A) CE
B) \[\frac{CE{{R}_{1}}}{{{R}_{2}}-r}\]
C) \[\frac{CE{{R}_{2}}}{{{R}_{2}}+r}\]
D) \[\frac{CE{{R}_{1}}}{{{R}_{1}}-r}\]
Correct Answer: C
Solution :
In steady state current drawn from the battery \[i=\frac{E}{({{R}_{2}}+r)}\] In steady state capacitor is fully charged hence No current will flow through line (2) Hence potential difference across line (1) is \[V=\frac{E}{({{R}_{2}}+r)}\times {{R}_{2}}\], the same potential difference appears across the capacitor, so charge on capacitor \[Q=C\times \frac{E{{R}_{2}}}{({{R}_{2}}+r)}\]You need to login to perform this action.
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