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 **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

**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.*

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