Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section C / C10 / Solution

For the supplied linear meter model let ${\omega }_0^2=C/I$ and $2\gamma ={\Phi }_0^2/(RI)$. If the applied step is $i_s$, the settled deflection is ${\theta }_{\infty }={\Phi }_0{i}_s/C$. An underdamped response overshoots and rings about that value; a critically damped response reaches it monotonically without ringing; an overdamped response is also monotone from these initial conditions but has a slower final approach. The original plot uses the common final value and natural-frequency time scale.

The critical step response of a galvanometer has $\gamma ={\omega }_0$ and repeated homogeneous root $-{\omega }_0$. Apply $\theta (0)=\dot{\theta }(0)=0$ to the general constant-force solution to obtainIts derivative is ${\theta }_{\infty }{\omega }_0^{2}t{e}^{-{\omega }_{0}t}$, with the sign of the step. It is the fastest nonoscillatory response within this second-order family at fixed ${\omega }_0$, avoiding a needle that repeatedly crosses the reading or an unnecessarily slow overdamped settling.

For sinusoidal drive choose $e^{i\omega t}$ phasors. Writing $F_0={\Phi }_0{i}_0/I$, the response amplitude isFor positive $F_0$, choose the continuous lag branch from zero to $-\pi$ as frequency increases; an ordinary single-argument arctangent without the quadrant would give the wrong high-frequency phase.

For the critical galvanometer transfer function, the denominator factors as $({\omega }_0+i\omega {)}^2$. ThusFor $\omega \ll {\omega }_0$, the meter follows the quasi-static deflection ${\Phi }_0{i}_0/C$ with a small lag $\varphi \simeq -2\omega /{\omega }_0$. For $\omega \gg {\omega }_0$, its amplitude is $F_0/{\omega }^2$ and its lag approaches $-\pi$.

The stated components give $\gamma =12.5$ s⁻¹ and, if critical, ${\omega }_0=12.5$ rad/s, or $f_0=1.99$ Hz. At 50 Hz, $\omega =314.16$ rad/s andThe needle barely responds compared with its DC sensitivity and its mean deflection for a pure sinusoid is zero in this linear model. It is unsuitable for directly measuring 50-Hz amplitude or RMS current without an additional rectifier or another readout method. The calculation treats ${\Phi }_0$ as the stipulated torque/back-emf coefficient of the given equation; it does not infer an unprovided winding geometry.