 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016/11/i/a/solution

A series connection has the same electric current through each element, so complex impedances add:

Z_{s} = R + i (ω L - \frac{1}{ω C}) .

For

R > 0

ϕ = ar g V - ar g I = ar g Z_{s}

. Then

ϕ = tan^{- 1} (\frac{ω L - 1/ ( ω C )}{R}) .

Above resonance, the circuit is inductive and electric potential difference leads current; below resonance it is capacitive and electric potential difference lags current. At

ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C},

the inductive and capacitive reactances cancel and

Z_{s} = R

is purely resistive. The same resonance condition holds for an ideal series circuit independent of the drive amplitude.

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