IPhO 1982 · Theory problem 1

Fluorescent lamp

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An alternating voltage of 50 Hz frequency is applied to the fluorescent lamp shown in the accompanying circuit diagram.

Figure 1

The following quantities are measured:

overall voltage (main voltage) \(U = 228.5\ \mathrm{V}\)
electric current \(I = 0.6\ \mathrm{A}\)
partial voltage across the fluorescent lamp \(U' = 84\ \mathrm{V}\)
ohmic resistance of the series reactor \(R_{\mathrm d} = 26.3\ \Omega\)

The fluorescent lamp itself may be considered as an ohmic resistor in the calculations.

a) What is the inductance \(L\) of the series reactor?

b) What is the phase shift \(\varphi\) between voltage and current?

c) What is the active power \(P_{\mathrm w}\) transformed by the apparatus?

d) Apart from limiting the current the series reactor has another important function. Name and explain this function!

Hint: The starter

Figure 2

includes a contact which closes shortly after switching on the lamp, opens up again and stays open.

e) In a diagram with a quantitative time scale sketch the time sequence of the luminous flux emitted by the lamp.

f) Why has the lamp to be ignited only once although the applied alternating voltage goes through zero in regular intervals?

g) According to the statement of the manufacturer, for a fluorescent lamp of the described type a capacitor of about \(4.7\ \mu\mathrm{F}\) can be switched in series with the series reactor. How does this affect the operation of the lamp and to what intent is this possibility provided for?

h) Examine both halves of the displayed demonstrator lamp with the added spectroscope. Explain the differences between the two spectra. You may walk up to the lamp and you may keep the spectroscope as a souvenir.

Official English text of the problem.

Answers

  • a) \(\boxed{L \approx 1.09\ \mathrm{H}}\).
  • b) \(\boxed{\varphi \approx 64.1^\circ}\), with the current lagging the mains voltage.
  • c) \(\boxed{P_{\mathrm w}\approx 59.9\ \mathrm{W}}\): \(50.4\ \mathrm{W}\) in the lamp and \(9.47\ \mathrm{W}\) in the reactor resistance.
  • d) The reactor produces the high-voltage ignition pulse by self-induction when the starter opens after preheating the electrodes.
  • e) The luminous flux pulsates at \(\boxed{100\ \mathrm{Hz}}\), repeating every \(\boxed{10\ \mathrm{ms}}\). An idealized sketch is \(\Phi(t)\propto\sin^2(2\pi\,50t-\varphi)\).
  • f) Residual ionization and hot electrodes allow the discharge to resume after each brief current zero without another cold-start ignition pulse.
  • g) The capacitor approximately preserves the current and mean brightness while making the current lead the voltage. For \(4.7\ \mu\mathrm{F}\), \(I_C\approx0.611\ \mathrm{A}\), leading by \(63.6^\circ\). Pairing this lamp with an unmodified lamp gives nearly unity overall power factor and reduces flicker and stroboscopic effects.
  • h) Assuming the demonstrator has uncoated and phosphor-coated halves: the uncoated half shows discrete mercury emission lines; the coated half additionally shows broad visible phosphor emission, produced by conversion of ultraviolet radiation into visible light.

Solution

The measured voltages and current are taken to be RMS values. During normal operation the starter is open. We use the stated ohmic approximation for the lamp and model the reactor as an ideal inductance in series with its winding resistance.

a) Inductance of the series reactor

The lamp resistance is \[ R_{\mathrm{lamp}}=\frac{U'}{I} =\frac{84}{0.6}=140\ \Omega. \] Thus the total series resistance is \[ R=R_{\mathrm{lamp}}+R_{\mathrm d} =140+26.3=166.3\ \Omega. \]

The resistive voltage is in phase with the current, while the inductive voltage leads the current by \(90^\circ\). Consequently, the RMS voltages obey the phasor relation \[ U^2=(IR)^2+(I\omega L)^2, \qquad \omega=2\pi f. \] The total resistive voltage is \[ IR=U'+IR_{\mathrm d} =84+0.6(26.3)=99.78\ \mathrm V. \] Hence the inductive voltage is \[ U_L=\sqrt{228.5^2-99.78^2} =205.56\ \mathrm V, \] and \[ L=\frac{U_L}{2\pi fI} =\frac{205.56}{2\pi(50)(0.6)} =\boxed{1.09\ \mathrm H}. \]

The corresponding inductive reactance is \[ X_L=\omega L=\boxed{342.6\ \Omega}. \]

b) Phase shift

For a series resistance and inductance, \[ \tan\varphi=\frac{X_L}{R}, \] so \[ \varphi=\tan^{-1}\!\left(\frac{342.6}{166.3}\right) =\boxed{64.1^\circ}. \] The current lags the applied voltage. Equivalently, \[ \cos\varphi=\frac{IR}{U} =\frac{99.78}{228.5} =0.4367. \]

With the time origin at a rising zero crossing of the mains voltage, \[ u(t)=\sqrt2\,U\sin\omega t, \qquad i(t)=\sqrt2\,I\sin(\omega t-\varphi). \]

c) Active power

An ideal inductance stores and returns energy, consuming no average power. The active power is therefore dissipated in the lamp and the winding resistance: \[ P_{\mathrm w}=I^2(R_{\mathrm{lamp}}+R_{\mathrm d}). \] Numerically, \[ P_{\mathrm w}=0.6^2(166.3) =59.868\ \mathrm W \approx\boxed{59.9\ \mathrm W}. \]

Its two contributions are \[ P_{\mathrm{lamp}}=U'I=84(0.6) =\boxed{50.4\ \mathrm W}, \] \[ P_{\mathrm d}=I^2R_{\mathrm d} =0.6^2(26.3) =\boxed{9.47\ \mathrm W}. \] This agrees with \(P_{\mathrm w}=UI\cos\varphi\).

d) Ignition by self-induction

When the starter closes, current flows through the reactor, the two electrode filaments, and the starter contact. This heats the electrodes, facilitating electron emission.

When the starter subsequently opens, the reactor opposes the rapid decrease of current. Faraday’s law gives its induced emf: \[ \mathcal E_L=-L\frac{di}{dt}. \] A rapid interruption therefore produces a large voltage pulse across the lamp. This pulse initiates electrical breakdown of the gas and establishes the discharge. Energy stored in the reactor’s magnetic field, \[ E_L=\frac12Li^2, \] is available during this transient.

Once the discharge is established, the starter remains open and the reactor limits the operating current. The magnitude of the ignition pulse cannot be calculated from the supplied data because the interruption time and transient circuit properties are unspecified.

e) Time dependence of the luminous flux

For an idealized sketch, assume constant luminous efficacy and neglect the phosphor’s emission delay. The luminous flux is then proportional to the instantaneous power delivered to the lamp: \[ p_{\mathrm{lamp}}(t) =R_{\mathrm{lamp}}i^2(t) =2R_{\mathrm{lamp}}I^2\sin^2(\omega t-\varphi). \] Here, \[ p_{\mathrm{lamp}}(t) =(100.8\ \mathrm W)\sin^2(\omega t-\varphi). \]

Thus \[ \boxed{\Phi(t)=\Phi_{\max}\sin^2(\omega t-\varphi)} =\frac{\Phi_{\max}}2 \left[1-\cos(2\omega t-2\varphi)\right]. \] Both current directions produce light, so the luminous-flux period is \[ T_{\Phi}=\frac{1}{2f} =\boxed{10\ \mathrm{ms}}, \] corresponding to \(\boxed{100\ \mathrm{Hz}}\).

For the sketch, choose a local time \(\tau=0\) at a current zero. Draw a smooth, nonnegative curve through these points and repeat it every \(10\ \mathrm{ms}\):

\(\tau\) (\(\mathrm{ms}\)) 0 2.5 5.0 7.5 10.0
\(\Phi/\Phi_{\max}\) 0 \(1/2\) 1 \(1/2\) 0

The curve has horizontal tangents at its minima and maxima. Its mean is \(\Phi_{\max}/2\).

Relative to the mains-voltage time origin used in part (b), the first current zero occurs at \[ t=\frac{\varphi}{\omega}=3.56\ \mathrm{ms}. \] Accordingly, ideal light minima occur at \(3.56,\ 13.56,\ldots\ \mathrm{ms}\), and maxima at \(8.56,\ 18.56,\ldots\ \mathrm{ms}\).

The absolute luminous flux is undetermined because no luminous efficacy is supplied. In a real lamp, phosphor persistence smooths and delays the light output, so its minima need not reach zero.

f) Why a fresh ignition pulse is unnecessary

The large starting voltage is needed to establish a discharge in the initially weakly ionized gas. During operation, the gas contains charged particles and the electrodes remain hot.

Around each current zero, the interruption is too brief for complete loss of ionization and cooling of the electrodes. The discharge can therefore resume in the next half-cycle at a much lower voltage than was required for the initial cold start.

Also, the source-voltage zeros and current zeros do not coincide because of the phase shift. Nevertheless, the current does cross zero every \(10\ \mathrm{ms}\); residual ionization and the heated electrodes explain why another starter pulse is unnecessary.

g) Effect and purpose of the series capacitor

With a capacitor in series, the impedance becomes \[ Z_C=R+j\left(\omega L-\frac{1}{\omega C}\right). \] For \(C=4.7\ \mu\mathrm F\), \[ X_C=\frac{1}{2\pi(50)(4.7\times10^{-6})} =677.3\ \Omega. \] The net reactance is therefore \[ X_{\mathrm{net}}=342.6-677.3 =-334.6\ \Omega. \] It is capacitive and has almost the same magnitude as the original inductive reactance.

The new current is \[ I_C=\frac{U}{\sqrt{R^2+X_{\mathrm{net}}^2}} =\frac{228.5}{\sqrt{166.3^2+334.6^2}} =\boxed{0.611\ \mathrm A}. \] The impedance phase is \[ \varphi_C=\tan^{-1}\!\left(\frac{-334.6}{166.3}\right) =-63.6^\circ, \] so the current now leads the voltage by \(\boxed{63.6^\circ}\).

The lamp power becomes \[ P_{\mathrm{lamp},C}=I_C^2R_{\mathrm{lamp}} \approx52.3\ \mathrm W, \] close to its original \(50.4\ \mathrm W\). Thus its mean brightness is approximately preserved.

To see why this capacitance was chosen, require the current magnitude to remain exactly unchanged: \[ (X_L-X_C)^2=X_L^2. \] The nonzero solution is \[ X_C=2X_L, \] giving \[ C_*=\frac{1}{2\omega X_L} =\frac{1}{2\omega^2L} =\boxed{4.65\ \mu\mathrm F}, \] which agrees with the manufacturer’s approximate \(4.7\ \mu\mathrm F\). At this value, the impedance changes from \(R+jX_L\) to \(R-jX_L\).

This permits a useful two-lamp arrangement: connect two lamp branches in parallel across the supply, one with the original reactor and one with the reactor plus series capacitor.

  • Power-factor improvement. For the ideal equal-current case, the two currents lag and lead by equal angles. Their reactive components cancel: \[ \underline I_{\mathrm{total}} =\frac{U}{R+jX_L}+\frac{U}{R-jX_L} =\frac{2UR}{R^2+X_L^2}. \] The total current is in phase with the voltage, giving unity power factor. With the nominal \(4.7\ \mu\mathrm F\), cancellation is nearly complete.

  • Reduced light modulation. The lamps reach their light minima at different times. Under the model of part (e), their combined flux is \[ \begin{aligned} \Phi_{\mathrm{pair}}(t) &=\Phi_{\max}\left[ \sin^2(\omega t-\varphi)+\sin^2(\omega t+\varphi)\right]\\ &=\Phi_{\max}\left[1-\cos(2\varphi)\cos(2\omega t)\right]. \end{aligned} \] Since \(|\cos(2\varphi)|\approx0.619<1\), the combined light never falls to zero in this model. Flicker and stroboscopic effects are reduced, although the light is not perfectly constant.

h) Spectra of the two halves

The expected interpretation is that the demonstrator has one uncoated half and one half coated with fluorescent phosphor.

In the uncoated half, the observed visible light comes directly from the mercury discharge. Excited mercury atoms emit photons at discrete transition energies, \[ h\nu=E_{\mathrm{upper}}-E_{\mathrm{lower}}, \] so the spectroscope shows separated emission lines. Much of the discharge radiation is ultraviolet and is therefore invisible to the eye.

In the phosphor-coated half, the coating absorbs ultraviolet radiation and re-emits part of its energy as visible light. This fluorescence produces broad visible emission bands, often appearing nearly continuous in a simple spectroscope. Direct mercury lines can remain superposed on this broader emission.

The coating therefore increases the useful visible output and changes its colour toward white. The spectral difference results from ultraviolet-to-visible conversion in the phosphor.

The solution was checked against the official solution.