A) 1
B) 0.8
C) 0.4
D) 0.6
Correct Answer: B
Solution :
\[V=\frac{{{\operatorname{I}}_{max}}\,-\,{{I}_{\min }}}{{{\operatorname{I}}_{max}}+{{I}_{\min }}}=\frac{\frac{{{\operatorname{I}}_{max}}}{{{I}_{\min }}}-1}{\frac{{{\operatorname{I}}_{max}}}{{{I}_{\min }}}+1}\] \[\frac{{{\operatorname{I}}_{max}}}{{{I}_{\min }}}={{\left( \frac{\frac{{{I}_{1}}}{{{I}_{2}}}+1}{\frac{{{I}_{1}}}{{{I}_{2}}}-1} \right)}^{2}}\] According to question \[\frac{{{I}_{1}}}{{{I}_{2}}}\,\,=\,\,\frac{1}{4}\] ...(3) From eqs. (2) and (3) \[\frac{{{I}_{\max }}}{{{I}_{\min }}}\,\,=\,\,{{\left( \frac{\sqrt{\frac{1}{4}}+1}{\sqrt{\frac{1}{4}}-1} \right)}^{2}}\,\,=\,\,\frac{\frac{9}{4}}{\frac{1}{4}}\,\,=\,\,9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(4)\] From eqs. (1) and (4) \[V=\frac{[9-1]}{[9+1]}\,\,=\,\frac{8}{10}\,\,=\,\,0.8\]
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