Saved Bookmarks
| 1. |
At temperature 0""^(@) Cand 100 ""^(@) C, currents passing through one conductor are resectively 1 A and 0.7 A . Find current through it when its temperature is 1200""^(@) C . (Voltage source is same). |
|
Answer» Solution :We have, `R_(t) = R_(0) { 1 + prop ( t - 0) } ` `THEREFORE R_(t) = R_(0) + R_(0) prop t ` `therefore Rt - R_(0) prop ` t `therefore prop = (R_(t) - R_(0))/(R_(0)t)` `therefore prop = ((V)/(I_(t)) - (V)/(I_(0)) )/((V)/(I_(0)) xx t) ` ` therefore prop = ((1)/(I_(t))-(1)/(I_(0)))/((t)/(I_(0)))= (I_(0))/(t) ((I_(0)-I_(t))/(I_(t) I_(0))) ` `therefore prop = (I_(0) - I_(t))/(I_(t) t)"" `... (1) From equation (1), `prop= (I_(0) - I_(t1))/(I_(t1) xx t_(1)) "" `.... (2) Similarly, `prop = (I_(0) - I_(t2))/(I_(t2) xx t_(2)) "" ` ... (3) From equations (2) and (3) , `(I_(0) - I_(t1))/(I_(t1) xx t_(1))= (I_(0) - I_(t2))/( I_(t2) xx t_(2)) "" `... (4) Substituting give values, `(1- 0.7 )/(0.7 xx 100) = (1 - I_(t2))/( I_(t2) xx 1200)` `therefore (0.3)/(0.7) - (I - I_(t2))/( I_(t2) xx 12)` `therefore 3.6 I_(t2) = 0.7 - 0.7 I_(t2)` 4.3 `I_(t2) = 0.7` `therefore I_(t2) = (0.7)/(4.3) = 0.1627 ` A Note: Dear students, if CURRENT in a given conductor is `I_(0)`at REFERENCE temperature `t_(0)`then relation between currents temperature `t_(1) and t_(2)`is given by, `I_(t2) = I_(t1) xx ( (I_(t1) - I_(0))/(I_(t2) - I_(0))) ( (t_(1) - t_(0))/(t_(2) - t_(0)) ) ` Above formula can be obtained by simplifying equation (4) and then by making `I_(t2)` as subject of equation. |
|