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 :
| [c] \[{{v}_{1}}=Rc\,{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] |
| \[{{v}_{1}}=Rc\,{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=Rc\,{{Z}^{2}}\] |
| \[{{v}_{2}}=Rc\,{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3\,Rc\,{{Z}^{2}}}{4}\] |
| \[{{v}_{3}}=Rc\,{{Z}^{2}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=\frac{2\,Rc\,{{Z}^{2}}}{4}\] |
| \[\therefore \] \[{{v}_{1}}-{{v}_{2}}={{v}_{3}}\] |
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