IPhO 1983 · Theory problem 2

Different kind of oscillation

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Let’s consider the electric circuit in the figure, for which \(L_1=10\,mH\), \(L_2=20\,mH\), \(C_1=10\,nF\), \(C_2=5\,nF\) and \(R=100\,k\Omega\). The switch \(K\) being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.

Figure 1

a. Find the ratio of frequency \(f_m\) for which the active power in circuit has the maximum value \(P_m\) and the frequency difference \(\Delta f=f_+-f_-\) of the frequencies \(f_+\) and \(f_-\) for which the active power in the circuit is half of the maximum power \(P_m\).

The switch \(K\) is now open. In the moment \(t_0\) immediately after the switch is open the intensities of the currents in the coils \(L_1\) and \(L_2\) are \(i_{01}=0{,}1\,A\) and \(i_{02}=0{,}2\,A\) (the currents flow as in the figure); at the same moment, the potential difference on the capacitor with capacity \(C_1\) is \(u_0=40\,V\) :

Figure 2

b. Calculate the frequency of electromagnetic oscillation in \(L_1C_1C_2L_2\) circuit;

c. Determine the intensity of the electric current in the \(AB\) conductor;

d. Calculate the amplitude of the oscillation of the intensity of electric current in the coil \(L_1\).

Neglect the mutual induction of the coils, and the electric resistance of the conductors. Neglect the fast transition phenomena occurring when the switch is closed or opened.

Official English text of the problem.

Answers

  • a. \(\displaystyle \boxed{\frac{f_m}{\Delta f}=150}\), with \(f_m=15.9\,\mathrm{kHz}\) and \(\Delta f=106\,\mathrm{Hz}\).
  • b. \(\boxed{f_0=15.9\,\mathrm{kHz}}\), corresponding to \(\omega_0=1.00\times10^5\,\mathrm{rad\,s^{-1}}\).
  • c. \(\boxed{I_{AB}=0.100\,\mathrm A}\), constant and directed from \(B\) to \(A\).
  • d. \(\boxed{I_{1,\mathrm{osc}}=0.204\,\mathrm A}\). The current oscillates about a mean of \(-0.100\,\mathrm A\), taking the indicated initial-current direction as positive.

Solution

a. Driven circuit: resonance and half-power bandwidth

With \(K\) closed, both capacitors, both inductors, and the resistor are connected in parallel. Their equivalent capacitance and inductance are \[ C=C_1+C_2=15\,\mathrm{nF}, \qquad L=\left(\frac1{L_1}+\frac1{L_2}\right)^{-1} =\frac{L_1L_2}{L_1+L_2} =6.6667\,\mathrm{mH}. \] The inductance formula uses the stipulated absence of mutual induction.

Let \(I\) be the constant rms source current and \(\omega=2\pi f\). The circuit admittance is \[ Y=\frac1R+\mathrm i\left(\omega C-\frac1{\omega L}\right). \] Only \(R\) dissipates active power. Since the rms voltage is \(I/|Y|\), \[ P(\omega) =\frac{I^2}{R|Y|^2} =\frac{I^2R} {1+R^2\left(\omega C-\frac1{\omega L}\right)^2}. \]

The maximum occurs when the reactive part of the admittance vanishes: \[ \omega_m C-\frac1{\omega_m L}=0. \] Consequently, \[ \omega_m=\frac1{\sqrt{LC}}, \qquad P_m=I^2R. \]

At either half-power frequency, \[ P=\frac{P_m}{2} \quad\Longrightarrow\quad \left|\omega C-\frac1{\omega L}\right|=\frac1R. \] The two positive solutions are \[ \omega_\pm =\sqrt{\omega_m^2+\frac1{4R^2C^2}} \pm\frac1{2RC}. \] Their difference is therefore exactly \[ \Delta\omega=\frac1{RC}, \qquad \Delta f=\frac1{2\pi RC}. \] No narrow-band approximation is needed.

Numerically, \[ LC=1.00\times10^{-10}\,\mathrm{s^2}, \] so \[ f_m=\frac{10^5}{2\pi} =1.59155\times10^4\,\mathrm{Hz}, \] and \[ \Delta f =\frac1{2\pi(10^5)(15\times10^{-9})} =106.103\,\mathrm{Hz}. \] Thus the requested ratio is \[ \boxed{\frac{f_m}{\Delta f} =\omega_m RC =R\sqrt{\frac CL} =150}. \]

b. Free electromagnetic oscillation

After the switch opens, the resistor and source no longer load the LC circuit. Under the stated idealizations, its oscillation is undamped.

Define \[ u(t)=V_A-V_C=V_B-V_C, \] and take both inductor currents \(i_1,i_2\) as positive from the left rail to the right rail. These agree with the indicated initial directions, so \[ i_1(t_0)=0.1\,\mathrm A,\qquad i_2(t_0)=0.2\,\mathrm A,\qquad u(t_0)=40\,\mathrm V. \]

The inductor voltage laws give \[ L_1\dot i_1=-u,\qquad L_2\dot i_2=-u. \] The total capacitor current from right to left is \(C\dot u\). Kirchhoff’s current law therefore gives \[ C\dot u=i_1+i_2. \] Differentiating, \[ C\ddot u =\dot i_1+\dot i_2 =-\left(\frac1{L_1}+\frac1{L_2}\right)u, \] or \[ \ddot u+\omega_0^2u=0, \qquad \omega_0^2=\frac1C\left(\frac1{L_1}+\frac1{L_2}\right) =\frac1{LC}. \] Hence \[ \boxed{\omega_0=1.00\times10^5\,\mathrm{rad\,s^{-1}}}, \qquad \boxed{f_0=\frac{\omega_0}{2\pi}=15.9\,\mathrm{kHz}}. \]

For the remaining parts, the initial voltage derivative is \[ \dot u(t_0)=\frac{i_{01}+i_{02}}C =\frac{0.3}{15\times10^{-9}} =2.00\times10^7\,\mathrm{V\,s^{-1}}. \] Writing \(\theta=\omega_0(t-t_0)\), the voltage is therefore \[ \boxed{u(t)=\bigl(40\cos\theta+200\sin\theta\bigr)\,\mathrm V}. \]

c. Current in conductor \(AB\)

Let \(j\) be the current in \(AB\), positive from \(A\) to \(B\). At \(A\), the incoming inductor current splits between \(C_1\) and \(AB\): \[ j=i_1-C_1\dot u. \] Its derivative is \[ \dot j =-\frac{u}{L_1}-C_1\ddot u =\left(C_1\omega_0^2-\frac1{L_1}\right)u. \]

For the given components, \[ L_1C_1=L_2C_2=1.00\times10^{-10}\,\mathrm{s^2}. \] This implies \[ \omega_0^2=\frac1{L_1C_1}=\frac1{L_2C_2}, \] and consequently \[ \dot j=0. \] Thus the current in \(AB\) is constant despite the oscillating capacitor voltage and coil currents.

Its value follows from the initial conditions: \[ j=i_{01}-C_1\frac{i_{01}+i_{02}}{C_1+C_2} =0.1-\frac{10}{15}(0.3) =-0.100\,\mathrm A. \] The negative sign reverses the chosen reference direction: \[ \boxed{I_{AB}=0.100\,\mathrm A,\quad\text{from }B\text{ to }A}. \]

d. Oscillation amplitude of the current in \(L_1\)

Using \(i_1=j+C_1\dot u\) and the results above, \[ \begin{aligned} i_1(t) &=-0.100 +C_1\omega_0\bigl(-40\sin\theta+200\cos\theta\bigr)\\ &=\bigl(-0.100+0.200\cos\theta-0.0400\sin\theta\bigr)\,\mathrm A. \end{aligned} \]

For an oscillatory term \(a\cos\theta+b\sin\theta\), the amplitude is \(\sqrt{a^2+b^2}\). Therefore, \[ \boxed{ I_{1,\mathrm{osc}} =\sqrt{(0.200)^2+(0.0400)^2} =0.20396\,\mathrm A \simeq0.204\,\mathrm A}. \]

The constant term represents a persistent circulating current supported by the ideal inductors. Thus \(0.204\,\mathrm A\) is the amplitude about the mean \(-0.100\,\mathrm A\); the signed current ranges from approximately \(-0.304\,\mathrm A\) to \(+0.104\,\mathrm A\).

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.