 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-2/5c/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

First construct the periodic response required in the introductory clause. At

ω_{0}^{2} = 1/ (L C)

, the terms

L I_{0}^{''} + I_{0} / C

cancel for a sinusoid, leaving

R I_{0}^{'} = ω_{0} cos (ω_{0} t)

. Thus

I_{0} (t) = \frac{1}{R} sin (ω_{0} t), I_{M} = \frac{1}{R} .

D = \frac{1}{R} \int_{0}^{T} sin^{2} (ω_{0} t) d t = \frac{T}{2 R} = \frac{π}{R ω _{0}} .

The Q factor, using the maximum magnetic energy

L I_{M}^{2} /2

, is consequently

Q = \frac{2 π}{D} \frac{L I _{M}^{2}}{2} = \frac{L ω _{0}}{R}, Q ω_{0} RC = L C ω_{0}^{2} = 1.

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