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State two requirements of shunts when used in the measurement of electrical quantities.

Physics

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Answer

0.0160 V\text{0.0160 V}

Step-by-step solution

7 steps

Huu ni tatizo kuhusu kipimo cha koili inayohamia (moving coil instrument) kinachotumika kama voltmeter, na jinsi mabadiliko ya halijoto yanavyoathiri usahihi wake.

  • Upinzani wa chombo, Rm=25 ΩR_m = 25 \ \Omega
  • Voltage ya mkengeuko kamili (full scale deflection), Vm=25 mV=0.025 VV_m = 25 \ mV = 0.025 \ V
  • Kiwango cha voltage kinachohitajika, V=10 VV = 10 \ V
  • Mgawo wa halijoto wa shaba (copper), αCu=0.004/ ∘C\alpha_{Cu} = 0.004/ \ ^{\circ}C
  • Mgawo wa halijoto wa manganin, αMn=0.00015/ ∘C\alpha_{Mn} = 0.00015/ \ ^{\circ}C
  • Ongezeko la halijoto, ΔT=10 ∘C\Delta T = 10 \ ^{\circ}C

Tunahitaji kuamua kosa linalosababishwa na ongezeko la halijoto.

  1. Kokotoa mkondo wa mkengeuko kamili (ImI_m) wa chombo.

    Im=VmRmI_m = \frac{V_m}{R_m} Im=0.025 V25 Ω=0.001 A=1 mAI_m = \frac{0.025 \ V}{25 \ \Omega} = 0.001 \ A = 1 \ mA
  2. Kokotoa upinzani wa kiongezeo (multiplier resistance, RsR_s) unaohitajika ili kupanua kiwango cha voltage hadi 10 V10 \ V.

    Kwa voltmeter, jumla ya upinzani ni Rtotal=VImR_{total} = \frac{V}{I_m}.

    Rs=Rtotal−Rm=VIm−RmR_s = R_{total} - R_m = \frac{V}{I_m} - R_m Rs=10 V0.001 A−25 Ω=10000 Ω−25 Ω=9975 ΩR_s = \frac{10 \ V}{0.001 \ A} - 25 \ \Omega = 10000 \ \Omega - 25 \ \Omega = 9975 \ \Omega
  3. Kokotoa upinzani mpya wa chombo (Rm′R_m') baada ya ongezeko la halijoto la 10 ∘C10 \ ^{\circ}C.

    Rm′=Rm(1+αCuΔT)R_m' = R_m (1 + \alpha_{Cu} \Delta T) Rm′=25 Ω(1+0.004/ ∘C×10 ∘C)R_m' = 25 \ \Omega (1 + 0.004/ \ ^{\circ}C \times 10 \ ^{\circ}C) Rm′=25 Ω(1+0.04)=25 Ω(1.04)=26 ΩR_m' = 25 \ \Omega (1 + 0.04) = 25 \ \Omega (1.04) = 26 \ \Omega
  4. Kokotoa upinzani mpya wa kiongezeo (Rs′R_s') baada ya ongezeko la halijoto la 10 ∘C10 \ ^{\circ}C.

    Rs′=Rs(1+αMnΔT)R_s' = R_s (1 + \alpha_{Mn} \Delta T) Rs′=9975 Ω(1+0.00015/ ∘C×10 ∘C)R_s' = 9975 \ \Omega (1 + 0.00015/ \ ^{\circ}C \times 10 \ ^{\circ}C) Rs′=9975 Ω(1+0.0015)=9975 Ω(1.0015)=9989.9625 ΩR_s' = 9975 \ \Omega (1 + 0.0015) = 9975 \ \Omega (1.0015) = 9989.9625 \ \Omega
  5. Kokotoa jumla ya upinzani mpya wa voltmeter (Rtotal′R_{total}') baada ya ongezeko la halijoto.

    Rtotal′=Rm′+Rs′R_{total}' = R_m' + R_s' Rtotal′=26 Ω+9989.9625 Ω=10015.9625 ΩR_{total}' = 26 \ \Omega + 9989.9625 \ \Omega = 10015.9625 \ \Omega
  6. Kokotoa voltage mpya ya mkengeuko kamili (V′V') kwa mkondo uleule wa ImI_m.

    V′=Im×Rtotal′V' = I_m \times R_{total}' V′=0.001 A×10015.9625 Ω=10.0159625 VV' = 0.001 \ A \times 10015.9625 \ \Omega = 10.0159625 \ V
  7. Kokotoa kosa linalosababishwa na ongezeko la halijoto.

    Kosa = Voltage mpya - Voltage iliyokusudiwa

    Kosa=V′−V\text{Kosa} = V' - V Kosa=10.0159625 V−10 V=0.0159625 V\text{Kosa} = 10.0159625 \ V - 10 \ V = 0.0159625 \ V

    Kosa linaweza kuzungushwa hadi sehemu nne za desimali.

    Kosa≈0.0160 V\text{Kosa} \approx 0.0160 \ V

    Kosa linalosababishwa na ongezeko la halijoto ni:

    0.0160 V\text{0.0160 V}
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