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 balancing length is 100 cm. What is the ratio of \[{{E}_{1}}\] and \[{{E}_{2}}\]?
A) 3/2
B) 2/3
C) 1/2
D) 1/3
Correct Answer: A
Solution :
In potentiometer Emf \[\propto \] length of potentiometer \[\therefore \] When connected between A and B \[{{E}_{1}}\propto 300\] \[{{E}_{1}}=K\,(300)\] ... (i) and when connected between A and C \[({{E}_{2}}-{{E}_{1}})\propto 100\] \[{{E}_{2}}-{{E}_{1}}=(100)\] ... (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{2}\]
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