A) \[\frac{V}{K+n}\]
B) \[V\]
C) \[\frac{(n+1)V}{(K+n)}\]
D) \[\frac{nV}{K+n}\]
Correct Answer: C
Solution :
[c] |
After fully charging, battery is disconnected |
Total charge of the system \[=CV+nCV\] |
\[=(n+1)CV\] |
After the insertion of dielectric of constant K |
New potential (common) |
\[{{V}_{C}}=\frac{total\,ch\arg e}{total\,capaci\,\tan ce}\] |
\[=\frac{\left( n+1 \right)CV}{KC+nC}=\frac{(n+1)V}{K+n}.\] |
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