Capacitance — Question 5

Question 5

Hard charge and voltage bounds give an asymmetric capacitance interval

Independent hard bounds are Q=50.0$\pm$2.0 nC and V=10.0$\pm$.5 V. Both quantities are positive.

1. Find nominal C and its exact worst-case interval.

2. Compare the interval with the first-order relative bound from adding the fractional errors.

Question 5 - Solution

Hard charge and voltage bounds give an asymmetric capacitance interval

(a) Method and calculation

$$C_{\mathrm{nominal}}=50.0/10.0=5.00\,\mathrm{nF}.$$$$C\in[48.0/10.5,52.0/9.5]=[4.57143,5.47368]\,\mathrm{nF}.$$

(b) Interpretation and further calculation

$$\frac{|\delta C|}{C}\lesssim\frac2{50}+\frac{.5}{10}=.0900.$$$$C_{\mathrm{linear}}\simeq5.00\pm.45\,\mathrm{nF}.$$This linearized interval is not the exact hard bound.

Physical check

The largest ratio combines the largest numerator with the smallest denominator; no statistical independence formula is being used.