard deviation as computed in MATLAB

>> mean(I)

ans =

22.5198

>> std(I)

ans =

0.5319

mean = 22.5198 and standard deviation = 0.5319.

The theoretical value of I = 22.5 which is equal to mean value of I when rounding to one decimal place.

D) Cutpoint of the smallest 2.5% = 21.5121

Cutpoint of the largest 2.5% = 23.5559

CI(95%) = (23.5559,21.5121)

This concept can be visualized in the following diagram

N=200000; % No. of random samples taken
R1=random('unif',1.9,2.1,[N,1]); % Generate N values of R1
R2=random('unif',1.9,2.1,[N,1]); % Generate N values of R2
R3=random('unif',3.8,4.2,[N,1]); % Generate N values of R3
Req_inv = 1./R1+1./R2+1./R3;
Req = 1./Req_inv;
V =random('unif',17.5,18.5,[N,1]); % Generate N values of V
I = V./Req; m = mean(I); s = std(I);

CL = prctile(I,2.5); %will provide the cut point between the lowest 2.5% of values and the remaining 97.5% of values
CU = prctile(I,97.5);%will provide the cut point between the lowest 97.5% of values and the remaining 2.5% of values