When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the F balancing length is 100 cm. The ratio \[\frac{{{E}_{1}}}{{{E}_{2}}}\] is
A) 3 : 1
B) 1 : 3
C) 2 : 3
D) 3 : 2
Correct Answer: D
Solution :
When potentiometer is connected between A and B, then it measures only \[{{E}_{1}}\] and when connected between A and C, then it measures\[{{E}_{1}}-{{E}_{2}}\]. \[\therefore \] \[\frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] \[\frac{{{E}_{1}}-{{E}_{2}}}{{{E}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\] \[\Rightarrow \] \[1-\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{100}{300}\] \[\Rightarrow \] \[\frac{{{E}_{2}}}{{{E}_{1}}}=1-\frac{1}{3}\] \[\Rightarrow \] \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{2}{3}\] \[\Rightarrow \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{2}\]
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