The electronic circuit of a scale which uses the capacitance of a parallel plate

Question

The electronic circuit of a scale which uses the capacitance of a parallel plate capacitor to measure weight is shown below: SW1 k .-spring with spring constant k dielectric Let the battery voltage V-Vo, and let the capacitance of the capacitor C-C 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 x 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

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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