## viernes, 14 de diciembre de 2012

### 6.002x Final Exam (I)

6.002x Final exam is very close. Last course Info, dated december 4, said: "The final exam will take place on December 20th, with full details below." Just one line below, they contradicted themself; quote:  "The final exam will be released on December 19th at 14:01 (2:01 pm) Boston time and will close on December 23rd at 23:59 (11:59 pm) Boston time." So you do not know if the exam is going to be opened on December 20th or 19th.

Last spring 6.002x final exam
I would like to write several post dedicated to last spring final exam. Our university closes its doors today. It will be reopened by January 16th. So this blog would be a way to help students who are willing to take the final exam. This first post is dedicated to the very first problem; which has been rewriten below. The solution of the problem attracted my attention a lot. Recently the civil engineering department acquired a new machine which is based on the same principle.

Strain
One way to measure the deformation of an object is with strain gauges. A strain gauge is a resistor whose resistance varies with the amount it is stretched or compressed. For example, an Omega SG-13/1000-LY11 has a nominal undeformed resistance of RX=1000Ω. The change in resistance is very small: if the piece of steel the gauge is bonded to stretches by Δll=0.001 the gauge increases resistance by only 2Ω.. To measure the strain we need to get a voltage proportional to the change in resistance from the nominal resistance of the gauge. This is arranged with a circuit like this:
Let's start by assuming that RA=1800Ω, RB=3600Ω, RC=1500Ω, and VS=26V. Note: In this problem it is necessary to compute your answers to within 1% of the correct value.

Q1: What is the output voltage v_o , in Volts, if the gauge is not deformed?

The circuit is know as bridge circuit. The problem is quite tricky. You can not use tension division directly to calculate v_o. Instead, one solution could be to calculate tension among all resistors. Then you have to figure out how to subtract the correct tensions. It would be ok to write a formula, because it would be helpful for solving next questions.
Q2: Now suppose the gauge stretches so its resistance changes to R_X=1002.0Ω. What is the change in the output voltage, in milliVolts?

Tension in R_C has changed. It must be recalculated. Then you must subtract this value from the one calculated in Q1.
Q3: We would really like the output voltage to be zero when the gauge is not deformed. Keeping the other resistors the same, what should we make the resistance RB to accomplish this? Express your answer in Ohms.

Solving Q1, we must concluded that v_o is the difference between two tensions. we are required v_o=0. So the only unknown value is the value of R_B.

Q4/Q5: Now we want to choose some resistances to get the maximum sensitivity: we want the biggest change in output voltage for a change in the gauge resistance. We also want the output voltage to be zero when the gauge is not deformed.

Assume that we are given RA=3000.0Ω, and remember that the nominal undeformed resistance of the gauge RX=1000Ω.

What value should we choose for RC, in Ohms?
What value should we choose for RB, in Ohms?
To find when it is maximized, we find the derivative of dv_o / dR_x  and set it to zero. It is quite tricky. From Q1, if you got a formula for v_o just derive it with respect to R_x. You are going to get an expresion like dv_o/dR_x = R_C V_S / (R_x+R_C)^2. It is the expression you have to derive and set to zero.