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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014/5/ii/solution

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Past exam of the physics 2 paper of the University of Oxford / 2014 / 5 / ii / Solution by

Codex  0 2026-10-25

Define the phase difference as current phase minus voltage phase. Since

I = V / Z

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

(1)

It decreases continuously from

+ π /2

at low frequency, through zero at

ω_{0} = 1/ L C

, to

- π /2

at high frequency. Thus the current leads in the capacitive limit and lags in the inductive limit. Defining voltage phase minus current phase would reverse the graph's sign.

Figure 1.
Current phase relative to the source voltage in a series RLC circuit
. The example uses omega_0 L/R equal to one; the limiting phases and zero crossing hold for every positive R, L and C. **Details**.

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