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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015/8/c/solution

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Past exam of the physics 2 paper of the University of Oxford / 2015 / 8 / c / Solution by

Codex  0 2026-10-25

Keep the source diagram's topology: the arm from the left supply node to

X

contains

R_{0}

and

C_{0}

in series; the arm from

X

to the right supply node is

C_{1}

; the arm from the left node to

Y

is

R_{1}

; and the arm from

Y

to the right node contains

R

and

C

in parallel. Their complex impedances are

Z_{0} = R_{0} + \frac{1}{iω C _{0}}, Z_{1} = \frac{1}{iω C _{1}}, Z_{p} = (\frac{1}{R} + iω C)^{- 1} .

For this capacitor bridge with series loss and a parallel standard arm, the detector carries no electric current at balance. Taking the right supply node as zero electric potential and the left one as the source voltage

V_{s}

, the two voltage dividers give

V_{X} = V_{s} Z_{1} / (Z_{0} + Z_{1})

and

V_{Y} = V_{s} Z_{p} / (R_{1} + Z_{p})

. Equality of these complex voltages means equality at every time, rather than merely equality of their magnitudes. Cross multiplication yields

R_{1} Z_{1} = Z_{0} Z_{p}

. Thus

R_{0} + \frac{1}{iω C _{0}} = R_{1} \frac{Z _{1}}{Z _{p}} = \frac{R _{1} C}{C _{1}} + \frac{R _{1}}{iω R C _{1}} .

Equating real and imaginary parts gives

R_{0} = \frac{R _{1} C}{C _{1}}, C_{0} = \frac{R C _{1}}{R _{1}} .

The two adjustable standards determine the two components of the unknown complex impedance. Treating the lower-right arm as a series combination would give a different and incorrect balance equation for the source circuit. The result also gives the dimensionless dielectric loss tangent

ω C_{0} R_{0} = ω RC

for the series-loss model.

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