A capacitor discharges through a resistor with

PHYSICS · Multi-part · Fermi University Physics · Scientific Practice & Experimental Methods · Mathematical Methods for Physics · Calculus of change and accumulation

Mathematical Methods for Physics, Ch The derivative as a physical instrument

Problem

A capacitor discharges through a resistor with V(t)=V0e−t/τV(t) = V_0 e^{-t/\tau}, where V0=5.00 VV_0 = 5.00\,\mathrm{V} and τ=2.00 s\tau = 2.00\,\mathrm{s} are known exactly. A digital voltmeter reads VV to the nearest 1 mV1\,\mathrm{mV}, so each individual reading carries a reading uncertainty of ±0.5 mV\pm 0.5\,\mathrm{mV}. The times at which the meter is triggered are set by an exact electronic clock and carry no uncertainty of their own. The rate dV/dt\mathrm{d}V/\mathrm{d}t at t=2.00 st = 2.00\,\mathrm{s} is to be estimated by the central difference [V(t+h)−V(t−h)]/(2h)[V(t+h) - V(t-h)]/(2h) for three choices of half-interval: h=0.500 sh = 0.500\,\mathrm{s}, h=0.0500 sh = 0.0500\,\mathrm{s} and h=0.00500 sh = 0.00500\,\mathrm{s}. ((a)) Find the six meter readings the difference needs, by evaluating the decay law and rounding each value to the nearest 1 mV1\,\mathrm{mV} as the meter would, and from them compute the central-difference estimate of dV/dt\mathrm{d}V/\mathrm{d}t at t=2.00 st=2.00\,\mathrm{s} for each of the three intervals. ((b)) For each interval, state the resolution error of the estimate, using the rule that the resolution error of a difference estimate is the reading error divided by the interval used (the two voltage readings each carry ±0.5 mV\pm 0.5\,\mathrm{mV}, so their difference carries a combined uncertainty of 1.0 mV1.0\,\mathrm{mV}). ((c)) Compute the exact derivative directly from the decay law, dV/dt=−(V0/τ)e−t/τ\mathrm{d}V/\mathrm{d}t = -(V_0/\tau)e^{-t/\tau}, and compare it with the three estimates. State which of the three intervals gives the most trustworthy estimate, and explain why choosing the smallest available interval is not the right rule here.

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