Question 5.EC.2: Black Box Apparatus and materials (1) A double beam oscillos......

(1) The type of the elements Z.

Adjust the oscilloscope to obtain the same gain for the two channels. Use the sine wave from the generator.

Use the circuit in Fig. 5 – 16 to determine the type of the elements. We find finally:

Z_{3}, Z_{1}^{\prime} and Z_{2}^{\prime} are resistors and Z_{1}^{\prime} = Z_{2}^{\prime}=2Z_{3}=(10.0\pm0.5)\ \mathrm{k\Omega},

Z_{1}, Z_{2} and Z_{3}^{\prime} are capacitors, and C_{1}=C_{2}=\frac{1}{2}C^{\prime}{}_{3}=\ (47\pm2)\ \mathrm{nF}.

(2) (a) The electric circuit of the black box DD’ A’ is shown In Fig. 5 -17.

(b) Apply a sine wave signal to connectors D and A’. Connect D and A’ to channel 1 (CH1) and D’ and A’ to channel 2 (CH2).

The ratio K={\frac{U_{D^{\prime} A^{\prime}}}{U_{\mathrm{DA^{\prime}}}}} is determined from the amplitudes of the signals U_{D^{\prime} A^{\prime}}

vá U_{D A^{\prime}}. The phase shift φ can be determined directly from the traces of the signals (or from the Lissajous patterns).

Tabulating K and φ versus f, we get for example:

Table–1–

The experimental error of f is ± 1 Hz, of φ is ± 5° and of K is ± 0.02. Here are the plots of K and φ as functions of f.

(c) The graphs possess a particular point at f_{0} = 330 ± 1 Hz, at which φ equals zero and K has a maximum of K = 0.33 ± 0.01. φ changes its sign from positive to negative with increasing of the frequency f across f_{0}.
The value of f_{0} may vary from 325 Hz to 335 Hz depending on the set of experiment, due to the deviation in the value of the resistance and capacitance in the set.
(d) The phasor diagram for the circuit is shown in Fig. 5 – 20, where u_{1} is the instantaneous voltage between D and D’ , and U_{1} – its amplitude.

We have tan \alpha_{1}=\frac{1}{\omega C R}\mathrm{~and}\,U_{1}=I_{1}\sqrt{\Bigl(\frac{R}{2}\Bigr)^{2}+\frac{1}{4C^{2}\omega^{2}}}.

For the D’ A’ parallel cicuit, I_{1} = I_{2} + I_{3}, and the phasor diagram is shown in Fig. 5 – 21. U_{2} is in phase with I_{3} •

tan \alpha_{2}=\frac{I_{2}}{I_{1}}=\omega\cdot 2C\cdot\frac{R}{2}=\omega C R. Let u_{2} the voltage between D’ and A’, we have I_{3}=\frac{U_{2}}{\frac{R}{2}};~I_{2}=U_{2}\cdot 2C\cdot\omega. Hence

By combining Fig. 5 – 20 and Fig. 5 – 21, we obtain Fig. 5 – 22, with U = U_{1} + U_{2} being the instantaneous voltage between D and A’ .

For \omega=2\pi f=\frac{1}{C R},\mathrm{~tan~}\alpha_{1}=\frac{1}{\omega C R}=\mathrm{tan~}\alpha_{2}=\omega C R. In this condition, U_{1} ,U_{2} and U are in phase, so that φ={\alpha}_{2}-{\alpha}_{1}\,=\,0. Hence K=\frac{U_{D^{\prime} A^{\prime}}}{U_{D A^{\prime}}}\,= {\frac{U_{2}}{U_{1}+U_{2}}}. Substituing  \omega={\frac{1}{C R}} into  U_{1} and U_{2} , we obtain K={\frac{1}{3}}.

That is, for   f_{0}\,=\,{\frac{\omega}{2\pi}}={\frac{1}{2\pi R C}}={\frac{1}{2\pi\cdot 10^{4}\cdot 48\cdot 10^{-9}}}\,=\,331{\mathrm{~Hz}},\;K {\frac{1}{3}}=0.33,\;\varphi=0\,,  which is observed in the experiment.

For  \omega\neq{\frac{1}{C R}}, U_{1} and U_{2} are out of phase, and K has values smaller than  {\frac{1}{3}}.