A) 4/3
B) 5/4
C) 5/3
D) 4/1
Correct Answer: A
Solution :
Consider, the diagram shown below \[({{V}_{p}}-{{V}_{\theta }})={{E}_{1}}\propto {{L}_{1}}\] ...(i) \[{{V}_{p}}-{{V}_{R}}=({{V}_{p}}-{{V}_{Q}})+({{V}_{Q}}-{{V}_{R}})\] \[=({{E}_{1}}+(-{{E}_{2}}))={{E}_{1}}-{{E}_{2}}\propto {{L}_{2}}\]... (ii) where,\[{{L}_{1}}\]and\[{{L}_{2}}\]are the lengths of potentiometer From Eqs. (i) and (ii), we get \[\frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{0.8}{0.2}\Rightarrow \frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{8}{2}=4\] \[\Rightarrow \] \[{{E}_{1}}=4{{E}_{1}}-4{{E}_{2}}\] \[\Rightarrow \] \[4{{E}_{2}}=3{{E}_{1}}\Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{4}{3}\]You need to login to perform this action.
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