A potentiometer wire AB having length L resistance 12 r is joined to a cell D of emf \[\varepsilon \]and internal resistance r. A cell C having emf \[\varepsilon \text{/2}\]and internal resistance 3r is connected. The length AJ galvanometer as shown in figure shows no deflection is: |
A) \[\frac{11}{12}L\]
B) \[\frac{11}{24}L\]
C) \[\frac{13}{24}L\]
D) \[\frac{5}{12}L\]
Correct Answer: C
Solution :
\[{{V}_{AJ}}=I{{R}_{AJ}}\] \[\frac{E}{3}=\left( \frac{E}{12r+3r} \right)\times \left( \frac{x}{L}\times 12r \right).\]You need to login to perform this action.
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