A) \[{{I}_{g}}=\frac{n}{\text{Current sensitivity}}=\frac{150}{10}\]
B) \[=15mA=15\times {{10}^{-3}}A\]
C) \[=\text{15}0\times \text{1}=\text{15}0\text{V}\]
D) \[R=\frac{V}{{{I}_{g}}}-G=\frac{150}{15\times {{10}^{-3}}}-5=9995\Omega \]
Correct Answer: B
Solution :
\[{{T}_{r+1}}{{=}^{8}}{{C}_{r}}{{[({{t}^{-1}}-1)x]}^{8-r}}{{[{{({{t}^{-1}}+1)}^{-1}}{{x}^{-1}}]}^{r}}\] \[{{=}^{8}}{{C}_{r}}{{({{t}^{-1}}-1)}^{8-r}}{{({{t}^{-1}}+1)}^{-r}}{{x}^{8-2r}}\] For independent of \[x,8-2r=0\] \[\Rightarrow \] \[r=4\] \[\therefore \] \[{{T}_{4+1}}{{=}^{8}}{{C}_{4}}{{({{t}^{-1}}-1)}^{4}}{{({{t}^{-1}}+1)}^{-4}}\] \[=70{{\left( \frac{1-t}{1+t} \right)}^{4}}\]
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