 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013/6/solution

L

–

R

–

C

branch and a separate resistor

R

in parallel. At

ω_{0} = 1/ L C

, the series branch is purely resistive, so its peak current is

I_{p} = V_{0} / R

. Its stored energies are

U_{L} = \frac{1}{2} L I_{p}^{2} sin^{2} ω_{0} t, U_{C} = \frac{1}{2} L I_{p}^{2} cos^{2} ω_{0} t .

Their sum is constant in sinusoidal steady state, with value

U_{m a x} = L V_{0}^{2} / (2 R^{2})

V_{0}^{2} / (2 R)

, including the resistor directly across the source. Over one cycle,

W_{diss} = \frac{2 π}{ω _{0}} \frac{V _{0}^{2}}{R}, \frac{U _{m a x}}{W _{diss}} = \frac{ω _{0} L}{4 π R} = \frac{1}{4 π R} \frac{L}{C} .

Omitting the parallel resistor would double this ratio. The expression concerns the driven steady state, not a switch-on transient.

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