You measure the capacitance of two capacitors with uncertainties, C1 = (1925 ± 30) μF and C2 = (673 ± 14) μF. You connect these capacitors in series. What is the uncertainty in the value for [math: 1C1] \frac{1}{C_1} ?
[math: Δ(1C1)=] \Delta (\frac{1}{C_1}) = Answer Question 8[input] 1/μF (3 sig. fig.) Note: this is asking for just the uncertainty, not the value, only enter the uncertainty into the box

Question

You measure the capacitance of two capacitors with uncertainties, C1 = (1925 ± 30) μF and C2 = (673 ± 14) μF. You connect these capacitors in series. What is the uncertainty in the value for [math: 1C1] \frac{1}{C_1} ?
[math: Δ(1C1)=] \Delta (\frac{1}{C_1}) =  Answer Question 8[input] 1/μF (3 sig. fig.) Note: this is asking for just the uncertainty, not the value, only enter the uncertainty into the box

BlackTom AI · Accepted Answer

To find the uncertainty in 1/C1 we use the propagation of uncertainty for a function y = 1/C. The differential form gives Δ(1/C) ≈ |d(1/C)/dC| · ΔC = (1/C^2) · ΔC, since d(1/C)/dC = -1/C^2 and we take the absolute value for the uncertainty.

Here, C1 is 1925 μF with an uncertainty ΔC = 30 μF. We compute C1^2 = (1925)^2 = 3,705,625 μF^2.

Then Δ(1/C1) ≈ ΔC / C1^2 = 30 / 3,705,625 ≈ 8.098... × 10^-6 1/μF.

Rounding to three significant figures gives 8.10 × 10^-6 1/μF. The units are per microfarad, as expected from the derivative with respect to C in μF. This approach assumes ΔC is small compared to C1 (valid here since 30 μF is a small fraction of 1925 μF), so the linear (first-order) approximation is appropriate.

The result is independent of the second capacitor for this particular calculation, since the question isolates the uncertainty in 1/C1. If you were combining the two capacitors in series, you would propagate uncertainties through the series formula for 1/C_total, but that is not required for finding Δ(1/C1) alone.

PHYS1221-PHYS1231-Physics 1B T1 2025

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