NBPhO 2026

Overview

The circuit is an AC bridge, observed not with a galvanometer but with an oscilloscope in $X$–$Y$ mode. Two filter branches sit in parallel between $A$ and $E$:

• a series $R_1C$ (low-pass), with the inter-element node $B$;
• a series $R_2L$ (high-pass), with the inter-element node $D$.

A potentiometer between the same two terminals gives a continuously adjustable real reference voltage $V_P$, with both probe channels referred to $P$ as the common ground:

$$V_x \;=\; V_B - V_P, \qquad V_y \;=\; V_D - V_P.$$

Two simple ideas drive the whole problem:

• A Lissajous figure degenerates into a line iff $V_x$ and $V_y$ are in phase (or anti-phase). Equivalently, the complex ratio $V_y/V_x$ must be real.
• The slider’s role is to be real. $V_P$ is in phase with the source $V$, while $V_B$ and $V_D$ are complex (each delayed by an $RC$ or $RL$ filter). Subtracting an adjustable real number from each is exactly what one needs to bring two phasors into alignment — like nulling a Wheatstone bridge with a variable resistor, but in the complex plane.

The line-segment condition will turn out to be a frequency-independent constraint on the components ($L = 2R_1R_2C$ for the given slider position), while the slope $\tan\alpha$ depends on the actual driving frequency through the dimensionless parameter $a \equiv \omega R_1 C$. The voltage we are asked to find, $|V_{BP}| = |V_x|$, is just the half-extent of the visible line segment along the screen’s $x$-axis.

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Phasor representation

Take $E$ as ground and let the source amplitude $V$ be real (this only fixes the global phase reference). With $a \equiv \omega R_1 C$ and $b \equiv \omega L/R_2$, the two filter nodes are ordinary voltage dividers between a resistor and a reactive impedance:

$$V_B \;=\; V\,\frac{1/(j\omega C)}{R_1 + 1/(j\omega C)} \;=\; \frac{V}{1 + ja},$$ $$V_D \;=\; V\,\frac{j\omega L}{R_2 + j\omega L} \;=\; \frac{V\,jb}{1 + jb}.$$

The slider, looking at the figure, divides $AE$ in the ratio $AP:PE = 1:2$ from $A$ (left) to $E$ (right). The potentiometer is a uniform resistor carrying a current driven by $V$, so its voltage drops linearly with arc length:

$$V_P \;=\; V_A - \frac{AP}{AE}\,(V_A - V_E) \;=\; V - \tfrac{1}{3}V \;=\; \tfrac{2}{3}V.$$

Crucially $V_P$ is purely real — the slider does not change phase, only amplitude. This is what makes the bridge nullable.

The two oscilloscope phasors

$$V_x \;=\; V_B - V_P \;=\; \frac{V}{1+ja} - \tfrac{2}{3}V \;=\; \frac{V\,(1 - 2ja)}{3(1+ja)},$$ $$V_y \;=\; V_D - V_P \;=\; \frac{V\,jb}{1+jb} - \tfrac{2}{3}V \;=\; \frac{-V\,(2 - jb)}{3(1+jb)}.$$

Both have the source amplitude $V$ pulled out front, the same factor $1/3$ from the slider position, and a complex factor that carries all the frequency dependence.

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The line-segment condition

A Lissajous figure traced by $(\mathrm{Re}\{V_x e^{j\omega t}\},\,\mathrm{Re}\{V_y e^{j\omega t}\})$ is in general an ellipse. It collapses to a line iff $V_x$ and $V_y$ are in phase — equivalently, iff the complex ratio $V_y/V_x$ is real. Form

$$\frac{V_y}{V_x} \;=\; -\,\frac{(2-jb)(1+ja)}{(1+jb)(1-2ja)}.$$

Multiplying out the brackets,

$$(2-jb)(1+ja) \;=\; (2+ab) + j(2a-b),$$ $$(1+jb)(1-2ja) \;=\; (1+2ab) + j(b-2a) \;=\; (1+2ab) - j(2a-b).$$

So writing $u = 2 + ab$, $w = 1 + 2ab$, $\delta = 2a - b$,

$$\frac{V_y}{V_x} \;=\; -\,\frac{u + j\delta}{w - j\delta}.$$

For this to be real, multiply numerator and denominator by $w + j\delta$ and read off the imaginary part of the new numerator:

$$\mathrm{Im}\bigl[-(u+j\delta)(w+j\delta)\bigr] \;=\; -\delta(u + w) \;=\; -\delta\,(3 + 3ab) \;=\; -3\delta\,(1 + ab).$$

Since $a,b > 0$, the factor $(1 + ab)$ never vanishes. The line condition therefore reduces to

$$\boxed{\;\delta \;=\; 2a - b \;=\; 0\;} \qquad\Longleftrightarrow\qquad b \;=\; 2a.$$

In dimensional form,

$$\frac{\omega L}{R_2} \;=\; 2\,\omega R_1 C \quad\Longrightarrow\quad \boxed{\;L \;=\; 2\,R_1\,R_2\,C\;}.$$

Why this is a Wheatstone-bridge balance

The balance condition is frequency-independent — $\omega$ has dropped out. This is the AC analogue of the classical Wheatstone null. To see the analogy, note that the line condition is

$$\frac{V_B - V_P}{V_D - V_P} \in \mathbb{R}.$$

Reading $V_P$ as a “tap” between the two impedance ladders, the balance reads

$$\frac{Z_C}{R_1} \;\propto\; \frac{Z_L}{R_2}\quad\text{(in a sense made precise by the calculation above)},$$

with the slider position controlling the proportionality constant. For $V_P = \tfrac{2}{3}V$, the ratio works out to $L/R_2 = 2 R_1 C$. For a general slider position $V_P = kV$, the line condition becomes $b/a = k/(1-k)$, which is what one would adjust the slider to achieve in practice. Here the experimental observation $k = 2/3$ feeds directly back to $L = 2 R_1 R_2 C$.

Educational remark — what the slider really does

Without the slider, $V_x = V_B$ and $V_y = V_D$ each have phases set by an $RC$ or $RL$ section, and these phases generally differ; the Lissajous is then an ellipse fixed by the circuit. Sliding $P$ adds a real offset to each phasor — equivalent to translating both $V_B$ and $V_D$ in the complex plane along the real axis by the same amount. Two phasors that are not collinear with the origin can be made collinear with the origin by such a real translation iff their imaginary parts are proportional to one specific real translation (the slider position). That is the content of the calculation: only one slider position $k$ exists, in general, for which the figure straightens.

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The slope of the line

With $b = 2a$, the imaginary parts of numerator and denominator both vanish (each contains $\delta=0$). The remaining real ratio is

$$\frac{V_y}{V_x} \;=\; -\,\frac{u}{w} \;=\; -\,\frac{2 + ab}{1 + 2ab} \;=\; -\,\frac{2 + 2a^2}{1 + 4a^2} \;=\; -\,\frac{2(1+a^2)}{1+4a^2}.$$

This real ratio is, by construction, the slope $\tan\alpha$ of the line on the oscilloscope:

$$\boxed{\;\tan\alpha \;=\; -\,\frac{2(1+a^2)}{1+4a^2},\qquad a = \omega R_1 C\;}$$

Physical reading of the slope

Two limits are immediate:

• $a\to 0$ (low frequency, $\omega R_1 C\ll 1$): $\tan\alpha\to -2$, so $\alpha \to \arctan(-2)\approx -63^\circ$.
• $a\to\infty$ (high frequency): $\tan\alpha\to -\tfrac{1}{2}$, so $\alpha \to \arctan(-\tfrac{1}{2})\approx -27^\circ$.

So the slope is always negative and bounded:

$$-2 \;<\; \tan\alpha \;<\; -\tfrac{1}{2}.$$

Why negative? At low frequency the capacitor is open-circuit-like, so $V_B \to V$, almost equal to $V_A$; correspondingly $V_x = V_B - V_P \to \tfrac{V}{3}$ (positive). At the same low frequency the inductor is short-circuit-like, so $V_D\to 0$ and $V_y = V_D - V_P \to -\tfrac{2V}{3}$ (negative). So $V_y/V_x < 0$ in this limit, and continuity (plus the no-zero-crossing of the ratio across $a\in(0,\infty)$) keeps it negative throughout. The high-frequency limit just swaps the roles of $C$ and $L$ and gives the mirror-image bound.

The fact that $|\tan\alpha|$ varies continuously between $\tfrac{1}{2}$ and $2$ as the frequency is swept means the experimentalist can read the frequency off the slope once the bridge is balanced — that is the practical use of the device.

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Amplitude of $V_{BP}$

The voltage between $B$ and $P$ is precisely the $x$-channel signal:

$$V_{BP} \;=\; V_B - V_P \;=\; V_x \;=\; \frac{V\,(1 - 2ja)}{3(1+ja)}.$$

Take its modulus:

$$|V_{BP}| \;=\; \frac{V}{3}\,\frac{|1 - 2ja|}{|1+ja|} \;=\; \frac{V}{3}\,\sqrt{\frac{1+4a^2}{1+a^2}}.$$

We need this in terms of the given angle $\alpha$, not the unknown frequency parameter $a$. From the slope formula,

$$\tan\alpha \;=\; -\,\frac{2(1+a^2)}{1+4a^2} \quad\Longrightarrow\quad \frac{1+4a^2}{1+a^2} \;=\; -\,\frac{2}{\tan\alpha}.$$

Substituting,

$$\boxed{\;|V_{BP}| \;=\; \frac{V}{3}\,\sqrt{-\,\frac{2}{\tan\alpha}} \;=\; \frac{V}{3}\,\sqrt{-2\cot\alpha}\;}$$

The radicand is positive because, as we saw, $\tan\alpha$ is necessarily negative; equivalently,

$$|V_{BP}| \;=\; \frac{V}{3}\,\sqrt{2\,|\cot\alpha|}.$$

Geometric reading

The whole Lissajous line segment, as the source goes through one cycle, spans $|V_x|$ along $x$ and $|V_y|$ along $y$:

$$|V_y| \;=\; |V_x|\cdot|\tan\alpha|,\qquad \text{full half-length}\;=\;\sqrt{|V_x|^2+|V_y|^2} \;=\; |V_x|\sqrt{1 + \tan^2\alpha} \;=\; \frac{|V_x|}{|\cos\alpha|}.$$

So $|V_{BP}|$ is the projection of the line segment’s half-length onto the $x$-axis — exactly what one would measure on the screen by reading off the $x$-extent of the bright trace.

Sanity checks

• Range of $|V_{BP}|/V$. As $a$ runs from $0$ to $\infty$, $|V_{BP}|/V$ runs monotonically from $\tfrac{1}{3}$ to $\tfrac{2}{3}$. From the boxed formula: at $\tan\alpha=-2$, $-2/\tan\alpha = 1$ and $|V_{BP}|/V = \tfrac{1}{3}$; at $\tan\alpha=-\tfrac{1}{2}$, $-2/\tan\alpha = 4$ and $|V_{BP}|/V = \tfrac{2}{3}$. ✓
• Direct check at $a=1/\sqrt{2}$. Then $\tan\alpha = -2(3/2)/3 = -1$ and the boxed formula gives $|V_{BP}| = V\sqrt{2}/3$. From the original: $\sqrt{(1+2)/(1+1/2)} = \sqrt{2}$, so $|V_{BP}| = (V/3)\sqrt{2}$. ✓
• Dimensions. $\sqrt{-2\cot\alpha}$ is dimensionless; $V$ is a voltage; $|V_{BP}|$ is therefore a voltage. ✓
• $V_P = 0$ (slider at $E$). Then $|V_{BP}| = |V_B| = V/\sqrt{1+a^2}$ — a standard low-pass amplitude. The boxed formula does not apply in that case because the line condition itself collapses to a different relation; it is specific to the experimentally observed slider position $k = 2/3$.

Educational remark — why the slider’s $1{:}2$ ratio matters quantitatively

The factor of $\tfrac{1}{3}$ in front of $\sqrt{-2\cot\alpha}$ is precisely $1 - k$, the fraction of the source amplitude on the $E$ side of the slider, with $k = 2/3$. Had the slider divided in ratio $1{:}1$ (slider at the centre, $k=\tfrac{1}{2}$), one would repeat the algebra with $V_P = V/2$ and find a balance condition $L = R_1 R_2 C$ (rather than $2R_1R_2C$), and a prefactor of $\tfrac{1}{2}$ rather than $\tfrac{1}{3}$. The $1{:}2$ ratio is therefore not just a number to plug in — it sets both which circuit can be balanced and the overall amplitude of the visible $x$-signal.

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Summary of results

Quantity	Result
Line-segment (balance) condition	$\;L = 2\,R_1\,R_2\,C\;$
Slope of the line	$\;\tan\alpha = -\dfrac{2(1+a^2)}{1+4a^2},\quad a=\omega R_1 C\in\bigl(\tfrac{1}{2},2\bigr)\text{ for slope range}\;$
Voltage between $B$ and $P$	$\;\boxed{\,\lvert V_{BP}\rvert \;=\; \dfrac{V}{3}\,\sqrt{-2\cot\alpha}\,}\;$

One sentence to take away. Adjusting the slider at a real-valued tap to align two filter-output phasors is the AC bridge’s defining trick: balance gives a frequency-independent component relation ($L = 2R_1R_2C$ for slider ratio $1{:}2$), while the residual slope of the resulting line encodes the driving frequency, and the on-screen $x$-extent of that line is $\,|V_{BP}| = (V/3)\sqrt{-2\cot\alpha}$.