A) 2A
B) 1A
C) 0.5A
D) 0.25A
Correct Answer: C
Solution :
[b] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\alpha {{I}_{1}}=\alpha {{I}_{2}}\] \[{{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}\] \[{{I}_{\max }}={{\left( \sqrt{\alpha {{I}_{2}}}+\sqrt{{{I}_{2}}} \right)}^{2}}=\alpha {{I}_{2}}+{{I}_{2}}+2\sqrt{\alpha }{{I}_{2}}\] \[{{I}_{\min }}={{\left( \sqrt{\alpha {{I}_{2}}}-\sqrt{{{I}_{2}}} \right)}^{2}}=\alpha {{I}_{2}}+{{I}_{2}}-2\sqrt{\alpha }{{I}_{2}}\] \[{{I}_{\max }}+{{I}_{\min }}=2\alpha {{I}_{1}}+2{{I}_{2}}=2{{I}_{2}}[1+\alpha ]\] \[{{I}_{\max }}-{{I}_{\min }}=4{{I}_{2}}\sqrt{\alpha }\] \[(1)\div (2)\] \[\frac{4{{I}_{2}}\sqrt{\alpha }}{2{{I}_{2}}[1+\alpha ]}=\frac{2\sqrt{\alpha }}{1+\alpha }\]
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