Uncertainty of capacitance. What is the uncertainty in the equivalent capacitance when C1 and C2 are connected in series?

Question

1)You measure the capacitance of two capacitors with uncertainties, C1 = (1548 ± 23) μF and C2 = (865 ± 14) μF. You connect these capacitors in series. What is the uncertainty in the value for 1/C1?

2)What is the uncertainty in the value of 1/C1 + 1/C2?

3) What is the uncertainty in the equivalent capacitance when C₁ and C₂ are connected in series? Give your answer to the nearest whole number as you would present the value of the equivalent capacitance to 3 significant figures, so the uncertainty should be given to the nearest whole number.

Answer 1

The easiest way to think of the uncertainty is to use percentage error for the original Capacitance and apply that same percentage to 1/C. So for C₁:

23/1548 = 1.49%

1/C₁ = 646 F^-1

so δC₁ = .0149 * 646 = 9.60 F^-1

And beyond there, you just need normal error propagation:

Z = X + Y

δZ = ( ( δX )² + ( δY )² )^1/2