A parallel-plate capacitor has plates of area a distance apart

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Electromagnetic induction and alternating-current circuits

Electricity, Magnetism and Electromagnetism, Ch AC circuits: phasors, complex impedance, resonance and power

Problem

A parallel-plate capacitor has plates of area

A = 0.50\,\mathrm{m^2}

a distance

d = 0.10\,\mathrm{mm}

apart. A loss-free slab of relative permittivity

\varepsilon_r = 4.0

and thickness

0.050\,\mathrm{mm}

lies flat against one plate, filling half the gap; the remaining

0.050\,\mathrm{mm}

is air, for which

\varepsilon_r = 1

. Edge effects are ignored, so the field is uniform within each layer. This capacitor is placed in series with an ideal inductor

L = 2.0\,\mathrm{mH}

and a resistor

R = 5.0\,\Omega

, which is the total resistance of the loop. The series combination is connected across a full-wave rectified and smoothed supply of negligible internal resistance whose output is a steady

15.9\,\mathrm{V}

together with a ripple at

100\,\mathrm{Hz}

that may be modelled as a sinusoid of amplitude

1.15\,\mathrm{V}

. The steady part merely charges the capacitor and drives no steady current, so only the ripple drives a current. Take

\varepsilon_0 = 8.854\times10^{-12}\,\mathrm{F\,m^{-1}}

. ((a)) Find the capacitance

C

. ((b)) Find the natural angular frequency

\omega_0

of the loop. ((c)) Find the complex impedance of the loop at

100\,\mathrm{Hz}

and its magnitude, and say whether the current leads or lags the ripple voltage. ((d)) Find the amplitude of the ripple current. ((e)) Find the mean power delivered by the ripple and the power factor of the loop, and say which element takes that mean power.

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