Past exam of the physics 2 paper of the University of Oxford / 2018 / 9

For this question use phasors with time factor $e^{i\omega t}$, so an electrical impedance is $Z=\tilde{V}/\tilde{I}$. The same electric current flows through all components of a series circuit, and their voltages add by Kirchhoff's voltage law. Hence the total series electrical impedance isThe individual steady sinusoidal currents and their phase relations are derived in the following three parts. The diagram normalizes each current by its own amplitude, so it compares phase differences without mixing unlike physical units.

For a resistor, Ohm's law givesThe electric current and voltage are in phase.

For a capacitor, $I_C=CdV/dt$, soThe electric current leads the voltage by $\pi /2$.

For an inductor, $V=Ld{I}_L/dt$. In the steady AC solution with no added constant electric current,The electric current lags the voltage by $\pi /2$. An ideal inductor can also sustain an arbitrary DC current, fixed by initial conditions; it is excluded from this steady sinusoidal comparison.

The output is measured across the capacitor. The voltage divider gives the RC low-pass filter transfer functionThereforeThe amplitude ratio tends to $1$ and the phase shift to zero as $\omega \to 0$. For $\omega \to \infty$, the amplitude ratio is asymptotic to $1/(\omega RC)$ and the phase shift tends to $-\pi /2$.

The half-power amplitude condition for the RC low-pass filter gives $1+(\omega RC{)}^2=2$. Since angular frequency is positive,At this cutoff angular frequency, the output amplitude is smaller by $1/\sqrt{2}$ and the phase shift is $-\pi /4$.

The RC low-pass filter satisfies $RC{\dot{V}}_{\mathrm{out}}+V_{\mathrm{out}}=V$ for $t>0$. With zero initial capacitor voltage, the first-order linear differential equation givesThe output is continuous at switching and approaches $V$ with time constant $RC$.

For explicit sketches take a symmetric, 50% duty cycle square wave with levels $\pm V_0$. The source does not specify levels or duty cycle; a unipolar 50% wave follows by adding its mean and rescaling. Let $\tau =RC$ and $M$ be the positive output extremum in the periodic steady state. The square-wave response of an RC low-pass filter hasextended periodically. The extremum formula follows by imposing $V_{\mathrm{out}}(0)=-M$ and $V_{\mathrm{out}}(T/2)=M$. It makes the three regimes below quantitative, including their finite amplitudes.

When $\omega RC\ll 1$, each half-cycle lasts many time constants, so $M\simeq V_0$. The output nearly follows the two input levels, with exponentially rounded transitions lasting order $RC$. It remains continuous rather than making ideal vertical jumps.

When $\omega RC\gg 1$, the capacitor voltage changes little during each half-cycle. Thus ${\dot{V}}_{\mathrm{out}}\simeq V_{\mathrm{in}}/(RC)$, giving an approximately triangular output. Its extrema areThe finite but small triangle is centered on zero for the chosen zero-mean input. A nonzero input mean shifts the output mean by the same amount.

When $\omega RC$ is of order one, the output consists of visibly curved exponential charging and discharging arcs, with neither long flat plateaus nor nearly straight sides. At $\omega RC=1$,The output is continuous and lags the abrupt input transitions; its precise shape is the periodic solution given in the parent part.