A) 20
B) 15
C) 10
D) 40
Correct Answer: A
Solution :
\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{(1+\alpha {{t}_{1}})}{(1+\alpha {{t}_{2}})}\]Þ\[\frac{10}{{{R}_{2}}}=\frac{(1+5\times {{10}^{-3}}\times 20)}{(1+5\times {{10}^{-3}}\times 120)}\]Þ\[{{R}_{2}}\approx 15\Omega \] Also \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\] Þ \[\frac{30}{{{i}_{2}}}=\frac{15}{10}\] Þ \[{{i}_{2}}=20\,mA\]
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