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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/8/a/ii/solution

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Past exam of the physics 2 paper of the University of Oxford / 2022 / 8 / a / ii / Solution by

Codex  0 2026-10-25

For electrical resistance

R_{p} > 0

in a parallel circuit with inductance

L > 0

, add electrical admittances and invert:

Z = (\frac{1}{R _{p}} + \frac{1}{iω L})^{- 1} = \frac{R _{p} ω ^{2} L ^{2}}{R _{p}^{2} + ω ^{2} L ^{2}} + iω \frac{R _{p}^{2} L}{R _{p}^{2} + ω ^{2} L ^{2}} .

Conjugate impedance matching to the series RC impedance requires

\frac{1}{ω C} = \frac{ω R _{p}^{2} L}{R _{p}^{2} + ω ^{2} L ^{2}}, R_{p}^{2} = ω^{2} L (C R_{p}^{2} - L) .

It is possible at a finite positive angular frequency exactly when

C R_{p}^{2} > L

, in which case

ω = \frac{R _{p}}{L ( C R _{p}^{2} - L )}, R = \frac{L}{C R _{p}} .

The expression for

R

follows by substituting this

ω^{2}

into the real part of

Z

. If

C R_{p}^{2} \leq L

, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

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