A) \[\frac{30E}{100.5}\]
B) \[\frac{30E}{100-0.5}\]
C) \[\frac{30(E-0.5i)}{100}\]where \[i\]is the current in the potentiometer wire
D) \[\frac{30E}{100}\,\]
Correct Answer: D
Solution :
(d )As \[V\propto l\] \[\therefore \] \[\frac{V}{E}=\frac{l}{L}\]where, \[l=\]balance point, \[L=\]length of potentiometer wire or \[V=\frac{l}{L}E\] \[V=\frac{30\times E}{100}=\frac{30}{100}E\]You need to login to perform this action.
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