A) \[{{\operatorname{f}}_{\,1}}-{{\operatorname{f}}_{\,2}}={{\operatorname{f}}_{\,3}}\]
B) \[{{\operatorname{f}}_{\,1}}={{\operatorname{f}}_{\,2}}-{{\operatorname{f}}_{\,3}}\]
C) \[\frac{1}{{{\operatorname{f}}_{\,2}}}=\frac{1}{{{\operatorname{f}}_{\,1}}}+\frac{1}{{{\operatorname{f}}_{\,3}}}\]
D) \[\frac{1}{{{\operatorname{f}}_{\,1}}}=\frac{1}{{{\operatorname{f}}_{\,2}}}+\frac{1}{{{\operatorname{f}}_{\,3}}}\]
Correct Answer: A
Solution :
frequency \[{{\operatorname{f}}_{0}}-Rc\left( \frac{1}{{{n}^{2}}_{1}}-\frac{1}{{{n}^{2}}_{2}} \right)\] \[{{\operatorname{f}}_{1}}-Rc\left( 1-\frac{1}{\infty } \right),\,\,\,\,\]c\[{{\operatorname{f}}_{1}}-Rc\left( 1-\frac{1}{\infty } \right),\,\,\,{{\operatorname{f}}_{2}}=\operatorname{Rc}\left( 1-\frac{1}{4} \right)=\frac{3}{4}Rc\], \[{{\operatorname{f}}_{3}}=Rc\left( \frac{1}{4}-\frac{1}{\infty } \right)=\frac{Rc}{4}\] \[{{\operatorname{f}}_{1}}-{{f}_{2}}={{f}_{3}}\]
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