Figure 2 shows a balanced Maxwell's inductance bridge. (i) Draw the phasor diagram. (ii) Determine the expression for inductance L1.

Mchoro wa fasori wa daraja la Maxwell huonyesha uhusiano wa voltage na mkondo katika kila mkono. Kwa daraja lililosawazishwa, mkondo kupitia kigunduzi (D) ni sifuri.

• Hatua ya 1: Andika vipingamizi vya kila mkono. $Z_1 = R_1 + j\omega L_1$ $Z_2 = R_2$ $Z_3 = R_3$ $Z_4 = \frac{1}{\frac{1}{R_4} + j\omega C_4} = \frac{R_4}{1 + j\omega C_4 R_4}$ (tukichukulia $R_L$ ni $R_4$ na $L_L$ ni $C_4$ kwa daraja la Maxwell)

• Hatua ya 2: Tumia masharti ya usawa wa daraja. Kwa daraja lililosawazishwa, bidhaa za vipingamizi vya mikono inayopingana ni sawa:

  $$Z_1 Z_3 = Z_2 Z_4$$

  Badilisha vipingamizi:

  $$(R_1 + j\omega L_1) R_3 = R_2 \left( \frac{R_4}{1 + j\omega C_4 R_4} \right)$$

• Hatua ya 3: Rahisisha na utenganishe sehemu halisi na za kufikirika. Panua pande zote mbili:

  $$(R_1 + j\omega L_1) R_3 (1 + j\omega C_4 R_4) = R_2 R_4$$ $$R_1 R_3 (1 + j\omega C_4 R_4) + j\omega L_1 R_3 (1 + j\omega C_4 R_4) = R_2 R_4$$ $$R_1 R_3 + j\omega C_4 R_4 R_1 R_3 + j\omega L_1 R_3 + (j\omega)^2 L_1 R_3 C_4 R_4 = R_2 R_4$$

  Kumbuka kuwa $(j\omega)^2 = j^2 \omega^2 = -\omega^2$:

  $$R_1 R_3 + j\omega C_4 R_4 R_1 R_3 + j\omega L_1 R_3 - \omega^2 L_1 R_3 C_4 R_4 = R_2 R_4$$

  Panga upya maneno katika sehemu halisi na za kufikirika:

  $$(R_1 R_3 - \omega^2 L_1 R_3 C_4 R_4) + j(\omega C_4 R_4 R_1 R_3 + \omega L_1 R_3) = R_2 R_4$$

• Hatua ya 4: Linganisha sehemu halisi na za kufikirika. Kwa usawa wa daraja, sehemu halisi na za kufikirika lazima ziwe sawa kwa pande zote mbili. Sehemu halisi:

  $$R_1 R_3 - \omega^2 L_1 R_3 C_4 R_4 = R_2 R_4$$

  Hii inatoa usemi kwa $R_1$:

  $$R_1 R_3 = R_2 R_4 + \omega^2 L_1 R_3 C_4 R_4$$ $$R_1 = \frac{R_2 R_4}{R_3} + \omega^2 L_1 C_4 R_4$$

  Sehemu za kufikirika:

  $$\omega C_4 R_4 R_1 R_3 + \omega L_1 R_3 = 0$$

  Gawanya kwa $\omega R_3$ (kwa kuwa $\omega \neq 0$ na $R_3 \neq 0$):

  $$C_4 R_4 R_1 + L_1 = 0$$

  Hii inatoa $L_1 = -C_4 R_4 R_1$, ambayo si sahihi kwa inductance chanya.

  Kuna makosa katika uelewa wa jinsi daraja la Maxwell linavyosawazishwa au jinsi vipingamizi vinavyopangwa. Katika daraja la Maxwell, mkono wa capacitor sambamba na upinzani mara nyingi huwekwa kinyume na mkono wa inductance isiyojulikana.

  Kurekebisha: Tumia Admittance kwa mkono wa capacitor.
  Wakati mwingine, daraja la Maxwell huandikwa kama $Z_1 Y_4 = Z_2 Y_3$ au $Z_1 Y_2 = Z_3 Y_4$ kulingana na mpangilio.
  Hebu tutumie mpangilio wa kawaida wa daraja la Maxwell ambapo $Z_1$ na $Z_3$ ni mikono inayopingana, na $Z_2$ na $Z_4$ ni mikono inayopingana.
  $Z_1 = R_1 + j\omega L_1$
  $Z_2 = R_2$
  $Z_3 = R_3$
  $Z_4 = \frac{1}{1/R_4 + j\omega C_4}$

  Balance condition: $Z_1 Z_3 = Z_2 Z_4$
  $(R_1 + j\omega L_1) R_3 = R_2 \left( \frac{1}{1/R_4 + j\omega C_4} \right)$
  $(R_1 + j\omega L_1) R_3 \left( \frac{1}{R_4} + j\omega C_4 \right) = R_2$
  $R_1 R_3 \left( \frac{1}{R_4} + j\omega C_4 \right) + j\omega L_1 R_3 \left( \frac{1}{R_4} + j\omega C_4 \right) = R_2$
  $\frac{R_1 R_3}{R_4} + j\omega C_4 R_1 R_3 + j\frac{\omega L_1 R_3}{R_4} + (j\omega)^2 L_1 R_3 C_4 = R_2$
  $\frac{R_1 R_3}{R_4} + j\omega C_4 R_1 R_3 + j\frac{\omega L_1 R_3}{R_4} - \omega^2 L_1 R_3 C_4 = R_2$

  Panga upya maneno:

  $$\left( \frac{R_1 R_3}{R_4} - \omega^2 L_1 R_3 C_4 \right) + j\omega \left( C_4 R_1 R_3 + \frac{L_1 R_3}{R_4} \right) = R_2$$

  Linganisha sehemu halisi:

  $$\frac{R_1 R_3}{R_4} - \omega^2 L_1 R_3 C_4 = R_2$$ $$R_1 = \frac{R_2 R_4}{R_3} + \omega^2 L_1 C_4 R_4$$

  Linganisha sehemu za kufikirika:

  $$\omega \left( C_4 R_1 R_3 + \frac{L_1 R_3}{R_4} \right) = 0$$

  Kwa kuwa $\omega \neq 0$:

  $$C_4 R_1 R_3 + \frac{L_1 R_3}{R_4} = 0$$

  Gawanya kwa $R_3$:

  $$C_4 R_1 + \frac{L_1}{R_4} = 0$$ $$L_1 = -C_4 R_1 R_4$$

  Bado tunapata thamani hasi kwa $L_1$. Hii inaonyesha kuwa kuna tatizo na jinsi daraja la Maxwell linavyoeleweka au jinsi mchoro unavyopaswa kutafsiriwa.

  Kurekebisha tena: Daraja la Maxwell-Wien.
  Daraja la Maxwell-Wien hutumika kupima inductance. Katika daraja hili, mkono wa nne (DA) una upinzani $R_4$ sambamba na capacitor $C_4$.
  $Z_1 = R_1 + j\omega L_1$
  $Z_2 = R_2$
  $Z_3 = R_3$
  $Z_4 = \frac{R_4}{1 + j\omega C_4 R_4}$

  Balance condition: $Z_1 Z_3 = Z_2 Z_4$
  $(R_1 + j\omega L_1) R_3 = R_2 \frac{R_4}{1 + j\omega C_4 R_4}$
  $(R_1 + j\omega L_1) R_3 (1 + j\omega C_4 R_4) = R_2 R_4$
  $R_1 R_3 + j\omega C_4 R_4 R_1 R_3 + j\omega L_1 R_3 - \omega^2 L_1 R_3 C_4 R_4 = R_2 R_4$

  Equating real parts:
  $R_1 R_3 - \omega^2 L_1 R_3 C_4 R_4 = R_2 R_4$

  Equating imaginary parts:
  $\omega C_4 R_4 R_1 R_3 + \omega L_1 R_3 = 0$
  $C_4 R_4 R_1 + L_1 = 0 \implies L_1 = -C_4 R_4 R_1$.

  Kuna uwezekano mkubwa kwamba mpangilio wa daraja la Maxwell katika swali hili unamaanisha kuwa mkono wa $Z_4$ (DA) una upinzani $R_4$ na capacitor $C_4$ katika mfululizo, au kwamba mkono wa $Z_2$ au $Z_3$ una capacitor.

  Hata hivyo, ufafanuzi wa kawaida wa "Maxwell's inductance bridge" unahusisha capacitor katika sambamba na upinzani katika mkono mmoja. Ikiwa matokeo ni hasi, inaweza kumaanisha kuwa mpangilio wa mikono katika fomula ya usawa unahitaji kurekebishwa, au kwamba daraja lililochorwa si daraja la Maxwell la kawaida.

  Hebu tuchukue mpangilio wa kawaida wa daraja la Maxwell ambapo $Z_1$ na $Z_4$ ni mikono inayopingana, na $Z_2$ na $Z_3$ ni mikono inayopingana.
  Katika mchoro, $Z_1$ (AB) na $Z_3$ (CD) ni mikono inayopingana. $Z_2$ (BC) na $Z_4$ (DA) ni mikono inayopingana.
  Hivyo, $Z_1 Z_3 = Z_2 Z_4$.

  Kuna aina nyingine ya daraja la Maxwell ambapo mkono wa $Z_2$ una $R_2$ sambamba na $C_2$.
  Ikiwa $Z_2 = \frac{R_2}{1 + j\omega C_2 R_2}$ na $Z_4 = R_4$.
  $(R_1 + j\omega L_1) R_3 = \frac{R_2}{1 + j\omega C_2 R_2} R_4$
  $(R_1 + j\omega L_1) R_3 (1 + j\omega C_2 R_2) = R_2 R_4$
  $R_1 R_3 + j\omega C_2 R_2 R_