Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section B / 8

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

The original diagram has a series circuit containing $R$ and $L$, in parallel with $C$; it is not three separate parallel components. The electrical admittance is thereforeFor $R>0$, voltage and current are in phase when the imaginary part of $Y$ vanishes. Excluding the zero-frequency solution, division by $\omega$ givesA nonzero lossy parallel coil resonance exists only if $L>R^{2}C$. With $R=10\Omega$, $L=0.150\mathrm{H}$ and $C=1.00\times 10^{-5}\mathrm{F}$,For zero electrical resistance, the ideal parallel inductor-capacitor cancellation occurs atThe fractional reduction relative to the zero-resistance coil isAt exactly $R=0$ and $\omega ={\omega }_0$, $Y=0$, so finite applied voltage draws zero net current in the ideal model. The quoted frequency is the limiting parallel-resonance value; the phase of an exactly zero current is undefined.

Let $C_T=C+C_1$ be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put ${\omega }_{\ast }=2\pi (50\mathrm{Hz})$ into the lossy parallel coil resonance condition:The additional capacitance is thereforeThis is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.

For a long solenoid, solenoid inductance obeys $L={\mu }_0{N}^{2}A/\ell$, where $N$ is the number of turns, $A$ the cross-sectional area and $\ell$ the axial winding length. The natural reading of half as many turns occupying half the axial length givesWith unchanged wire material and gauge, halving the wire length also halves its electrical resistance: $R^{\prime }=R/2$. With the total capacitance $C_T$ fixed, the new angular frequency satisfiesThusNeglecting the small resistive correction gives the estimate $\sqrt{2}$.

There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L^{\prime }=L/4$ while $R^{\prime }=R/2$, givingThis corresponds to treating “length of the winding” as wire length while preserving axial geometry. The examiner's report explicitly accepted either inductance scaling, provided the resistance change was included. The first interpretation follows the statement that the axial winding length is halved; the second records the accepted ambiguity rather than silently choosing a different geometry.^[ref]