A) \[1R\Omega \]
B) \[8R\Omega \]
C) \[3R\Omega \]
D) \[4R\Omega \]
Correct Answer: D
Solution :
From relation, \[{{R}_{1}}={{R}_{0}}\,(1+\alpha \Delta t)\] Case I: \[5R={{R}_{0}}(1+\alpha \times 50)\] ?..(i) Case II: \[6R={{R}_{0}}(1+\alpha \times 100)\] .....(ii) On dividing equation (i) by equation (ii) we get. \[\frac{5}{6}=\frac{1+50\,\alpha }{1+100\,\alpha }\] \[\Rightarrow \] \[5\times (1+100\alpha )=6\times (1+50\alpha )\] \[\Rightarrow \] \[5+500\alpha =6+300\alpha \] \[\Rightarrow \] \[200\alpha =1\] \[\alpha =\frac{1}{200}\Omega {{/}^{o}}C\] On substituting in equation (i) we get, \[5R={{R}_{0}}\,\left( 1+\frac{1}{4} \right)\] \[\Rightarrow \] \[5R=\frac{5}{4}{{R}_{0}}\] \[\Rightarrow \] \[{{R}_{0}}=4R\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
