NBPhO 2026, Теория и експеримент — Задача 2. Lissajous bridge

— Solution by Jaan Kalda. Set $V_E = 0$ and $V_A = V$ (phasor). Consider the RC branch: since $R_1$ and $C$ carry the same current, their voltages are in the impedance ratio $Z_C/Z_{R_1} = 1/(\mathrm{i}\omega R_1 C) = -\mathrm{i}/(\omega R_1 C)$. The factor $-\mathrm{i}$ is a $90^{\circ}$ clockwise rotation in the phasor plane, so $V_{BE}$ is $V_{AB}$ rotated $90^{\circ}$ clockwise — in particular, $V_{AB} \perp V_{BE}$, and $V_B$ lies in the lower half-plane ($\mathrm{Im}\, V_B < 0$).

For the RL branch, $Z_L/Z_{R_2} = \mathrm{i}\omega L/R_2$: the factor $\mathrm{i}$ is a $90^{\circ}$ counterclockwise rotation, so $V_{DE}$ is $V_{AD}$ rotated $90^{\circ}$ counterclockwise, giving $V_{AD} \perp V_{DE}$ and $V_D$ in the upper half-plane ($\mathrm{Im}\, V_D > 0$).

By Thales’ theorem applied to each branch, both $V_B$ and $V_D$ lie on the circle of diameter $\overline{AE}$ (centre $V/2$, radius $V/2$), on opposite semicircles.

The oscilloscope displays $V_{PB}$ on $x$ and $V_{PD}$ on $y$; the Lissajous figure degenerates to a line when these phasors are parallel or antiparallel — equivalently, when $P$, $B$, $D$ are collinear in the phasor plane. Since $V_B$ and $V_D$ are on opposite sides of the real axis, and $V_P$ is real (a fraction of $V$), $V_P$ lies between them on the line, so $V_{PB}$ and $V_{PD}$ are antiparallel. Hence the displayed line has negative slope and

The potentiometer is split 1:2 from left to right; depending on the orientation this gives $V_P = V/3$ or $V_P = 2V/3$. The power of point $P$ with respect to the Thales circle,

takes the value $-2V^2/9$ in both cases (by symmetry of the quadratic about $V_P = V/2$). Since $P$ is inside the circle and the line $BPD$ cuts it at two points,

Combining with $|\tan\alpha| = |V_{PD}|/|V_{PB}|$:

Grading: (6 points total)

• Showing that $V_{AB} \perp V_{BE}$ 0.5 pts
• Showing that $V_{AD} \perp V_{DE}$ 0.5 pts
• Correctly identifying that $V_B$ lies in the lower half-plane ($\mathrm{Im}\, V_B < 0$) ($\mathrm{Im}\, V_B > 0$ if $V_A = 0$) 0.3 pts
• Correctly identifying that $V_D$ lies in the upper half-plane ($\mathrm{Im}\, V_D > 0$) ($\mathrm{Im}\, V_D < 0$ if $V_A = 0$) 0.3 pts
• Identifying that $V_B$ lies on a circle with $AE$ as a chord 0.3 pts
• Identifying that $V_D$ lies on a circle with $AE$ as a chord 0.3 pts
• Identifying $AE$ as the diameter of the (common) circle 0.3 pts
• Recognising that Lissajous degeneracy requires $B$, $P$, $D$ collinear in the phasor plane 1.5 pts
• Applying power of a point at $V_P$ to get $|PB| \cdot |PD| = 2V^2/9$ 1.0 pts
• Correctly relating $|\tan\alpha|$ to the voltage amplitude ratio $|V_{PD}|/|V_{PB}|$ (only if phasors considered) 0.5 pts
• Combining with the power-of-a-point result to obtain the final answer 0.5 pts

Since the condition the line segment makes an angle $\alpha$ with the $x$-axis could be understood in different ways, we did not deduct marks for failing to realize that the slope is negative or using $\tan\alpha$ without absolute value.