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State three commonly used detectors for a.c bridge measurements. With the aid of a labelled diagram, derive the balance equation for an inductive impedance a.c bridge in terms of rectangular co-ordinates.

Physics

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2 parts
a

C1=C2R2R3R2R4−R1R3C_1 = C_2 \frac{R_2 R_3}{R_2 R_4 - R_1 R_3}

b

f=2500π Hz≈795.77 Hzf = \frac{2500}{\pi} \, Hz \approx 795.77 \, Hz

Step-by-step solution

7 steps

Haya Engineer, twende kazi.

Part a· 4 steps

**Derive the expression for capacitor C1C_1

  1. Andika vizuizi (impedances) vya kila mkono wa daraja.

    Kwa daraja la Wein lililosawazishwa, tunatumia masharti ya usawa wa daraja la a.c.
    Vizuizi vya mikono ni:

    • Mkono AD: ZAD=R1+1jωC1Z_{AD} = R_1 + \frac{1}{j\omega C_1}
    • Mkono AB: ZAB=R2⋅1jωC2R2+1jωC2=R21+jωR2C2Z_{AB} = \frac{R_2 \cdot \frac{1}{j\omega C_2}}{R_2 + \frac{1}{j\omega C_2}} = \frac{R_2}{1 + j\omega R_2 C_2}
    • Mkono BC: ZBC=R4Z_{BC} = R_4
    • Mkono DC: ZDC=R3Z_{DC} = R_3
  2. Tumia sharti la usawa wa daraja ZADZDC=ZABZBCZ_{AD} Z_{DC} = Z_{AB} Z_{BC}.

    (R1+1jωC1)R3=(R21+jωR2C2)R4\left(R_1 + \frac{1}{j\omega C_1}\right) R_3 = \left(\frac{R_2}{1 + j\omega R_2 C_2}\right) R_4

    Panua na panga upya:

    R1R3−jR3ωC1=R2R41+jωR2C2R_1 R_3 - j\frac{R_3}{\omega C_1} = \frac{R_2 R_4}{1 + j\omega R_2 C_2}

    Ili kutenganisha sehemu halisi na dhahania, zidisha upande wa kulia kwa kiambishi tamati (conjugate) cha denomineta:

    R1R3−jR3ωC1=R2R4(1−jωR2C2)1+(ωR2C2)2R_1 R_3 - j\frac{R_3}{\omega C_1} = \frac{R_2 R_4 (1 - j\omega R_2 C_2)}{1 + (\omega R_2 C_2)^2} R1R3−jR3ωC1=R2R41+(ωR2C2)2−jωR22R4C21+(ωR2C2)2R_1 R_3 - j\frac{R_3}{\omega C_1} = \frac{R_2 R_4}{1 + (\omega R_2 C_2)^2} - j\frac{\omega R_2^2 R_4 C_2}{1 + (\omega R_2 C_2)^2}
  3. Linganisha sehemu halisi na dhahania.

    Kulinganisha sehemu halisi:

    R1R3=R2R41+(ωR2C2)2(Eq. 1)R_1 R_3 = \frac{R_2 R_4}{1 + (\omega R_2 C_2)^2} \quad \text{(Eq. 1)}

    Kulinganisha sehemu dhahania:

    R3ωC1=ωR22R4C21+(ωR2C2)2(Eq. 2)\frac{R_3}{\omega C_1} = \frac{\omega R_2^2 R_4 C_2}{1 + (\omega R_2 C_2)^2} \quad \text{(Eq. 2)}
  4. Pata usemi wa C1C_1 kutoka kwa milinganyo hii.

    Kutoka Eq. 1, tunaweza kuandika:

    1+(ωR2C2)2=R2R4R1R3(Eq. 3)1 + (\omega R_2 C_2)^2 = \frac{R_2 R_4}{R_1 R_3} \quad \text{(Eq. 3)}

    Badilisha Eq. 3 kwenye Eq. 2:

    R3ωC1=ωR22R4C2R2R4R1R3\frac{R_3}{\omega C_1} = \frac{\omega R_2^2 R_4 C_2}{\frac{R_2 R_4}{R_1 R_3}} R3ωC1=ωR1R2C2R3\frac{R_3}{\omega C_1} = \omega R_1 R_2 C_2 R_3

    Gawanya pande zote mbili kwa R3R_3:

    1ωC1=ωR1R2C2\frac{1}{\omega C_1} = \omega R_1 R_2 C_2

    Hii inatoa usemi wa C1C_1 unaotegemea ω\omega:

    C1=1ω2R1R2C2C_1 = \frac{1}{\omega^2 R_1 R_2 C_2}

    Ili kupata usemi wa C1C_1 usiotegemea ω\omega, tunaweza kutumia Eq. 1 na Eq. 2 tena.
    Badilisha ω2=1R1R2C1C2\omega^2 = \frac{1}{R_1 R_2 C_1 C_2} (kutoka hapo juu) kwenye Eq. 1:

    R1R3=R2R41+(1R1R2C1C2)R22C22R_1 R_3 = \frac{R_2 R_4}{1 + \left(\frac{1}{R_1 R_2 C_1 C_2}\right) R_2^2 C_2^2} R1R3=R2R41+R2C2R1C1R_1 R_3 = \frac{R_2 R_4}{1 + \frac{R_2 C_2}{R_1 C_1}} R1R3(1+R2C2R1C1)=R2R4R_1 R_3 \left(1 + \frac{R_2 C_2}{R_1 C_1}\right) = R_2 R_4 R1R3+R1R3R2C2R1C1=R2R4R_1 R_3 + \frac{R_1 R_3 R_2 C_2}{R_1 C_1} = R_2 R_4 R1R3+R3R2C2C1=R2R4R_1 R_3 + \frac{R_3 R_2 C_2}{C_1} = R_2 R_4

    Panga upya ili kupata C1C_1:

    R3R2C2C1=R2R4−R1R3\frac{R_3 R_2 C_2}{C_1} = R_2 R_4 - R_1 R_3 C1=R3R2C2R2R4−R1R3C_1 = \frac{R_3 R_2 C_2}{R_2 R_4 - R_1 R_3} C1=C2R2R3R2R4−R1R3C_1 = C_2 \frac{R_2 R_3}{R_2 R_4 - R_1 R_3}

    Usemi wa capacitor C1C_1 ni:

    C1=C2R2R3R2R4−R1R3\boxed{C_1 = C_2 \frac{R_2 R_3}{R_2 R_4 - R_1 R_3}}

Part b

f=2500π Hz≈795.77 Hzf = \frac{2500}{\pi} \, Hz \approx 795.77 \, Hz

**Determine the frequency at balance.

  1. Tumia usemi wa masafa ya usawa wa daraja la Wein.

    Kutoka hatua ya 4 ya sehemu (a), tuligundua usemi wa ω2\omega^2:

    ω2=1R1R2C1C2\omega^2 = \frac{1}{R_1 R_2 C_1 C_2}

    Masafa ff yanatolewa na f=ω2πf = \frac{\omega}{2\pi}, hivyo:

    f=12πR1R2C1C2f = \frac{1}{2\pi\sqrt{R_1 R_2 C_1 C_2}}
  2. Badilisha thamani zilizotolewa kutoka mchoro.

    R1=100 ΩR_1 = 100 \, \Omega
    C1=1 μF=1×10−6 FC_1 = 1 \, \mu F = 1 \times 10^{-6} \, F
    R2=200 ΩR_2 = 200 \, \Omega
    C2=2 μF=2×10−6 FC_2 = 2 \, \mu F = 2 \times 10^{-6} \, F

  3. Kokotoa masafa ff.

    f=12π(100 Ω)(200 Ω)(1×10−6 F)(2×10−6 F)f = \frac{1}{2\pi\sqrt{(100 \, \Omega)(200 \, \Omega)(1 \times 10^{-6} \, F)(2 \times 10^{-6} \, F)}} f=12π100×200×1×2×10−12f = \frac{1}{2\pi\sqrt{100 \times 200 \times 1 \times 2 \times 10^{-12}}} f=12π40000×10−12f = \frac{1}{2\pi\sqrt{40000 \times 10^{-12}}} f=12π4×104×10−12f = \frac{1}{2\pi\sqrt{4 \times 10^4 \times 10^{-12}}} f=12π4×10−8f = \frac{1}{2\pi\sqrt{4 \times 10^{-8}}} f=12π⋅(2×10−4)f = \frac{1}{2\pi \cdot (2 \times 10^{-4})} f=14π×10−4f = \frac{1}{4\pi \times 10^{-4}} f=1044πf = \frac{10^4}{4\pi} f=100004πf = \frac{10000}{4\pi} f=2500π Hzf = \frac{2500}{\pi} \, Hz

    Kukokotoa thamani ya namba:

    f≈25003.14159f \approx \frac{2500}{3.14159} f≈795.77 Hzf \approx 795.77 \, Hz

    Masafa katika usawa ni:

    f=2500π Hz≈795.77 Hz\boxed{f = \frac{2500}{\pi} \, Hz \approx 795.77 \, Hz}
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