A capacitor of capacitance discharges through a resistor of resistance

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Electricity, Magnetism & Electromagnetism · Foundations of mathematical language

Mathematical Methods for Physics, Ch Series, limits and controlled approximation

Problem

A capacitor of capacitance CC discharges through a resistor of resistance RR, with no other loss in the circuit, so the voltage across it obeys V(t)=V0e−t/τV(t) = V_0 e^{-t/\tau}, where τ=RC\tau = RC and V0V_0 is the voltage at t=0t = 0. To find how long the capacitor has been discharging, someone measures the fractional drop y=1−V/V0y = 1 - V/V_0 and uses the linear estimate t≈τy.t \approx \tau y. (a) Show that the exact elapsed time is t=−τln⁡(1−y)t = -\tau\ln(1-y). Using Taylor's theorem with the Lagrange remainder taken to the linear term, bound the error t/τ−yt/\tau - y above and below for 0<y<10 < y < 1, saying where on the interval from 00 to yy the second derivative must be taken for each bound. Hence give the regime, as a bound on yy, within which the linear estimate of t/τt/\tau is guaranteed to be in error by no more than ε\varepsilon. (b) Take R=1.0×103 ΩR = 1.0\times10^3\,\Omega, C=2.2×10−6 FC = 2.2\times10^{-6}\,\mathrm{F} and ε=0.5×10−2\varepsilon = 0.5\times10^{-2}. State τ\tau, the largest fractional drop ymax⁡y_{\max} that your regime allows (to two significant figures), and the linear estimate τymax⁡\tau y_{\max} in milliseconds. (c) Check part (b) by computing the exact error −ln⁡(1−y)−y-\ln(1-y) - y at y=ymax⁡y = y_{\max}, and confirm it lies at or below ε\varepsilon. Then show, by the same exact computation, that the regime y≤2εy \le \sqrt{2\varepsilon}, which a bound taken with the second derivative at y=0y = 0 would give, is not safe.

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