For a copper-iron and a chromel-alumel thermocouple, the plots between thermoelectric emf and the temperature `theta` of the hot junction (when the cold junction is at `0^(@)C`)are found to satisfy approximately the parabola equation `V=alpha theta+(1)/(2)betatheta^(2)` with `alpha=14muV^(@)C^(-1) beta=-0.04V^(@)C^(-2)` (copper-iron) `alpha=41muCV^(@)C^(-1),beta=-0.002V^(@)C^(-2)` (chromel-alumel) Which of the two thermocouples would you use to measure temperature in the range of a about `500^(@)C` to `600^(@)C`?

For a copper-iron and a chromel-alumel thermocouple, the plots between

1.	For a copper-iron and a chromel-alumel thermocouple, the plots between thermoelectric emf and the temperature `theta` of the hot junction (when the cold junction is at `0^(@)C`)are found to satisfy approximately the parabola equation `V=alpha theta+(1)/(2)betatheta^(2)` with `alpha=14muV^(@)C^(-1) beta=-0.04V^(@)C^(-2)` (copper-iron) `alpha=41muCV^(@)C^(-1),beta=-0.002V^(@)C^(-2)` (chromel-alumel) Which of the two thermocouples would you use to measure temperature in the range of a about `500^(@)C` to `600^(@)C`?
Answer» The temperature `theta_(n)` (neutral temperature) corresponding to maximum emf is given by `(dV)/(dtheta)=0,i.e., alpha+beta theta_(n)=0` or `theta_(n)=-(alpha)/(beta)` For a copper-iron therm ocouple, `theta_(n)=-(14)/(-0.04).^(@)C=350^(@)C`

Discussion

No Comment Found

Related InterviewSolutions

The effective between A&B in the given circuit isA. `7Omega`B. `2Omega`C. `6Omega`D. `5Omega`
A source of e.m.f. `E = 15V` and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as `i = 1.2 t +3` Then, the total charge that will flow in first five second will beA. `10C`B. `20C`C. `30C`D. `40C`
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. The battery that runs the potentiometer should have voltage of 8 VB. The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 VC. The first portion of 50 cm of wire itself should have a potential drop of 10 VD. Potentiometer is usually used for comparing resistance and not voltages
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. The battery that runs the potentionmeter should have voltage of 8 VB. The battery of potentionmeter can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10 VC. The first portion of 50cm of wire itself should have a potential drop of 10VD. Potentiometer is usually used for comparing resistances and not voltages
Two cells of e.m.f. 10V & 15V are connected in parallel to each other between points A&B. The cell of e.m.f. 10V is ideal but the cell of e.m.f. 15V has internal resistance `1Omega`. The equivalent e.m.f. between A and B is: A. `25/2V`B. not definedC. 15 VD. 10 V
Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm.A. Toe battery that runs the potentiometer should have voltage of 8 V.B. Toe battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.C. The first portion of 50 cm of wire itself should have a potential drop of 10 VD. Potentiometer is usually used for comparing resistances and not voltages.
In the `R_(1) = 10 Omega ,R_(2) = 20 Omega, R_(3) = 40 Omega R_(4) = 80 Omega` and `V_(A) = 5V,V_(B) = 10V,V_(c) = 20V,V_(D) = 15V` The current in the resistance `R_(1)` will be A. `0.4 A` toward OB. `0.4A` away from OC. `0.8A` toward OD. `0.8A` toward O
The amount of charge Q passed in time t through a cross-section of a wire is `Q=5t^(2)+3t+1`. The value of current at time t=5 s isA. 9AB. 49AC. 53AD. None of these
The charge flowing through a conductor varies with time as `q= 8 t - 3t^(2)+ 5t^(3)` Find (i) the initial current (ii) time after which the current reaches a maximum value of current
The current in a conductor varies with time `t` as `I=3t+4t^(2)` Where I in amp and t in sec. The electric charge flows through the section of the conductor between `t=1s` and `t=3s`A. `(100)/(3)C`B. `(127)/(4)C`C. `(140)/(3)C`D. `(150)/(3)C`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment