A) \[\frac{30\,E}{100.5}\]
B) \[\frac{30\,E}{100-05}\]
C) \[\frac{30\,(E-0.5i)}{100}\], where \[i\] is the current in the potentiometer wire
D) \[\frac{30\,E}{100}\]
Correct Answer: D
Solution :
| [d] Potential difference of wire |
| \[V\propto l\] |
| \[\therefore \] \[\frac{V}{E}=\frac{l}{L}\] |
| where, \[l=\] length of balance point of potentiometer-wire |
| L = length of potentiometer wire |
| or \[V=\frac{l}{L}E\] |
| \[V=\frac{30\times E}{100}=\frac{30}{100}E\] |
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