Physics, asked by isaacherutich, 9 months ago

A thermocouple circuit uses Chrome - Alumel which gives an e.m.f of 33.3 mV when
measuring a temperature of 800 oC with reference temperature of 0 oC. The resistance of
the metal coil is Rm = 50 Ω and current of 0.1 mA gives a full scale deflection. The
resistance of the junction and leads Re = 12 Ω. Determine,
i. Resistance of the series resistance if a temperature of 800 oC is to give full scale
deflection
ii. The approximate error due to rise of 1Ω in Re
iii. The approximate error due to rise of 10 oC in a copper coil of the meter. The
resistance temperature coefficient of the coil is 0.00426 /oC [10]

Answers

Answered by sp6559568
2

Answer:

A thermocouple circuit uses a chromel- alumel thermocouple which gives an emf of 33.3 v ... of 800oc with reference temperature 0oc. the resistance of the meter coil,

Answered by dreamrob
4

Given :

emf = 33.3mV

Rm = 50Ω

I = 0.1mA

Re = 12Ω

∝ = 0.00426 /°C

To find :

i) Resistance of the series resistance if a temperature of 800°C is to give full scale  deflection

ii) The approximate error due to rise of 1Ω in Re

iii) The approximate error due to rise of 10°C in a copper coil of the meter.

Solution :

i) Let series resistance be Rs

emf = I (Rm + Rs + Re)

Rs = (emf / I) - Rm - Re

Rs = (33.3 × 10⁻³) / (0.1 × 10⁻³) - 50 - 12

Rs = 333 - 62

Rs = 271Ω

ii) Due to rise of 1Ω in Re

Now new Re is Re' = 1 + 12 = 13Ω

emf = I' (Rm + Re' + Rs)

I' = emf / (Rm + Re' + Rs)

I' = (33.3 × 10⁻³) / (50 + 271 + 13)

I' = 0.0997mA

Approximate error in temperature = [(I' - I) / I] × 100

                                                         = [(0.0997 - 0.1) / 0.1] × 800

                                                         = - 0.3%

iii) Change in resistance with a temperature increase of 10°C

ΔRm = Rm × ∝ × 10°C

        = 50 × 0.00426 × 10

        = 2.13Ω

New Rm' = Rm + ΔRm

Rm' = 50 + 2.13

Rm' = 52.13Ω

Now current I'' = emf / (Rm' + Rs +Re)

I'' = 0.09936mA

Therefore approximate error in temperature = ([I'' - I) / I] × 100

                                                                         = -0.64%

Similar questions