Thermo-couple RatioWhen We Have to Find An UnKnown Quantity from the given:If we have 1st thermocouple at:(4.00V)Junction 1: 0 Degree CJunction 2: 100 Degree CAnd 2nd Thermocouple at: (-1.50V) Junction 1: 0 Degree CJunction 2: ? Degree C

Then following formula gives us the answer: ?-0 -1.50----- = ------- 100-0 4.00

-1.50Which Is: ? = ------- x 100 4.00 = -37.5 Degree C

Now Coming To Question 2 (ii): We Have all Readings but The New One is Unknown:(a)1st Junction is in Ice and Water and is always 0 degree C(b)The temperature in 2nd Junction (furnace) varies!

If we have 1st range at:(7.70V)Junction 1: 0 Degree CJunction 2: 800 Degree CAnd 2nd range at: (6.20V) Junction 1: 0 Degree CJunction 2: 750 Degree C

Therefore: (800-0) - (750-0) 50------------------- = Ratio = ------- = 33.333..... 7.70 - 6.20 1.50

We Need Our Third Temperature By Comparison;

So You Can Take Either Range:

(?-0)-(750-0) ------------- = 33.333..... 6.80 - 6.20

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Thermo-couple Ratio ?-750 = 33.333... x 0.6 ? = 20 + 750 ? = 770 degree CelciusOr:

(?-0)-(800-0) ------------- = 33.333..... 6.80 - 7.70

?-800 = 33.333... x -0.9 ? = -30 + 750 ? = 770 degree Celcius

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