SOLUTION: Please help, I'm stuck! (Modeling) In electronics, circuit gain is modeled by G = Bt/R + R(base)t Where R is the value of the resistor, t is temperture, R(base)t is the val

Algebra ->  Systems-of-equations -> SOLUTION: Please help, I'm stuck! (Modeling) In electronics, circuit gain is modeled by G = Bt/R + R(base)t Where R is the value of the resistor, t is temperture, R(base)t is the val      Log On


   

Question 127357This question is from textbook
: Please help, I'm stuck!
(Modeling)
In electronics, circuit gain is modeled by
G = Bt/R + R(base)t
Where R is the value of the resistor, t is temperture, R(base)t is the value of R at room temperture t and B is constant. The sensitivity of the circuit to temperture is modeled by
S = BR/(R + R (base)t)^2
If B = 3.7 and t is 90 K, find the values of R and R (base)t that will make G = .4 and S = .001.
I huge thank you to anyone who can get me goingon the right track. This question is from textbook

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
I hope that I correctly interpret your two given equations (one for gain and the other for
sensitivity) as being:
.
G+=+%28B%2At%29%2F%28R+%2B+R%5Bt%5D%29
.
For this one I'm guessing on what the right side of this equation should be. You wrote it as:
.
G+=+%28B%2At%29%2FR+%2B+R%5Bt%5D
.
but I'm guessing that my version above is you intended. [If I'm wrong, you may want to repost your
problem and we'll try again.]
.
and
.
S+=+%28B%2AR%29%2F%28R+%2B+R%5Bt%5D%29%5E2
.
You are given that B = 3.7, t = 90, G = 0.4, S = 0.001
.
Substitute these values into the two equations and they become:
.
0.4+=+%283.7%2A90%29%2F%28R+%2B+R%5Bt%5D%29 <=== call this equation #1
.
and
.
0.001+=+%283.7%2AR%29%2F%28R+%2B+R%5Bt%5D%29%5E2 <=== call this equation #2
.
Go to equation #1 and multiply both sides by the quantity R+%2B+R%5Bt%5D. When you do,
on the right side this multiplication cancels the denominator and you are left with equation #1
becoming:
.
0.4%2A%28R+%2B+R%5Bt%5D%29+=+3.7%2A90
.
Multiply out the right side and you have:
.
0.4%2A%28R+%2B+R%5Bt%5D%29+=+333
.
Finally solve for R+%2B+R%5Bt%5D by dividing both sides of this equation by 0.4 to get:
.
R+%2B+R%5Bt%5D+=+333%2F0.4+=+832.5
.
You now have a numerical value of 832.5 ohms for R + R[t] and you can substitute that value in
for the denominator in equation #2 to make equation #2 become:
.
0.001+=+%283.7%2AR%29%2F%28832.5%29%5E2
.
Square out the denominator on the right side and the equation becomes:
.
0.001+=+%283.7%2AR%29%2F693056.25
.
Get rid of the denominator on the right side by multiplying both sides of this equation by
693056.25 and you get:
.
0.001%2A693056.25+=+3.7%2AR
.
Multiply out the left side and you reduce the equation to
.
693.05625+=+3.7%2AR
.
Finally solve for R by dividing both sides by 3.7 and you have:
.
187.3125+=+R
.
[If only resistors could be manufactured to that precision ...]
.
Now that we have a value for R we can return to either of the original equations and substitute
this value of R and solve the resulting equation for R%5Bt%5D.
.
Let's return to equation #1 and set R+=+187.3125. This makes equation #1 become:
.
0.4+=+%283.7%2A90%29%2F%28187.3125+%2B+R%5Bt%5D%29
.
Again multiply out the numerator on the right side and you have:
.
0.4+=+333%2F%28187.3125+%2B+R%5Bt%5D%29
.
Get rid of the denominator on the right side of this equation by multiplying both sides
by 187.3125+%2B+R%5Bt%5D and the equation then is changed to:
.
0.4%28187.3125+%2B+R%5Bt%5D%29+=+333
.
Get rid of the multiplier 0.4 by dividing both sides of the equation by 0.4 and the equation
is reduced to:
.
187.3125+%2B+R%5Bt%5D+=+832.5
.
Get rid of the constant on the left side by subtracting 187.3125 from both sides and you are
left with the answer of:
.
R%5Bt%5D+=+645.1875
.
So the answers obtained for this problem is R+=+187.3125 ohms and R%5Bt%5D+=+645.1875 ohms
.
Check the math above to make sure I didn't make some "fat finger" error on the calculator or
in the process.
.
Hope this helps to clarify the general process a little, and I hope my initial assumption
regarding the Gain formula was correct.
.