em Measure Weigh The electronic circuit of a scale which uses the capacitance of
ID: 1413319 • Letter: E
Question
em Measure Weigh The electronic circuit of a scale which uses the capacitance of a parallel plate capacitor to measure weight is shown below: SW1 Mechanical k spring with spring constant k Compressible dielectric Let the battery voltage V= Vo, and let the capacitance of the capacitor C= Co when no weight is applied to the top plate of the capacitor, C C when weight is applied to the top plate. Ignore any inductive effects of the spring. The spring has spring constant k and the compressible dielectric has dielectric constant K. a) (5 points) While the switch remains open, and no weight is placed on the upper plate of the capacitor, what is the potential in volts at points B, C, and D? What is the value of the current, i?Explanation / Answer
a) Potential = V
current = V/R
b) I = V/R * ( 1 - e -t/RC)
=> Vr = IR
c) V(t) = V * ( 1 - e -t/RC)
After long time current is zero .
d) Co = Eo * A/d
e) Decrease in distance , x = mg/Co
f) C1 = Eo * A/(d - x)
g) Here, V = Ed
h) voltage across capacitor is unchanged ..
V(t) = V * ( 1 - e -t/RC)
i) V = IR + Q/C - L * dI/dt
j) I = V/R * ( 1 - e -tR/L)
It gains additional charge .
k) ratiio of charge gained = mg(d + x)/EoA
l) resolution = 210
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