A) \[{{v}_{3}}=\frac{1}{2}({{v}_{1}}-{{v}_{3}})\]
B) \[{{v}_{2}}-{{v}_{1}}={{v}_{3}}\]
C) \[{{v}_{1}}-{{v}_{2}}={{v}_{3}}\]
D) \[{{v}_{1}}+{{v}_{2}}={{v}_{3}}\]
Correct Answer: C
Solution :
\[{{v}_{1}}=Rc{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[{{v}_{1}}=Rc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}}\, \right)=Rc{{Z}^{2}}\] \[{{v}_{2}}=Rc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3Rc{{Z}^{2}}}{4}\] \[{{v}_{3}}=Rc{{Z}^{2}}\left( \frac{1}{2_{{}}^{2}}-\frac{1}{\infty _{{}}^{2}} \right)=\frac{2Rc{{Z}^{2}}}{4}\] \[\therefore \] \[{{v}_{1}}-{{v}_{2}}={{v}_{3}}\]
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